HCF and LCM Word Problems

SSC CGL · CHSL · CPO — method by method

4 Part49 worked examples150 practice questions

In the previous chapter, “HCF and LCM”, we learnt to find the HCF and LCM and the relations between them. In this chapter we use the same tools in stories. The first job is always the same: spot what the story is doing — are we splitting something into equal pieces (HCF), or looking for the time or size at which things meet again or build up (LCM)?

In the story	Method	Where
greatest number leaving the same remainder	HCF of the differences	Remainders
greatest number leaving given remainders	HCF of (number − remainder)	Remainders
smallest number leaving remainder $r$	$L CM + r$ (or $L CM - g$ )	Remainders
bells, lights, runners, gears — together again	LCM of the intervals	Together
largest tile, longest piece, fewest groups	HCF	Cutting
smallest square/cube, least distance or capacity	LCM	Cutting

Spot itRemember two phrases: “the greatest … that divides all” ⟹ HCF. “the smallest … that all divide” ⟹ LCM. The rest is just costume.

Remainder problems

The greatest and the smallest number

The trick of cancelling the remainder to make a number exactly divisible, and all its forms.

Remainder questions look scary but they all use one trick: cancel the remainder — by taking a difference, or subtracting, or adding — so that the number becomes exactly divisible. After that the HCF (for the greatest number) or LCM (for the smallest) applies directly.

Greatest number: the same remainder

N = H CF

(of the differences)

The greatest divisor that leaves the same remainder for several numbers.

The idea

Suppose a number $N$ leaves remainder $r$ both times when dividing $a$ and $b$ : $a = N q_{1} + r$ and $b = N q_{2} + r$ . Subtract: $a - b = N (q_{1} - q_{2})$ — the remainder $r$ has cancelled, and the difference is an exact multiple of $N$ . So $N$ divides the difference. With three numbers it divides both differences. For the greatest $N$ we need the HCF of the differences.

∑ Formula

N = H CF (∣ a - b ∣, ∣ b - c ∣, \dots)

The greatest number that leaves the same remainder every time

$N$: the number asked (the divisor)
$a, b, c$: the numbers being divided
$∣ a - b ∣$: the difference of two numbers (bigger − smaller)

NoteWith only two numbers the difference itself is the greatest

N

(its HCF is itself). If the remainder is wanted, divide any number by

N

once you have it.

CheckA remainder must always be smaller than the divisor. The HCF of the differences will give a remainder smaller than

N

; check with one division.

Steps

Put the numbers in order and find the neighbouring differences.
Take the HCF of these differences — that is $N$ .
If the remainder is asked, divide any number by $N$ .
Check: every number gives the same remainder and it is smaller than $N$ .

Example 1 · Basic

Find the greatest number which, on dividing 1232, 1771 and 2555, leaves the same remainder every time.

Solution, step by step

1Let the number be $N$ and the common remainder $r$ . Every given number is $N \times (quotient) + r$ . Taking the difference of two such numbers cancels $r$ , so the difference is a multiple of $N$ .
2The differences: 1771−1232=539, 2555−1771=784. $N$ divides all of them and we want the greatest $N$ , so $N = H CF (539, 784)$ .
3 $539 = 7^{2} \times 11$ , $784 = 2^{4} \times 7^{2}$ ⟹ HCF $= 49$ .
4Check: $1232 = 49 \times 25 + 7$ , $1771 = 49 \times 36 + 7$ , $2555 = 49 \times 52 + 7$ — the remainder is 7 every time, and 7 < 49 (a remainder must be smaller than the divisor).

Answer 49

Not the differences 539 and 784 themselves — 49 is the greatest of their common factors.

Example 2 · Standard

N

is the greatest number which, on dividing 1121, 1750 and 2416, leaves the same remainder every time. Find that remainder.

Solution, step by step

1Let the number be $N$ and the common remainder $r$ . Every given number is $N \times (quotient) + r$ . Taking the difference of two such numbers cancels $r$ , so the difference is a multiple of $N$ .
2The differences: 1750−1121=629, 2416−1750=666. $N$ divides all of them and we want the greatest $N$ , so $N = H CF (629, 666)$ .
3 $629 = 17 \times 37$ , $666 = 2 \times 3^{2} \times 37$ ⟹ HCF $= 37$ .
4Check: $1121 = 37 \times 30 + 11$ , $1750 = 37 \times 47 + 11$ , $2416 = 37 \times 65 + 11$ — the remainder is 11 every time, and 11 < 37 (a remainder must be smaller than the divisor).
5The remainder $= 1121 mod 37 = 11$ .

Answer 11

The remainder is smaller than

N

— 11 < 37 — so

N

is valid.

Example 3 · Standard

The greatest number which, on dividing 1016, 1811 and 3030, leaves the same remainder each time is

N

. Find the sum of the digits of

N

Solution, step by step

1Let the number be $N$ and the common remainder $r$ . Every given number is $N \times (quotient) + r$ . Taking the difference of two such numbers cancels $r$ , so the difference is a multiple of $N$ .
2The differences: 1811−1016=795, 3030−1811=1219. $N$ divides all of them and we want the greatest $N$ , so $N = H CF (795, 1219)$ .
3 $795 = 3 \times 5 \times 53$ , $1219 = 23 \times 53$ ⟹ HCF $= 53$ .
4Check: $1016 = 53 \times 19 + 9$ , $1811 = 53 \times 34 + 9$ , $3030 = 53 \times 57 + 9$ — the remainder is 9 every time, and 9 < 53 (a remainder must be smaller than the divisor).
5The digit sum of $N = 53$ is $5 + 3 = 8$ .

Answer 8

When a question asks for the digit sum of

N

, first find

N

Example 4 · Advanced

Find the greatest number which, on dividing 1456 and 2128, leaves the same remainder every time.

Solution, step by step

1Let the number be $N$ and the common remainder $r$ . Every given number is $N \times (quotient) + r$ . Taking the difference of two such numbers cancels $r$ , so the difference is a multiple of $N$ .
2The differences: 2128−1456=672. $N$ divides all of them and we want the greatest $N$ , so $N = H CF (672)$ .
3 $672 = 2^{5} \times 3 \times 7$ ⟹ HCF $= 672$ .
4Check: $1456 = 672 \times 2 + 112$ , $2128 = 672 \times 3 + 112$ — the remainder is 112 every time, and 112 < 672 (a remainder must be smaller than the divisor).

Answer 672

With two numbers the difference itself became

N

(672), since the remainder 112 is smaller.

BasicTry it yourself first

Find the greatest number which, on dividing 498, 826 and 1195, leaves the same remainder every time.

BasicTry it yourself first

N

is the greatest number which, on dividing 622, 1028 and 1463, leaves the same remainder every time. Find that remainder.

StandardTry it yourself first

Find the greatest number which, on dividing 716, 1325 and 2108, leaves the same remainder every time.

StandardTry it yourself first

The greatest number which, on dividing 715, 1323 and 1703, leaves the same remainder each time is

N

. Find the sum of the digits of

N

AdvancedTry it yourself first

Find the greatest number which, on dividing 1245, 2001 and 2865, leaves the same remainder every time.

AdvancedTry it yourself first

N

is the greatest number which, on dividing 959, 1584 and 2459, leaves the same remainder every time. Find that remainder.

StandardTry it yourself first

Find the greatest number which, on dividing 647, 1214 and 1718, leaves the same remainder every time.

AdvancedTry it yourself first

Find the greatest number which, on dividing 1601 and 2105, leaves the same remainder every time.

Greatest number: given remainders

N = H CF (a - r_{1}, b - r_{2}, \dots)

When a different remainder is given for each number: subtract it and take the HCF.

The idea

If dividing $a$ by $N$ leaves remainder $r_{1}$ , then $N$ divides $a - r_{1}$ exactly. So does it divide $b - r_{2}$ and $c - r_{3}$ . That is, once each remainder is subtracted every number becomes a multiple of $N$ . Now the question is plain: the greatest common divisor of the reduced numbers — the HCF.

∑ Formula

N N = H CF (a - r_{1}, b - r_{2}, c - r_{3}) > max (r_{1}, r_{2}, r_{3})

With the given remainders

$N$: the number asked
$a, b, c$: the numbers being divided
$r_{1}, r_{2}, r_{3}$: their given remainders, in order

CheckThe divisor must be bigger than the largest remainder. If the HCF comes out smaller than or equal to it, no such number exists. Exam questions usually satisfy this.

RememberIf the remainder is 0 (“divides exactly”) subtraction changes nothing and this is just the HCF of the numbers.

Steps

Subtract each number’s own remainder from it.
Find the HCF of the reduced numbers.
Check that the HCF is bigger than the largest remainder.
Divide each number by $N$ and match the remainders.

Example 5 · Basic

Find the greatest number which, on dividing 1084, 1885 and 2874, leaves the remainders 3, 5, 7 respectively.

Solution, step by step

1If dividing by $N$ leaves the remainders 3, 5, 7 in turn, then subtracting each remainder from its number makes it divisible by $N$ exactly. So 1084−3=1081, 1885−5=1880, 2874−7=2867 are all multiples of $N$ .
2The greatest $N$ is the HCF of these numbers: $1081 = 23 \times 47$ , $1880 = 2^{3} \times 5 \times 47$ , $2867 = 47 \times 61$ ⟹ $N = 47$ .
3One thing to make sure of: the divisor must be bigger than the remainders. 47 > 7 ✓. Check: $1084 = 47 \times 23 + 3$ , $1885 = 47 \times 40 + 5$ , $2874 = 47 \times 61 + 7$ .

Answer 47

All three reduced numbers became multiples of 47 (1081, 1880, 2867) — that is the test.

Example 6 · Standard

N

is the greatest number which, on dividing 1007, 1717 and 2430, leaves the remainders 4, 6, 11 respectively. Find the digit sum of

N

Solution, step by step

1If dividing by $N$ leaves the remainders 4, 6, 11 in turn, then subtracting each remainder from its number makes it divisible by $N$ exactly. So 1007−4=1003, 1717−6=1711, 2430−11=2419 are all multiples of $N$ .
2The greatest $N$ is the HCF of these numbers: $1003 = 17 \times 59$ , $1711 = 29 \times 59$ , $2419 = 41 \times 59$ ⟹ $N = 59$ .
3One thing to make sure of: the divisor must be bigger than the remainders. 59 > 11 ✓. Check: $1007 = 59 \times 17 + 4$ , $1717 = 59 \times 29 + 6$ , $2430 = 59 \times 41 + 11$ .
4The digit sum of $N = 59$ is $14$ .

Answer 14

The real solution ends at

N

; the digit sum is one more step after it.

Example 7 · Advanced

N

is the greatest number which, on dividing 563, 1031 and 1469, leaves the remainders 5, 8, 12 respectively. Find the sum of all the quotients.

Solution, step by step

1If dividing by $N$ leaves the remainders 5, 8, 12 in turn, then subtracting each remainder from its number makes it divisible by $N$ exactly. So 563−5=558, 1031−8=1023, 1469−12=1457 are all multiples of $N$ .
2The greatest $N$ is the HCF of these numbers: $558 = 2 \times 3^{2} \times 31$ , $1023 = 3 \times 11 \times 31$ , $1457 = 31 \times 47$ ⟹ $N = 31$ .
3One thing to make sure of: the divisor must be bigger than the remainders. 31 > 12 ✓. Check: $563 = 31 \times 18 + 5$ , $1031 = 31 \times 33 + 8$ , $1469 = 31 \times 47 + 12$ .
4Quotients: 558÷31, 1023÷31, 1457÷31 ⟹ 18, 33, 47; their sum is $98$ .

Answer 98

Each reduced number ÷

N

gives the quotient that the real division would have given.

BasicTry it yourself first

Find the greatest number which, on dividing 715, 1016 and 1546, leaves the remainders 2, 4, 5 respectively.

BasicTry it yourself first

Find the greatest number which, on dividing 995 and 1573, leaves the remainders 7, 15 respectively.

StandardTry it yourself first

Find the greatest number which, on dividing 944, 1819 and 2626, leaves the remainders 6, 10, 13 respectively.

StandardTry it yourself first

N

is the greatest number which, on dividing 1005, 1594 and 2593, leaves the remainders 9, 17, 20 respectively. Find the digit sum of

N

StandardTry it yourself first

N

is the greatest number which, on dividing 663, 1163 and 1659, leaves the remainders 3, 19, 31 respectively. Find the sum of all the quotients.

AdvancedTry it yourself first

Find the greatest number which, on dividing 1639, 2786 and 4712, leaves the remainders 14, 36, 87 respectively.

AdvancedTry it yourself first

Find the greatest number which, on dividing 788, 1484, 2272 and 3351, leaves the remainders 12, 29, 41, 53 respectively.

StandardTry it yourself first

Find the greatest number which, on dividing 1499, 2427 and 3711, leaves the remainders 8, 13, 19 respectively.

AdvancedTry it yourself first

N

is the greatest number which, on dividing 985, 2062 and 3404, leaves the remainders 6, 15, 22 respectively. Find the digit sum of

N

Smallest number: $L CM + r$ and $L CM - g$

Same remainder, or the same shortfall

For the smallest number: add the remainder to the LCM if it is the same, subtract the shortfall if the shortfall is the same.

The idea

If every divisor leaves remainder $r$ , then $N - r$ is divisible by all of them exactly. So $N - r$ is a common multiple, and for the smallest $N$ it is the least common multiple — the LCM. Hence $N = L CM + r$ . In the other form, if each remainder falls short of its divisor by exactly the same $g$ ( $d - r = g$ ), then $N + g$ is divisible by all and $N = L CM - g$ .

∑ Formula

N N = L CM + r = L CM - g

First: remainder

r

every time; second: divisor − remainder

= g

every time

$N$: the smallest number asked
$L CM$: the LCM of all the divisors
$r$: the same remainder each time
$g$: the same shortfall each time (divisor − remainder)

The remainders are	The number
all equal ( $r$ )	$L CM + r$
divisor − remainder all equal ( $g$ )	$L CM - g$
no remainder	a multiple of the $L CM$
“divisible when $g$ is added”	$L CM - g$
“divisible when $g$ is taken away”	$L CM + g$
different shortfalls	next node: the stepping method

Here: $N$ = the number asked, $r$ = the remainder each time, $g$ = the shortfall each time (divisor − remainder).

ConventionThe remainder

r

must be smaller than every divisor. By convention the “smallest number” is

L CM + r

— the number

r

itself technically leaves remainder

r

, but it is not counted because it is smaller than the divisors.

Steps

See: are the remainders equal, or is divisor − remainder equal?
Find the LCM of all the divisors.
The answer is $L CM + r$ or $L CM - g$ .
Check: divide by each divisor and match the remainders.
If the “next number” is asked add one more LCM (the numbers are $L CM \times k + r$ ).

Example 8 · Basic

Find the smallest number which leaves remainder 9 when divided by each of 14, 16, 21 and 24.

Solution, step by step

1The remainder is $9$ for every divisor, so $N - 9$ is divisible exactly by every divisor — it is a common multiple of the divisors.
2For the smallest $N$ take $N - 9$ as the least common multiple, the LCM. $14 = 2 \times 7$ , $16 = 2^{4}$ , $21 = 3 \times 7$ , $24 = 2^{3} \times 3$ ⟹ $L CM = 336$ .
3 $N = 336 + 9 = 345$ . Check: 345÷14 leaves 9; 345÷16 leaves 9; 345÷21 leaves 9; 345÷24 leaves 9.

Answer 345

The remainder 9 is smaller than the smallest divisor 14, so it is valid.

Example 9 · Standard

Find the smallest number which leaves the remainders 5, 9, 17, 27 when divided by 8, 12, 20 and 30 respectively.

Solution, step by step

1The remainders differ, but look at divisor − remainder: 8−5=3, 12−9=3, 20−17=3, 30−27=3 — it is 3 every time. So at every division the number falls short by exactly 3.
2So $N + 3$ is divisible exactly by every divisor. The smallest $N + 3$ is the LCM $= 120$ ( $8 = 2^{3}$ , $12 = 2^{2} \times 3$ , $20 = 2^{2} \times 5$ , $30 = 2 \times 3 \times 5$ ).
3 $N = 120 - 3 = 117$ . Check: 117÷8 leaves 5; 117÷12 leaves 9; 117÷20 leaves 17; 117÷30 leaves 27.

Answer 117

The remainders 5, 9, 17, 27 look different, but the shortfall is 3 every time — that is the trick.

Example 10 · Standard

What is the least number to add to 9000 so that the sum is exactly divisible by 16, 24 and 36?

Solution, step by step

1A number divisible by all of 16, 24 and 36 is a multiple of their LCM. $L CM = 144$ .
2Dividing 9000 by 144: $9000 = 144 \times 62 + 72$ — the remainder is 72.
3To reach the next multiple we must add $144 - 72 = 72$ : $9000 + 72 = 9072 = 144 \times 63$ .

Answer 72

The number to add = LCM − remainder; the number to subtract is just the remainder.

Example 11 · Standard

Find the smallest number which becomes exactly divisible by 15, 20 and 24 when 8 is added to it.

Solution, step by step

1When 8 is added to the number $N$ , the result $N + 8$ must be divisible exactly by every divisor, so it is a common multiple. For the smallest $N$ take $N + 8$ as the least common multiple, the LCM.
2 $15 = 3 \times 5$ , $20 = 2^{2} \times 5$ , $24 = 2^{3} \times 3$ ⟹ $L CM = 120$ .
3 $N + 8 = 120$ ⟹ $N = 120 - 8 = 112$ . Check: 112+8=120, and dividing 120 by each divisor gives 8, 6, 5 — all whole numbers.

Answer 112

“Divisible when 8 is added” is the shortfall form in disguise:

N + 8

= LCM ⟹

N = L CM - 8

. In the “taken away” form

N - 8

= LCM ⟹

N = L CM + 8

Example 12 · Advanced

How many numbers from 200 to 2000 (both included) leave remainder 5 when divided by each of 9, 12 and 15?

Solution, step by step

1Every such number has the form $180 k + 5$ , where $L CM = 180$ .
2 $200 \leq 180 k + 5 \leq 2000$ ⟹ $1.08 \leq k \leq 11.08$ , so the whole-number values of $k$ run from 2 to 11.
3There are 11−2+1=10 such values of $k$ , so 10 numbers.

Answer 10

All such numbers are

180 k + 5

, so the count is a count of values of

k

BasicTry it yourself first

Find the smallest number which leaves remainder 4 when divided by each of 10, 15 and 24.

BasicTry it yourself first

Find the smallest number which leaves the remainders 5, 9, 13 when divided by 6, 10 and 14 respectively.

StandardTry it yourself first

Find the smallest number which leaves remainder 9 when divided by each of 14, 21, 28 and 35.

StandardTry it yourself first

What is the least number to subtract from 12345 so that the result is exactly divisible by 8, 12 and 15?

StandardTry it yourself first

Find the second smallest number which leaves remainder 4 when divided by each of 6, 8, 9 and 12.

AdvancedTry it yourself first

Find the smallest number which leaves the remainders 5, 11, 17, 23 when divided by 9, 15, 21 and 27 respectively.

AdvancedTry it yourself first

How many numbers from 500 to 5000 (both included) leave remainder 3 when divided by each of 8, 10 and 12?

StandardTry it yourself first

Find the smallest number which becomes exactly divisible by 12, 18 and 21 when 7 is added to it.

StandardTry it yourself first

Find the smallest number which becomes exactly divisible by 10, 16 and 24 when 9 is subtracted from it.

An extra condition and different shortfalls

N = L k \pm

and trying

k

When the number must also be divisible by something else, or the shortfalls all differ: a step-by-step search.

The idea

The previous method gives a whole family of numbers: $N = L k + r$ (or $L k - g$ ), $k = 1, 2, 3, \dots$ — each number is $L$ ahead of the last. If another condition comes (like “divisible by $m$ ”) we put $k = 1, 2, 3, \dots$ in this family and find the first member that works. Divide $L$ by $m$ and note the remainder $ρ$ ; for division by $m$ the number then behaves like $ρ k + r$ ; so at most $m$ tries are needed.

∑ Formula

N N m = L k + r = L k - g ∣ N

try

k = 1, 2, 3, \dots

; the first

k

with

m ∣ N

$N$: the number asked
$L$: the LCM of the divisors
$k$: the counter $1, 2, 3, \dots$
$r, g$: the same remainder or the same shortfall
$m$: the extra number by which $N$ must be exactly divisible

When the shortfalls differ (say remainder 3 on dividing by $7$ and 5 on dividing by $8$ ) the LCM − something trick fails. Then list the numbers of the bigger divisor’s condition ( $8 j + 5$ : 5, 13, 21, …) and test the other condition on them. The first hit is the smallest; all later ones come by adding the LCM.

TrapAlways start from

k = 1

, not

k = 0

(that would give the remainder itself). If

ρ

and

m

share a factor there may be no solution at all — then look at the HCF of

m

and the LCM.

Steps

Write the general form: $L k + r$ (or $L k - g$ ).
Find $L mod m$ and, putting $k = 1, 2, 3, \dots$ , look at the remainder on dividing by $m$ each time.
The first $k$ at which the remainder is 0 gives the answer: $N = L k \pm \dots$ .
With different shortfalls, list the bigger divisor’s numbers and test the smaller divisor’s condition.

Example 13 · Basic

Find the smallest number which leaves remainder 3 when divided by each of 4, 6, 9 and 10 but is exactly divisible by 11.

Solution, step by step

1The remainder is 3 for every divisor, so the number has the form $180 k + 3$ with $L CM (4, 6, 9, 10) = 180$ ( $k = 1, 2, \dots$ ).
2It must also be divisible by 11. 180 leaves remainder 4 on division by 11, so for division by 11 the number $180 k + 3$ behaves like $4 k + 3$ . We test $k = 1, 2, 3, \dots$ one by one.
3 $k = 1$ : 183 ÷ 11 leaves 7; $k = 2$ : 363 ÷ 11 leaves 0 ✓.
4The first exact division comes at $k = 2$ , so the smallest number is $180 \times 2 + 3 = 363$ .

Answer 363

It took 2 tries — even with

m = 11

at most 11 would be needed.

Example 14 · Standard

Find the smallest number which leaves the remainders 4, 6, 10 when divided by 6, 8 and 12 respectively but is exactly divisible by 7.

Solution, step by step

1Divisor − remainder is 2 every time, so the number has the form $24 k - 2$ with $L CM (6, 8, 12) = 24$ ( $k = 1, 2, \dots$ ).
2It must also be divisible by 7. 24 leaves remainder 3 on division by 7, so for division by 7 the number $24 k - 2$ behaves like $3 k - 2$ . We test $k = 1, 2, 3, \dots$ one by one.
3 $k = 1$ : 22 ÷ 7 leaves 1; $k = 2$ : 46 ÷ 7 leaves 4; $k = 3$ : 70 ÷ 7 leaves 0 ✓.
4The first exact division comes at $k = 3$ , so the smallest number is $24 \times 3 - 2 = 70$ .

Answer 70

In the shortfall form the number is

L k - g

— do not forget to subtract.

Example 15 · Standard

Find the smallest number which leaves remainder 3 when divided by 7 and remainder 5 when divided by 8.

Solution, step by step

1Here divisor − remainder differs (7−3=4, 8−5=3), so the “LCM − something” trick does not work. We list the numbers of one condition and test the other.
2List the numbers of the bigger divisor’s condition: 5, 13, 21, … ( $8 j + 5$ ) — this keeps the list short.
3Divide them by 7 looking for remainder 3: 5→5, 13→6, 21→0, 29→1, 37→2, 45→3 ✓.
4The smallest number is 45. (All later ones come by adding multiples of $L CM = 56$ : 45, 101, …)

Answer 45

Using the bigger divisor 8 keeps the list short: at most 7 numbers have to be tested.

Example 16 · Advanced

Find the smallest number which leaves remainder 4 when divided by each of 5, 8 and 12 but is exactly divisible by 13.

Solution, step by step

1The remainder is 4 for every divisor, so the number has the form $120 k + 4$ with $L CM (5, 8, 12) = 120$ ( $k = 1, 2, \dots$ ).
2It must also be divisible by 13. 120 leaves remainder 3 on division by 13, so for division by 13 the number $120 k + 4$ behaves like $3 k + 4$ . We test $k = 1, 2, 3, \dots$ one by one.
3 $k = 1$ : 124 ÷ 13 leaves 7; $k = 2$ : 244 ÷ 13 leaves 10; $k = 3$ : 364 ÷ 13 leaves 0 ✓.
4The first exact division comes at $k = 3$ , so the smallest number is $120 \times 3 + 4 = 364$ .

Answer 364

On division by 13,

120 mod 13 = 3

; so

3 k + 4

must be divisible by 13.

BasicTry it yourself first

Find the smallest number which leaves remainder 2 when divided by each of 6, 9 and 15 but is exactly divisible by 7.

BasicTry it yourself first

Find the smallest number which leaves remainder 3 when divided by 5 and remainder 4 when divided by 7.

StandardTry it yourself first

Find the smallest number which leaves the remainders 7, 9, 12 when divided by 10, 12 and 15 respectively but is exactly divisible by 7.

StandardTry it yourself first

Find the smallest number which leaves remainder 4 when divided by 9 and remainder 7 when divided by 11.

AdvancedTry it yourself first

Find the smallest number which leaves remainder 5 when divided by each of 8, 12 and 18 but is exactly divisible by 13.

AdvancedTry it yourself first

Find the smallest number which leaves the remainders 5, 8, 14 when divided by 9, 12 and 18 respectively but is exactly divisible by 11.

AdvancedTry it yourself first

Find the smallest number which leaves remainder 7 when divided by 12 and remainder 9 when divided by 13.

StandardTry it yourself first

Find the smallest number which leaves remainder 7 when divided by each of 12, 15 and 18 but is exactly divisible by 13.

AdvancedTry it yourself first

Find the smallest number which leaves remainder 6 when divided by 13 and remainder 11 when divided by 15.

$n$ -digit numbers

Largest / smallest

n

-digit

The largest or smallest number with a fixed number of digits that meets an LCM condition, and a count of such numbers.

The idea

The largest $n$ -digit number is $99 \dots 9$ and the smallest is $100 \dots 0$ . The multiples of $L$ are spread out like a ladder: $L, 2 L, 3 L, \dots$ . For the largest, divide $99 \dots 9$ by $L$ and subtract the remainder (the top multiple). For the smallest, take the first multiple at or above $100 \dots 0$ (if there is a remainder, add $L$ minus it). For a number with remainder $r$ add $r$ to that multiple (or subtract $g$ if the shortfall is $g$ ) — and check that the number still has $n$ digits.

Asked	What to do
largest, exactly divisible	$9 \dots 9$ − ( $9 \dots 9 mod L$ )
smallest, exactly divisible	the first multiple of $L$ at or above $1 \dots 0$
largest, remainder $r$	the top multiple $+ r$ (subtract $L$ if too big)
smallest, shortfall $g$	the first multiple $- g$ (add $L$ if too small)
how many	$⌊ 9 \dots 9/ L ⌋ - ⌊ 9 \dots 9_{(n - 1)} / L ⌋$

Here: $L$ = the LCM of the divisors, $r$ = the remainder, $g$ = the shortfall, $9 \dots 9_{(n - 1)}$ = the largest number of $n - 1$ digits.

TrapThe commonest slip: adding

r

to the multiple pushes the number above

99 \dots 9

(or subtracting

g

drops it below

100 \dots 0

) — then move back/forward by one

L

Steps

Find the LCM $L$ of the divisors.
Divide the largest (or smallest) $n$ -digit number by $L$ and see the remainder.
Get the nearest multiple: by subtracting from the top / adding from the bottom.
Add $r$ or subtract $g$ ; if it crosses the limit, move by one $L$ .

Example 17 · Basic

Find the largest 4-digit number which is exactly divisible by 10, 14 and 21.

Solution, step by step

1The number must be divisible by all of 10, 14 and 21, that is a multiple of $L CM = 210$ ( $10 = 2 \times 5$ , $14 = 2 \times 7$ , $21 = 3 \times 7$ ).
2The biggest 4-digit number is 9999. Dividing by 210: $9999 = 210 \times 47 + 129$ .
3The biggest multiple of 210 not above 9999 is $9999 - 129 = 9870$ .
4So the answer is 9870.

Answer 9870

Largest: subtract the leftover 129 from 9999 — that gives the top multiple.

Example 18 · Standard

Find the smallest 5-digit number which leaves remainder 3 when divided by each of 8, 10 and 12.

Solution, step by step

1The remainder is 3 every time, so the number has the form $120 k + 3$ , where $L CM = 120$ ( $8 = 2^{3}$ , $10 = 2 \times 5$ , $12 = 2^{2} \times 3$ ).
2The smallest 5-digit number is 10000. $10000 = 120 \times 83 + 40$ .
3The smallest multiple of 120 that is not below 10000 is $10000 + 80 = 10080$ .
4Add the remainder: $10080 + 3 = 10083$ . It is not below 10000, so this is the answer.

Answer 10083

Smallest: add the gap 80 to reach the next multiple from 10000.

Example 19 · Standard

Find the largest 4-digit number which leaves remainder 5 when divided by each of 9 and 11.

Solution, step by step

1The remainder is 5 every time, so the number has the form $99 k + 5$ , where $L CM = 99$ ( $9 = 3^{2}$ , $11 = 11$ ).
2The biggest 4-digit number is 9999. Dividing by 99: $9999 = 99 \times 101 + 0$ .
3The biggest multiple of 99 not above 9999 is $9999 - 0 = 9999$ .
4Adding the remainder gives $9999 + 5 = 10004$ , which is above 9999 — too big for 4 digits. So step down by one 99: $10004 - 99 = 9905$ .

Answer 9905

The multiple is 9999; adding 5 leaves 4 digits — so step back by 99.

Example 20 · Advanced

Find the smallest 4-digit number which leaves the remainders 6, 23 when divided by 8 and 25 respectively.

Solution, step by step

1Divisor − remainder is 2 every time, so the number has the form $200 k - 2$ , where $L CM = 200$ ( $8 = 2^{3}$ , $25 = 5^{2}$ ).
2The smallest 4-digit number is 1000. $1000 = 200 \times 5 + 0$ .
3The smallest multiple of 200 that is not below 1000 is 1000.
4Subtracting 2 gives $1000 - 2 = 998$ , which is below 1000 — no longer 4 digits. So take the next multiple: $1200 - 2 = 1198$ .

Answer 1198

The first multiple is 1000; subtracting 2 gives 998 — only 3 digits. So take the next multiple 1200.

BasicTry it yourself first

Find the largest 3-digit number which is exactly divisible by 8, 12 and 15.

BasicTry it yourself first

Find the smallest 4-digit number which is exactly divisible by 9, 14 and 21.

StandardTry it yourself first

How many 4-digit numbers are exactly divisible by 12, 18 and 30?

StandardTry it yourself first

Find the largest 4-digit number which leaves the remainders 10, 14, 18 when divided by 16, 20 and 24 respectively.

StandardTry it yourself first

Find the smallest 5-digit number which leaves the remainders 6, 23 when divided by 8 and 25 respectively.

AdvancedTry it yourself first

Find the largest 5-digit number which leaves remainder 11 when divided by each of 14, 21 and 35.

AdvancedTry it yourself first

How many 5-digit numbers are exactly divisible by 14, 21 and 35?

StandardTry it yourself first

Find the smallest 3-digit number which leaves the remainders 7, 13, 19 when divided by 12, 18 and 24 respectively.

AdvancedTry it yourself first

Find the largest 4-digit number which leaves remainder 9 when divided by each of 12, 16 and 20.

Perfect squares and cubes

Make every power a multiple of 2 (or 3)

What to multiply or divide the LCM or a number by to make it a perfect square or cube.

The idea

In a perfect square every prime has an even power ( $36 = 2^{2} \times 3^{2}$ ), and in a perfect cube every power is a multiple of 3 ( $216 = 2^{3} \times 3^{3}$ ). So “the smallest perfect square divisible by these numbers” is: first take the LCM, then multiply by exactly what is needed to make every power in it even. In “by what should we multiply/divide to get a square” it is the same — complete the incomplete primes, or remove the extra.

∑ Formula

N M = \prod p^{e} = \prod p^{(- e) mod s}

Multiplying to make a perfect square/cube

$p$: a prime factor of $N$
$e$: the power of $p$ in $N$
$s$: 2 for a square, 3 for a cube
$M$: the smallest multiplier that makes $N \times M$ a perfect square/cube
$(- e) mod s$: how much more is needed to make the power a multiple of $s$

To make it by dividing, remove the extra power $e mod s$ . For “the smallest $n$ -digit perfect square divisible by these”: find the smallest such square $S_{0} = L CM \times (\dots)$ ; then all such squares are $S_{0} \times j^{2}$ ( $j = 1, 2, 3, \dots$ ) — find the first $j$ that gives $n$ digits.

TrapIf the LCM is already a perfect square (like 36) it is the answer — the multiplier is 1. And a lone prime (

2^{1}

) needs one more 2 to make a square, i.e.

2^{2}

, not

2^{3}

Steps

Write the LCM (or the given number) in prime powers.
Look at each power: even for a square, a multiple of 3 for a cube.
Build the multiplier for what is missing (or the divisor for the extra) and multiply (or divide).
For the $n$ -digit condition raise $j$ in $S_{0} \times j^{2}$ .

Example 21 · Basic

Find the smallest perfect square that is exactly divisible by 6, 9 and 14.

Solution, step by step

1A number divisible by all of them is a multiple of the LCM. $L CM = 126 = 2 \times 3^{2} \times 7$ .
2In a perfect square every prime has a power that is a multiple of 2. Here the power of 2 is 1 — 1 more needed; the power of 3 is 2 — fine; the power of 7 is 1 — 1 more needed.
3So multiply the LCM by $2 \times 7 = 14$ : $126 \times 14 = 1764 = 4 2^{2}$ .

Answer 1764

In the LCM 126 the power of 3 was already 2, so we did not multiply by 3.

Example 22 · Standard

By what smallest number should 1176 be multiplied so that the result is a perfect square?

Solution, step by step

1 $1176 = 2^{3} \times 3 \times 7^{2}$ . In a perfect square every prime power must be a multiple of 2.
2Primes whose power is not a multiple of 2 must be multiplied by what is missing: 2: power 3, missing 1; 3: power 1, missing 1; 7: power 2, missing 0.
3The multiplier $= 2 \times 3 = 6$ ; the result is $7056 = 8 4^{2}$ .

Answer 6

7^{2}

was already complete; the powers 3 of 2 and 1 of 3 were incomplete.

Example 23 · Standard

Find the smallest perfect cube that is exactly divisible by 6, 10 and 15.

Solution, step by step

1A number divisible by all of them is a multiple of the LCM. $L CM = 30 = 2 \times 3 \times 5$ .
2In a perfect cube every prime has a power that is a multiple of 3. Here the power of 2 is 1 — 2 more needed; the power of 3 is 1 — 2 more needed; the power of 5 is 1 — 2 more needed.
3So multiply the LCM by $2^{2} \times 3^{2} \times 5^{2} = 900$ : $30 \times 900 = 27000 = 3 0^{3}$ .

Answer 27000

For a cube every power must be a multiple of 3:

2^{1} \to 2^{3}

, so

2^{2}

more is needed.

Example 24 · Advanced

Find the smallest 4-digit perfect square that is exactly divisible by 6, 8 and 9.

Solution, step by step

1 $L CM = 72 = 2^{3} \times 3^{2}$ ; to make it a perfect square multiply by $2$ : the smallest such square is $72 \times 2 = 144 = 1 2^{2}$ .
2All such squares are $144 \times j^{2}$ ( $j = 1, 2, 3, \dots$ ): a square times a square is a square, and no other square multiple is missed.
3To have 4 digits it must be at least 1000: $j = 1$ : 144; $j = 2$ : 576; $j = 3$ : 1296.
4The first one that reaches 1000 or more: $1296 = 3 6^{2}$ .

Answer 1296

The smallest square 144 has three digits; in the family of square multiples 1296 at

j = 3

is the first with four digits.

BasicTry it yourself first

By what smallest number should 540 be multiplied so that the result is a perfect square?

BasicTry it yourself first

By what smallest number should 2352 be divided so that the result is a perfect square?

StandardTry it yourself first

Find the smallest perfect square that is exactly divisible by 14, 15 and 21.

StandardTry it yourself first

By what smallest number should 1500 be multiplied so that the result is a perfect cube?

AdvancedTry it yourself first

Find the smallest 5-digit perfect square that is exactly divisible by 6, 10 and 12.

AdvancedTry it yourself first

By what smallest number should 5488 be divided so that the result is a perfect cube?

AdvancedTry it yourself first

Find the smallest perfect cube that is exactly divisible by 8, 12, 18 and 27.

StandardTry it yourself first

Find the smallest perfect square that is exactly divisible by 12, 16 and 27.

AdvancedTry it yourself first

By what smallest number should 1470 be divided so that the result is a perfect square?

Together again

Bells · runners · gears · calendars

When things repeating at their own rhythms are first together again — the LCM.

All these questions tell one story: several things repeat at their own rhythms and we ask when they all meet at the same moment for the first time. Each acts at multiples of its own interval, so the moment of togetherness is a common multiple of the intervals — and the first such moment is their LCM.

Bells, lights and clocks

The LCM, the clock time, and counting

The next time bells or lights act together and how many times they do so in a period.

The idea

Three bells ring at intervals of 4, 6 and 9 seconds and rang together at time 0. The first rings at 4, 8, 12, …, the second at 6, 12, 18, …, the third at 9, 18, 27, …. In the figure each row is one bell and the dots are its ringing moments. For the first time the dots of all three meet on one vertical line, at 36 seconds — and $36 = L CM (4, 6, 9)$ .

◭ Diagram

Three bells: intervals 4, 6, 9 seconds — first together at 36 seconds

∑ Formula

G n = L CM (a, b, c, \dots) = ⌊ \frac{T}{G} ⌋ (+ 1)

The gap between togethernesses and the count

$G$: the time until they are first together again (the gap)
$a, b, c$: the interval of each (in one unit)
$T$: the total period (in the same unit)
$n$: the number of togethernesses (not counting the start; $+ 1$ if the start is counted)
$⌊ ⌋$: the whole-number part (drop the decimal)

TrapMake the unit one — if the intervals are in seconds and the period in hours, turn the hours into seconds (

1

hour

= 3600

seconds). And is the start counted or not — “they rang together at the start, how many more times in the next hour?” takes

⌊ T / L ⌋

; “how many times in all (counting the start)” takes

+ 1

RememberIf the question asks “at what time” the LCM is a duration — add it to the starting clock time. Watch the 60s when converting to hours and minutes.

Steps

Write all intervals in one unit.
Find their LCM — the gap between two togethernesses.
“At what time”: turn the LCM into hours-minutes-seconds and add it to the start time.
“How many times”: the whole part of total period ÷ LCM; $+ 1$ if the start counts.

Example 25 · Basic

The bells of three temples ring at intervals of 24, 36 and 60 seconds respectively. They did so together at 9:15:00 am. When will they next ring together?

Solution, step by step

1They were all together at 9:15:00 am. After that each one acts only at multiples of its own interval: the first after 24, 48, 72, … seconds, and so on. They coincide only at a time which is a multiple of every interval; the first such time is their LCM.
2 $24 = 2^{3} \times 3$ , $36 = 2^{2} \times 3^{2}$ , $60 = 2^{2} \times 3 \times 5$ ⟹ $L CM = 360$ seconds.
3360 seconds = 6 min.
4Adding 6 min to 9:15:00 am gives 9:21:00 am.

Answer 9:21:00 am

The 360 seconds were divided by 60 to get 6 minutes before adding to the clock.

Example 26 · Standard

Three alarm clocks ring at intervals of 25, 40 and 60 minutes respectively. They did so together at 6:30 am. When will they next ring together?

Solution, step by step

1They were all together at 6:30 am. After that each one acts only at multiples of its own interval: the first after 25, 50, 75, … minutes, and so on. They coincide only at a time which is a multiple of every interval; the first such time is their LCM.
2 $25 = 5^{2}$ , $40 = 2^{3} \times 5$ , $60 = 2^{2} \times 3 \times 5$ ⟹ $L CM = 600$ minutes.
3600 minutes = 10 h.
4Adding 10 h to 6:30 am gives 4:30 pm.

Answer 4:30 pm

600 minutes

= 10

hours; 6:30 am + 10 hours = 4:30 pm.

Example 27 · Standard

The bells of three temples ring at intervals of 18, 24 and 30 seconds respectively. They start together. How many times do they ring together in 2 hours (counting the start)?

Solution, step by step

1The gap between two togethernesses is $L CM (18, 24, 30)$ : $18 = 2 \times 3^{2}$ , $24 = 2^{3} \times 3$ , $30 = 2 \times 3 \times 5$ ⟹ 360 seconds = 6 min.
2The total time 2 hours $= 7200$ seconds.
3Times they come together after the start in this period: $⌊ \frac{7200}{360} ⌋ = 20$ (it divides exactly, so the last time falls right at the end).
4The starting moment (time 0) is also to be counted: $20 + 1 = 21$ .

Answer 21

7200 ÷ 360 is exactly 20, so the 20th togetherness is right at the end of 2 hours. Counting the start makes 21.

Example 28 · Advanced

Three traffic lights at a crossing change colour at intervals of 28, 42 and 70 seconds respectively. They start together. How many times do they change together in 3 hours (counting the start)?

Solution, step by step

1The gap between two togethernesses is $L CM (28, 42, 70)$ : $28 = 2^{2} \times 7$ , $42 = 2 \times 3 \times 7$ , $70 = 2 \times 5 \times 7$ ⟹ 420 seconds = 7 min.
2The total time 3 hours $= 10800$ seconds.
3Times they come together after the start in this period: $⌊ \frac{10800}{420} ⌋ = 25$ .
4The starting moment (time 0) is also to be counted: $25 + 1 = 26$ .

Answer 26

Here 10800 ÷ 420 is not exact (25.7…), so the whole part 25 was taken — the next togetherness does not fit in the leftover 300 seconds.

BasicTry it yourself first

The bells of three temples ring at intervals of 16, 24 and 40 seconds respectively. They did so together at 10:00:00 am. When will they next ring together?

BasicTry it yourself first

Two traffic lights at a crossing change colour at intervals of 30 and 45 seconds respectively. They did so together at 7:45:00 am. When will they next change together?

StandardTry it yourself first

Three lighthouses flash at intervals of 48, 60 and 72 seconds respectively. They start together. How many times do they flash together in 1 hour (counting the start)?

StandardTry it yourself first

Three alarm clocks ring at intervals of 45, 60 and 90 minutes respectively. They did so together at 8:00 am. When will they next ring together?

StandardTry it yourself first

Three lighthouses flash at intervals of 20, 30 and 45 minutes respectively. They start together. How many times do they flash together in 24 hours (not counting the start)?

AdvancedTry it yourself first

The bells of three temples ring at intervals of 42, 56 and 63 seconds respectively. They start together. How many times do they ring together in 2 hours (counting the start)?

AdvancedTry it yourself first

Four traffic lights at a crossing change colour at intervals of 36, 48, 54 and 72 seconds respectively. They did so together at 6:50:30 am. When will they next change together?

StandardTry it yourself first

The bells of three temples ring at intervals of 40, 56 and 70 seconds respectively. They did so together at 5:45:10 am. When will they next ring together?

AdvancedTry it yourself first

Three alarm clocks ring at intervals of 90, 120 and 150 seconds respectively. They start together. How many times do they ring together in 5 hours (counting the start)?

Circular tracks and gears

Lap times and numbers of teeth

Runners meeting again at the start of a circular track, and gear teeth meeting again.

The idea

Track: each runner is back at the start after his lap time $\frac{D}{v}$ . So all the runners are together at the start only when the total time is a common multiple of all lap times — the LCM. Lap times are often fractions, and then the LCM of fractions $\frac{L CM ( numerators )}{H CF ( denominators )}$ is used. Gears: two meshed wheels turn through equal numbers of teeth. The same two teeth meet again only when both wheels have made whole turns — at the LCM of the numbers of teeth.

∑ Formula

T w = L CM (\frac{D}{v _{1}}, \frac{D}{v _{2}}, \dots) = \frac{L CM ( t _{1} , t _{2} )}{t _{2}}

Track and gears

$T$: the first time all are together at the start again
$D$: the length of the track
$v_{1}, v_{2}, \dots$: the speeds of the runners
$t_{1}, t_{2}$: the numbers of teeth of the two wheels ( $t_{2}$ the bigger)
$w$: turns of the bigger wheel between two meetings of the same teeth

TrapThis method is only for “together at the starting point”. If the question asks when they first meet anywhere on the track, it is a relative-speed question, not an LCM one.

RememberIn gears the bigger wheel turns fewer times:

L CM \div t_{2}

for the big and

L CM \div t_{1}

for the small. These two turn-counts are co-prime to each other.

Steps

Track: write each runner’s lap time $\frac{D}{v}$ (as a simple fraction).
Take the LCM of the lap times: LCM of numerators ÷ HCF of denominators.
Laps of each runner: speed × time ÷ $D$ — check that all are whole numbers.
Gears: the LCM of the teeth, then the turns, then total turns ÷ turns per meeting.

Example 29 · Basic

Three runners start together from one point of a circular track 600 m long. Their speeds are 40, 50, 75 m/min. After how many minutes will they first be together again at the starting point?

Solution, step by step

1Time for one lap $= \frac{length of track}{speed}$ : $\frac{600}{40} = \frac{15}{1}$ , $\frac{600}{50} = \frac{12}{1}$ , $\frac{600}{75} = \frac{8}{1}$ minutes.
2They are together at the starting point only when each has finished a whole number of laps — so the time must be the LCM of all the lap times. LCM of fractions $= \frac{LCM of numerators}{HCF of denominators} = \frac{L CM ( 15 , 12 , 8 )}{H CF ( 1 , 1 , 1 )} = 120$ .
3Time $= 120$ minutes.

Answer 120 minutes

The lap times came out whole (15, 12, 8), so an ordinary LCM worked.

Example 30 · Standard

Three runners start together from one point of a circular track 15 km long. Their speeds are 6, 10, 9 km/h. After how many hours will they first be together again at the starting point?

Solution, step by step

1Time for one lap $= \frac{length of track}{speed}$ : $\frac{15}{6} = \frac{5}{2}$ , $\frac{15}{10} = \frac{3}{2}$ , $\frac{15}{9} = \frac{5}{3}$ hours.
2They are together at the starting point only when each has finished a whole number of laps — so the time must be the LCM of all the lap times. LCM of fractions $= \frac{LCM of numerators}{HCF of denominators} = \frac{L CM ( 5 , 3 , 5 )}{H CF ( 2 , 2 , 3 )} = 15$ .
3Time $= 15$ hours.

Answer 15 hours

The lap times

\frac{5}{2}, \frac{3}{2}, \frac{5}{3}

are fractions — so the LCM of numerators is 15 and the HCF of denominators 1.

Example 31 · Standard

Two meshed gear wheels have 18 and 30 teeth. The bigger wheel makes 4 turns per hour. At the start two particular teeth, one of each, are in contact. How many times in 15 hours will the same two teeth meet again after the start?

Solution, step by step

1At the start two particular teeth (one of each wheel) are in contact. The two wheels turn through equal numbers of teeth (the gears are meshed), so the same two teeth meet again only when both wheels have made whole turns — at the LCM of the numbers of teeth.
2 $L CM (18, 30) = 90$ teeth. In this the bigger wheel (30 teeth) makes $90 \div 30 = 3$ turns and the smaller one (18 teeth) $90 \div 18 = 5$ turns.
3In 15 hours the bigger wheel turns $4 \times 15 = 60$ times.
4The same teeth meet after every 3 turns, so after the start they meet $60 \div 3 = 20$ times.

Answer 20

The starting contact was not counted — the question said “after the start”.

Example 32 · Advanced

Three runners start together from one point of a circular track 21 km long. Their speeds are 6, 9, 14 km/h. By the time they are first together again, how many laps will the fastest runner have run?

Solution, step by step

1Time for one lap $= \frac{length of track}{speed}$ : $\frac{21}{6} = \frac{7}{2}$ , $\frac{21}{9} = \frac{7}{3}$ , $\frac{21}{14} = \frac{3}{2}$ hours.
2They are together at the starting point only when each has finished a whole number of laps — so the time must be the LCM of all the lap times. LCM of fractions $= \frac{LCM of numerators}{HCF of denominators} = \frac{L CM ( 7 , 7 , 3 )}{H CF ( 2 , 3 , 2 )} = 21$ .
3Time $= 21$ hours.
4Laps of each $= speed \times time \div 21$ : $6 \times 21 \div 21 = 6$ , $9 \times 21 \div 21 = 9$ , $14 \times 21 \div 21 = 14$ . The fastest has run 14 laps.

Answer 14

The HCF of the laps 6, 9, 14 is 1 — that confirms 21 hours is the very first time.

BasicTry it yourself first

Three runners start together from one point of a circular track 400 m long. Their speeds are 25, 40, 50 m/min. After how many minutes will they first be together again at the starting point?

BasicTry it yourself first

Two meshed gear wheels have 20 and 28 teeth. The bigger wheel makes 5 turns per hour. At the start two particular teeth, one of each, are in contact. How many times in 12 hours will the same two teeth meet again after the start?

StandardTry it yourself first

Three runners start together from one point of a circular track 36 km long. Their speeds are 5, 8, 10 km/h. After how many hours will they first be together again at the starting point?

StandardTry it yourself first

Two meshed gear wheels have 24 and 40 teeth. The bigger wheel makes 3 turns per hour. How many turns does the smaller wheel make in 20 hours?

AdvancedTry it yourself first

Three runners start together from one point of a circular track 18 km long. Their speeds are 4, 6, 9 km/h. By the time they are first together again, how many laps will the fastest runner have run?

AdvancedTry it yourself first

Two meshed gear wheels have 24 and 36 teeth. The bigger wheel makes 5 turns per hour. At the start two particular teeth, one of each, are in contact. How many times in 18 hours will the same two teeth meet again after the start?

AdvancedTry it yourself first

Three runners start together from one point of a circular track 420 m long. Their speeds are 35, 45, 60 m/min. After how many minutes will they first be together again at the starting point?

StandardTry it yourself first

Three runners start together from one point of a circular track 30 km long. Their speeds are 9, 12, 20 km/h. After how many hours will they first be together again at the starting point?

AdvancedTry it yourself first

Two meshed gear wheels have 21 and 35 teeth. The bigger wheel makes 6 turns per hour. At the start two particular teeth, one of each, are in contact. How many times in 10 hours will the same two teeth meet again after the start?

Dates, weekdays and different starts

The LCM in days, and the remainder mod 7

Jobs repeating every so many days being together again: the date, the weekday, and the first common day when the starts differ.

The idea

Two jobs repeat every $a$ and $b$ days and were together on one day. They are next together after $L CM (a, b)$ days. For the date count that many days ahead. For the weekday, since a week has 7 days, move ahead only $L CM mod 7$ days. If the starts differ (one begins on day 3, the other on day 5) we must first reach the first common day: list one job’s days and see which also belongs to the other; after that a common day comes every $L CM$ days.

∑ Formula

s d_{k} = L CM (a, b) mod 7 = d_{1} + k L

When both jobs start on the same day, or from the first common day

d_{1}

$a, b$: the intervals of the two jobs (in days)
$L$: $L CM (a, b)$
$d_{1}$: the first common day
$s$: the weekday shift in days
$d_{k}$: the $k$ -th common day ( $k = 0, 1, 2, \dots$ )
$mod 7$: the remainder on dividing by 7

TrapWhen moving a date ahead mind the lengths of the months (28/29/30/31). And in “how many times in the first

N

days” count the first common day too: the count

= ⌊ \frac{N - d _{1}}{L} ⌋ + 1

Steps

Find the LCM of the intervals (in days).
Date: add the LCM days to the starting date. Weekday: move $L CM mod 7$ days ahead.
If the starts differ: list one job’s days and test the other’s condition; the first common day is $d_{1}$ .
Then a common day every $L$ days; for a count use $⌊ \frac{N - d _{1}}{L} ⌋ + 1$ .

Example 33 · Basic

One doctor visits a hospital every 6 days and another every 8 days. Both were there on 5 March (Friday). On what date and weekday will they next be there together?

Solution, step by step

1One job happens every 6 days and another every 8 days. Both happened on 5 March (Friday). They coincide only when the time passed is a multiple of both 6 and 8 — the first time after $L CM (6, 8)$ days.
2 $6 = 2 \times 3$ , $8 = 2^{3}$ ⟹ $L CM = 24$ days.
324 days after 5 March: 29 March.
4A week has 7 days, so the weekday moves by $24 mod 7 = 3$ — 3 days after Friday is Monday.

Answer 29 March, Monday

We did not count the 24 days one by one — 5 March + 24 is straight 29 March.

Example 34 · Standard

One volunteer helps at a library every 9 days and another every 12 days. Both were there on 1 May (Saturday). On what date will they next be there together?

Solution, step by step

1One job happens every 9 days and another every 12 days. Both happened on 1 May (Saturday). They coincide only when the time passed is a multiple of both 9 and 12 — the first time after $L CM (9, 12)$ days.
2 $9 = 3^{2}$ , $12 = 2^{2} \times 3$ ⟹ $L CM = 36$ days.
336 days after 1 May: 6 June.

Answer 6 June

May has 31 days: 1 May + 30 = 31 May, then 6 more days = 6 June.

Example 35 · Standard

One agency cleans a park every 4 days, its first cleaning being on day 3 of the month. Another agency cleans every 6 days, its first cleaning being on day 5. On which day of the month will both clean on the same day for the first time?

Solution, step by step

1The first group’s days: 3, 7, 11 … (every 4 days from day 3). Find the first of these that is also one of the second group’s days $5, 11, 17, \dots$ ( $5 + 6 j$ ).
23: before day 5 ✗; 7: 2 ÷ 6 leaves 2 ✗; 11: 6 after 5, a multiple of 6 ✓.
3The first common day is 11. After that a common day comes every $L CM (4, 6) = 12$ days.

Answer 11

The first match is at 11; later matches follow every 12 days: 11, 23, 35, ….

Example 36 · Advanced

One agency cleans a park every 6 days, its first cleaning being on day 2 of the month. Another agency cleans every 9 days, its first cleaning being on day 5. In the first 120 days, on how many days do both clean on the same day?

Solution, step by step

1The first group’s days: 2, 8, 14 … (every 6 days from day 2). Find the first of these that is also one of the second group’s days $5, 14, 23, \dots$ ( $5 + 9 j$ ).
22: before day 5 ✗; 8: 3 ÷ 9 leaves 3 ✗; 14: 9 after 5, a multiple of 9 ✓.
3The first common day is 14. After that a common day comes every $L CM (6, 9) = 18$ days.
4Common days up to day 120: $14, 32, \dots$ — in all $⌊ \frac{120 - 14}{18} ⌋ + 1 = 6$ .

Answer 6

The first common day is 14 and the step is 18; up to 120 it is

⌊ 106 \div 18 ⌋ + 1 = 6

BasicTry it yourself first

One guard goes on night patrol every 5 days and another every 6 days. Both were on patrol on 10 July (Saturday). On what date will they next be on patrol together?

BasicTry it yourself first

One doctor visits a hospital every 4 days and another every 10 days. Both were there on 3 September (Friday). On which day of the week will they next be there together?

StandardTry it yourself first

One truck unloads at a warehouse every 8 days and another every 12 days. Both arrived on 7 October (Thursday). On what date and weekday will they next arrive together?

StandardTry it yourself first

One agency cleans a park every 5 days, its first cleaning being on day 2 of the month. Another agency cleans every 7 days, its first cleaning being on day 4. On which day of the month will both clean on the same day for the first time?

StandardTry it yourself first

One agency cleans a park every 4 days, its first cleaning being on day 1 of the month. Another agency cleans every 10 days, its first cleaning being on day 7. In the first 200 days, on how many days do both clean on the same day?

AdvancedTry it yourself first

One volunteer helps at a library every 14 days and another every 21 days. Both were there on 20 February (Saturday). On what date and weekday will they next be there together?

AdvancedTry it yourself first

One agency cleans a park every 9 days, its first cleaning being on day 4 of the month. Another agency cleans every 12 days, its first cleaning being on day 7. In the first 365 days, on how many days do both clean on the same day?

StandardTry it yourself first

One guard goes on night patrol every 15 days and another every 20 days. Both were on patrol on 2 September (Thursday). On which day of the week will they next be on patrol together?

AdvancedTry it yourself first

One agency cleans a park every 8 days, its first cleaning being on day 5 of the month. Another agency cleans every 12 days, its first cleaning being on day 9. On which day of the month will both clean on the same day for the first time?

Cutting, dividing and building up

Ropes · tiles · posts · bricks · steps

Questions on splitting things into equal pieces (HCF) and building a bigger size from small units (LCM).

The story here is of “cutting” or of “building”. In cutting / dividing (ropes, tiles, groups) the piece divides every length exactly — the HCF. In building (squares, cubes, steps, capacities) the finished size is a multiple of every unit — the LCM. Once you tell the two apart, half the work is done.

Longest piece, biggest vessel, fewest groups

Piece = HCF, count = sum (length ÷ HCF)

Questions on equal pieces, vessels, groups and packets with nothing left over.

The idea

Ropes are to be cut into equal pieces with nothing left over, each piece as long as possible. A piece divides every rope length exactly, so its length is a common divisor of all the lengths; “longest” means the greatest common divisor — the HCF. Once the piece length is known, the pieces from each rope = length ÷ HCF and the total is their sum. The same reasoning works for a vessel (the biggest that measures every quantity), for groups (fewest groups ⟹ biggest groups) and for packets (most packets).

∑ Formula

p n = H CF (l_{1}, l_{2}, \dots) = \frac{l _{1}}{p} + \frac{l _{2}}{p} + \dots

The longest piece and the total number of pieces

$l_{1}, l_{2}, \dots$: all the lengths (in one unit)
$p$: the length of a piece = the HCF of all the lengths
$n$: the total number of pieces

Story	What the HCF gives	Then
pieces of rope / rod / ribbon	length of a piece	pieces = sum of length ÷ HCF
a vessel for milk / oil	capacity of the vessel	fillings = sum of quantity ÷ HCF
fewest groups (one kind, equal size)	size of a group	groups = sum of count ÷ HCF
most identical packets (every item in each)	the number of packets	in each packet = count ÷ HCF

Here: $l_{i}$ = the $i$ -th length (or quantity, or count).

TrapOne unit: turn 2 m 40 cm into 240 cm. And note that in a group question “fewest groups” means the biggest size; whereas in a packet question where “every item goes into every packet” it is the number of packets that is the HCF, not the size.

Steps

Write all the lengths/quantities in one unit.
Find their HCF — the piece / vessel / group size / number of packets.
Divide each by the HCF to get the counts; add them if the total is wanted.
Check that every length divided exactly (nothing left).

Example 37 · Basic

Iron rods of lengths 2 m 70 cm, 4 m 5 cm and 6 m 30 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. Find the length of each piece (in cm).

Solution, step by step

1The length of a piece must divide every length exactly (nothing left over), and we want the longest piece. So the length of a piece = the HCF of the lengths.
2Write all in one unit (cm): 2 m 70 cm $= 270$ cm, 4 m 5 cm $= 405$ cm, 6 m 30 cm $= 630$ cm.
3 $270 = 2 \times 3^{3} \times 5$ , $405 = 3^{4} \times 5$ , $630 = 2 \times 3^{2} \times 5 \times 7$ ⟹ $H CF = 45$ cm.

Answer 45 cm

Mixed lengths like 2 m 70 cm were first changed into cm.

Example 38 · Standard

Ropes of lengths 2 m 40 cm, 4 m 20 cm and 6 m 60 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. How many pieces will there be in all?

Solution, step by step

1The length of a piece must divide every length exactly (nothing left over), and we want the longest piece. So the length of a piece = the HCF of the lengths.
2Write all in one unit (cm): 2 m 40 cm $= 240$ cm, 4 m 20 cm $= 420$ cm, 6 m 60 cm $= 660$ cm.
3 $240 = 2^{4} \times 3 \times 5$ , $420 = 2^{2} \times 3 \times 5 \times 7$ , $660 = 2^{2} \times 3 \times 5 \times 11$ ⟹ $H CF = 60$ cm.
4Pieces from each rope: 240÷60=4, 420÷60=7, 660÷60=11. In all $4 + 7 + 11 = 22$ pieces.

Answer 22

The number of pieces differs from rope to rope; for the total add them — do not just divide by the HCF.

Example 39 · Standard

A dairy has three tanks holding 120, 168 and 216 litres of milk. They are to be measured with one vessel, leaving nothing over. Using the biggest such vessel, how many times in all must it be filled to measure everything?

Solution, step by step

1The vessel must measure each quantity exactly and be as big as possible, so its capacity = the HCF of the quantities.
2 $120 = 2^{3} \times 3 \times 5$ , $168 = 2^{3} \times 3 \times 7$ , $216 = 2^{3} \times 3^{3}$ ⟹ $H CF = 24$ litres.
3The vessel is filled 120÷24=5, 168÷24=7, 216÷24=9 times for the three tanks. In all $5 + 7 + 9 = 21$ times.

Answer 21

Pause at the vessel size (24) first — then count the fillings.

Example 40 · Advanced

A shop has 264 mango trees, 396 guava trees, 462 lemon trees, 594 jamun trees. It wants to make as many identical gift packets as possible, each holding every item, with nothing left over. How many mango trees will each packet hold?

Solution, step by step

1Every packet must hold every item and all packets must be alike, so the number of packets must divide every count. For the most packets, the number = the HCF.
2 $264 = 2^{3} \times 3 \times 11$ , $396 = 2^{2} \times 3^{2} \times 11$ , $462 = 2 \times 3 \times 7 \times 11$ , $594 = 2 \times 3^{3} \times 11$ ⟹ $H CF = 66$ packets.
3Each packet holds: 264÷66=4 mango trees, 396÷66=6 guava trees, 462÷66=7 lemon trees, 594÷66=9 jamun trees.

Answer 4

With four items it is the same: the HCF of all four. The number of packets is 66, each holding 4, 6, 7, 9.

BasicTry it yourself first

Ribbons of lengths 1 m 80 cm and 3 m are to be cut into equal pieces with nothing left over, each piece as long as possible. Find the length of each piece (in cm).

BasicTry it yourself first

A dairy has three tanks holding 96, 144 and 240 litres of water. They are to be measured with one vessel, leaving nothing over. What is the capacity (in litres) of the biggest such vessel?

StandardTry it yourself first

Pipes of lengths 210 cm, 350 cm and 490 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. How many pieces will there be in all?

StandardTry it yourself first

120 NCC cadets, 168 scouts, 264 guides take part in a parade. They are to be split into squads; a squad cannot mix two kinds and all squads must be equally strong. What is the least number of squads?

StandardTry it yourself first

A dairy has four tanks holding 250, 350, 550 and 750 litres of water. They are to be measured with one vessel, leaving nothing over. Using the biggest such vessel, how many times in all must it be filled to measure everything?

AdvancedTry it yourself first

A shop has 168 books, 252 notebooks, 336 pens. It wants to make as many identical gift packets as possible, each holding every item, with nothing left over. How many packets can be made?

AdvancedTry it yourself first

Ropes of lengths 6 m 30 cm, 8 m 40 cm, 10 m 50 cm and 14 m 70 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. How many pieces will there be in all?

StandardTry it yourself first

144 children, 180 parents, 252 teachers take part in a parade. They are to be split into squads; a squad cannot mix two kinds and all squads must be equally strong. What is the least number of squads?

AdvancedTry it yourself first

Pipes of lengths 4 m 20 cm, 7 m 35 cm and 10 m 50 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. Find the length of each piece (in cm).

Tiles, cubes and posts

Side = HCF; count = area ÷ side²

The biggest square tile for a floor, the biggest cubes from a solid, and posts at equal gaps along a boundary.

The idea

A rectangular floor is to be covered with square tiles without cutting any. The side of a tile must divide both the length and the width — a common divisor. The biggest tile has side $= H CF (l, b)$ , and then the fewest tiles are needed. The count = floor area ÷ area of one tile. In the figure a $30 \times 18$ floor with tiles of side 6: $30 \div 6 = 5$ and $18 \div 6 = 3$ , in all 15 tiles.

◭ Diagram

30 \times 18

floor, tiles of side

H CF (30, 18) = 6

5 \times 3 = 15

∑ Formula

s s = H CF (l, b), n = \frac{l \times b}{s ^{2}} = H CF (a, b, c), n = \frac{a b c}{s ^{3}}

Tiles (squares) and cubes

$s$: the side of the tile or cube
$l, b$: the length and breadth of the floor
$a, b, c$: the three edges of the solid block
$n$: the number of tiles or cubes

Posts: posts at equal gaps along the boundary of a field, one at every corner. The gap divides every side exactly — the HCF. On a closed boundary the posts are as many as the gaps: $\frac{perimeter}{gap}$ (for a rectangle the sides are $l, b, l, b$ ).

Spot itThe biggest tile ⟹ the fewest tiles. If “the least number of tiles” is asked, it is still the HCF. And if the tile is in cm and the room in metres, first make one unit.

Steps

Write all measures in one unit.
The side of the tile/cube or the gap of the posts = the HCF of the measures.
Tiles: (length ÷ side) × (breadth ÷ side); cubes: the product of the three quotients.
Posts: perimeter ÷ gap. For a cost, number × rate.

Example 41 · Basic

A hall floor is 12 m 60 cm long and 5 m 40 cm wide. It is to be covered with square tiles of one size, without cutting any tile. What is the side (in cm) of the biggest possible tile?

Solution, step by step

1The square tiles must fit without cutting, so the side must divide both the length and the width. For the biggest tile the side $= H CF (1260, 540)$ .
2 $1260 = 2^{2} \times 3^{2} \times 5 \times 7$ , $540 = 2^{2} \times 3^{3} \times 5$ ⟹ $H CF = 180$ cm = 1 m 80 cm.

Answer 180 cm

The metre-centimetre lengths were first put in cm: 1260 and 540.

Example 42 · Standard

A hall floor is 21 m long and 15 m wide. It is to be covered with square tiles of one size, without cutting any tile. What is the least number of tiles needed?

Solution, step by step

1The square tiles must fit without cutting, so the side must divide both the length and the width. For the biggest tile the side $= H CF (2100, 1500)$ .
2 $2100 = 2^{2} \times 3 \times 5^{2} \times 7$ , $1500 = 2^{2} \times 3 \times 5^{3}$ ⟹ $H CF = 300$ cm = 3 m.
37 tiles along the length ( $2100 \div 300$ ) and 5 along the width ( $1500 \div 300$ ), so $7 \times 5 = 35$ tiles in all (the biggest tiles need the fewest tiles).

Answer 35

Floor area ÷ tile area would also give the same

7 \times 5

Example 43 · Standard

A hall floor is 12 m 60 cm long and 5 m 40 cm wide. It is to be covered with square tiles of one size, without cutting any tile. Using the least number of tiles at ₹85 per tile, find the total cost.

Solution, step by step

1The square tiles must fit without cutting, so the side must divide both the length and the width. For the biggest tile the side $= H CF (1260, 540)$ .
2 $1260 = 2^{2} \times 3^{2} \times 5 \times 7$ , $540 = 2^{2} \times 3^{3} \times 5$ ⟹ $H CF = 180$ cm = 1 m 80 cm.
37 tiles along the length ( $1260 \div 180$ ) and 3 along the width ( $540 \div 180$ ), so $7 \times 3 = 21$ tiles in all (the biggest tiles need the fewest tiles).
4Cost $= 21 \times 85 = 1785$ rupees.

Answer ₹1785

Cost = tiles × rate; the biggest tiles mean fewer tiles and so a lower cost (if the rate per tile is the same).

Example 44 · Advanced

A solid wooden block measures 120 cm × 180 cm × 300 cm. It is to be cut into equal cubes with no waste. Cutting the biggest possible cubes, how many cubes are got?

Solution, step by step

1Identical cubes are to be cut with no waste, so the side of a cube must divide all three edges. For the biggest cube the side $= H CF (120, 180, 300)$ .
2 $120 = 2^{3} \times 3 \times 5$ , $180 = 2^{2} \times 3^{2} \times 5$ , $300 = 2^{2} \times 3 \times 5^{2}$ ⟹ $H CF = 60$ cm.
3Cubes: 2 along the length, 3 along the width, 5 along the height; in all $2 \times 3 \times 5 = 30$ .

Answer 30

For cubes the HCF of the three edges; the count is the product of the three quotients.

Example 45 · Advanced

A rectangular field measures 96 m × 72 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. With the greatest gap, how many posts are needed in all?

Solution, step by step

1The posts are at equal gaps with one at every corner, so the gap must divide every side. The biggest gap = the HCF of the sides.
2 $96 = 2^{5} \times 3$ , $72 = 2^{3} \times 3^{2}$ ⟹ $H CF = 24$ metres.
3Perimeter $= 96 + 72 + 96 + 72 = 336$ m. On a closed boundary the number of posts = the number of gaps $= 336 \div 24 = 14$ .

Answer 14

A rectangle has four sides, but for the HCF only the two different measures (96, 72) are needed.

BasicTry it yourself first

A hall floor is 7 m 20 cm long and 4 m 80 cm wide. It is to be covered with square tiles of one size, without cutting any tile. What is the side (in cm) of the biggest possible tile?

BasicTry it yourself first

A triangular plot has sides 60 m, 84 m, 108 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. What is the greatest possible gap (in metres) between two posts?

StandardTry it yourself first

A hall floor is 10 m 50 cm long and 6 m 30 cm wide. It is to be covered with square tiles of one size, without cutting any tile. What is the least number of tiles needed?

StandardTry it yourself first

A solid wooden block measures 150 cm × 210 cm × 90 cm. It is to be cut into equal cubes with no waste. What is the side (in cm) of the biggest cube?

StandardTry it yourself first

A rectangular field measures 150 m × 90 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. With the greatest gap, how many posts are needed in all?

AdvancedTry it yourself first

A hall floor is 19 m 80 cm long and 12 m 60 cm wide. It is to be covered with square tiles of one size, without cutting any tile. Using the least number of tiles at ₹120 per tile, find the total cost.

AdvancedTry it yourself first

A solid wooden block measures 180 cm × 240 cm × 420 cm. It is to be cut into equal cubes with no waste. Cutting the biggest possible cubes, how many cubes are got?

StandardTry it yourself first

A hall floor is 13 m 20 cm long and 9 m 90 cm wide. It is to be covered with square tiles of one size, without cutting any tile. What is the side (in cm) of the biggest possible tile?

AdvancedTry it yourself first

A rectangular field measures 135 m × 105 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. What is the greatest possible gap (in metres) between two posts?

Squares and cubes from bricks, steps and capacities

Size = LCM; count = volume ÷ brick

The smallest square or cube from identical bricks, the least common distance for steps, and the least capacity.

The idea

Now the story is reversed — we are building. Tiles of $18 \times 24$ are laid the same way to make a square. The length along one side comes in multiples of 18 and along the other in multiples of 24; the side of the square is both, so a common multiple, and for the smallest it is the LCM $= 72$ . Then the count: $\frac{72}{18} \times \frac{72}{24} = 4 \times 3 = 12$ . For a cube (bricks with three edges) it is the LCM of the three edges. In steps each person walks whole steps, so the distance is a multiple of each step — the LCM. In capacities each cup fills a whole number of times — the LCM.

∑ Formula

S S = L CM (a, b), n = \frac{S ^{2}}{ab} = L CM (a, b, c), n = \frac{S ^{3}}{ab c}

The smallest square and the smallest cube

$S$: the side of the square or cube
$a, b, c$: the edges of the brick (when laid the same way)
$n$: the total number of bricks

Story	What the LCM gives	Then
smallest square from tiles	side of the square	tiles $= S^{2} \div ab$
smallest cube from bricks	side of the cube	bricks $= S^{3} \div ab c$
people with different steps	least common distance	steps = distance ÷ step
cups of different sizes	least capacity of the tank	cups = capacity ÷ cup

Here: $S$ = the side of the square/cube built, $a, b, c$ = the edges of a brick, $n$ = the number of bricks.

ConditionRemember: the bricks are not turned — all lie the same way, which is why we speak of multiples of

a

on one side and of

b

on the other. If the bricks may be turned, the question changes.

Steps

Recognise: we are building (square/cube/distance/capacity) ⟹ LCM.
Write all the units in one measure and find the LCM.
Count: divide each edge of the finished size by the brick’s edge in that direction and multiply.
Check: volume of the finished size = bricks × volume of a brick.

Example 46 · Basic

Rectangular tiles of 18 cm × 24 cm, all laid the same way, are to make the smallest possible solid square. How many tiles will it need in all?

Solution, step by step

1Laid all the same way, the length along one side comes in multiples of 18 and along the other in multiples of 24 (without turning a tile). The side of the square must be a multiple of both, and we want the smallest square — the side = the LCM.
2 $18 = 2 \times 3^{2}$ , $24 = 2^{3} \times 3$ ⟹ $L CM = 72$ cm.
372÷18=4 tiles along one side and 72÷24=3 along the other; in all $4 \times 3 = 12$ tiles.

Answer 12

The tile area is

18 \times 24 = 432

and the square’s

7 2^{2} = 5184

;

5184 \div 432 = 12

— the same answer.

Example 47 · Standard

Solid bricks of 10 cm × 15 cm × 24 cm, all laid the same way, are to make the smallest possible cube. How many bricks will be needed in all?

Solution, step by step

1For a cube, the length in each direction must be a multiple of the brick’s edge in that direction: $10, 15, 24$ . The side is a common multiple of the three and the smallest one is the LCM.
2 $10 = 2 \times 5$ , $15 = 3 \times 5$ , $24 = 2^{3} \times 3$ ⟹ $L CM = 120$ cm.
3Bricks: 120÷10=12, 120÷15=8, 120÷24=5; in all $12 \times 8 \times 5 = 480$ .

Answer 480

The quotients differ in each direction (12, 8, 5); their product is 480.

Example 48 · Standard

Three persons take steps of 54, 63 and 72 cm respectively. They start from one point with the same foot. After the least distance (in cm) have all of them taken whole steps and arrived together?

Solution, step by step

1A person walks whole steps, so the distance covered is always a multiple of his step length. The distance must be a common multiple of all, and we want the least — the LCM.
2 $54 = 2 \times 3^{3}$ , $63 = 3^{2} \times 7$ , $72 = 2^{3} \times 3^{2}$ ⟹ $L CM = 1512$ cm = 15 m 12 cm.

Answer 1512 cm

Two slips are common: taking the HCF 9, or taking the product — both are wrong.

Example 49 · Advanced

A tank can be filled exactly, with no spilling or left-over, by any one of three cups of 18, 24 and 27 litres. For the tank of least capacity, how many times must the biggest cup be used to fill it?

Solution, step by step

1The tank must be filled by each cup with nothing left over, so its capacity is a multiple of every cup. The least capacity = the LCM.
2 $18 = 2 \times 3^{2}$ , $24 = 2^{3} \times 3$ , $27 = 3^{3}$ ⟹ $L CM = 216$ litres.
3Number of cups: 216÷18=12, 216÷24=9, 216÷27=8. The biggest cup (27 litres) is used 8 times.

Answer 8

The biggest cup is used the fewest times — 8, while the smallest is used 12 times.

BasicTry it yourself first

Rectangular tiles of 12 cm × 20 cm, all laid the same way, are to make the smallest possible solid square. What is the side (in cm) of the square?

BasicTry it yourself first

Three persons take steps of 40, 60 and 90 cm respectively. They start from one point with the same foot. After the least distance (in cm) have all of them taken whole steps and arrived together?

StandardTry it yourself first

Solid bricks of 12 cm × 18 cm × 30 cm, all laid the same way, are to make the smallest possible cube. What is the side (in cm) of the cube?

StandardTry it yourself first

A bottle can be filled exactly, with no spilling or left-over, by any one of three cups of 8, 12 and 20 millilitres. What is the least possible capacity of the bottle (in millilitres)?

StandardTry it yourself first

Rectangular tiles of 15 cm × 35 cm, all laid the same way, are to make the smallest possible solid square. How many tiles will it need in all?

AdvancedTry it yourself first

Four persons take steps of 45, 50, 54 and 60 cm respectively. They start from one point with the same foot. Over that least common distance, how many steps does the person with the shortest step take?

AdvancedTry it yourself first

A tank can be filled exactly, with no spilling or left-over, by any one of four cups of 14, 21, 28 and 35 litres. For the tank of least capacity, how many times must the biggest cup be used to fill it?

StandardTry it yourself first

Solid bricks of 14 cm × 21 cm × 35 cm, all laid the same way, are to make the smallest possible cube. What is the side (in cm) of the cube?

AdvancedTry it yourself first

Three persons take steps of 42, 56 and 70 cm respectively. They start from one point with the same foot. After the least distance (in cm) have all of them taken whole steps and arrived together?

Summary, mixed practice, revision

Every method in one place

A one-page table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

Story · method · formula

What to recognise in every story and which formula or trick to use.

Story	Recognise	Key formula / trick
same remainder, greatest number	“the same remainder each time”	$N = H CF (∣ a - b ∣, ∣ b - c ∣, \dots)$
given remainders, greatest number	“remainders … respectively”	$N = H CF (a - r_{1}, b - r_{2}, \dots)$ , $N > r_{m a x}$
smallest number	same remainder $r$ / shortfall $g$	$L CM + r$ ; $L CM - g$
add / subtract	“least to add/subtract”	remainder $= X mod L$ ; subtract: remainder, add: $L -$ remainder
extra condition / different shortfalls	“and also divisible by $m$ ”	$L k \pm$ , try $k = 1, 2, \dots$ ; list the bigger divisor
$n$ -digit numbers	“largest/smallest 4-digit”	the top/bottom multiple of $L$ , then $+ r$ or $- g$
perfect square / cube	“smallest perfect square divisible by …”	every power in the LCM a multiple of 2 (or 3); $S_{0} j^{2}$
bells, lights	“together again”	LCM; count $⌊ T / L ⌋$ ( $+ 1$ for the start)
track, gears	“at the start again”	fraction-LCM of the lap times; LCM of the teeth
date, weekday	“after how many days, which weekday”	LCM days; weekday: $L CM mod 7$
rope, vessel, group, packet	“longest / biggest / fewest groups”	HCF; count $= \sum l_{i} / H CF$
tiles, cubes, posts	“without cutting, equal”	$s = H CF$ ; $n = \frac{l b}{s ^{2}}$ , $\frac{ab c}{s ^{3}}$ , $\frac{P}{s}$
bricks to square/cube, steps, capacity	“smallest square/cube/distance/capacity”	LCM; $n = \frac{S ^{2}}{ab}$ , $\frac{S ^{3}}{ab c}$

Letters in the table: $N$ = the number asked; $a, b, c$ = the given numbers; $r, r_{1}, r_{2}$ = remainders; $g$ = the shortfall each time (divisor − remainder); $L$ = the LCM; $X$ = the number to add to or subtract from; $m$ = the extra divisor; $k$ = the counter; $S_{0}$ = the smallest perfect square and $j$ = its multiplier; $T$ = the total time; $l_{i}$ = the $i$ -th length/quantity/count; $s$ = the side of a tile or cube or the gap of the posts; $l, b$ = the length and breadth of the floor; $P$ = the perimeter; $S$ = the side of the square/cube built.

RememberThree questions in every story: (1) am I splitting/cutting (HCF) or meeting/building (LCM)? (2) are all measures in one unit? (3) in a remainder question, have I cancelled the remainder (a difference, a subtraction or an addition)?

Mixed practice (exam style)

All stories mixed

Mixed word problems like SSC CGL/CHSL/CPO — first spot the story, then solve.

These questions do not name the method — first spot whether the story is about splitting or meeting. Every question has a full hidden solution; look at the hint if you are stuck.

BasicTry it yourself first

Ribbons of lengths 350 cm, 525 cm and 875 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. How many pieces will there be in all?

BasicTry it yourself first

A dairy has three tanks holding 84, 126 and 210 litres of milk. They are to be measured with one vessel, leaving nothing over. What is the capacity (in litres) of the biggest such vessel?

BasicTry it yourself first

Find the smallest number which leaves remainder 6 when divided by each of 14, 16, 24 and 28.

BasicTry it yourself first

Find the largest 4-digit number which is exactly divisible by 14, 18 and 21.

BasicTry it yourself first

How many 3-digit numbers are exactly divisible by 9, 14 and 21?

BasicTry it yourself first

Three lighthouses flash at intervals of 50, 75 and 100 seconds respectively. They did so together at 11:30:00 am. When will they next flash together?

BasicTry it yourself first

A triangular plot has sides 45 m, 60 m, 75 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. With the greatest gap, how many posts are needed in all?

BasicTry it yourself first

Three persons take steps of 30, 40 and 50 cm respectively. They start from one point with the same foot. After the least distance (in cm) have all of them taken whole steps and arrived together?

StandardTry it yourself first

Find the greatest number which, on dividing 1004, 1679 and 2309, leaves the same remainder every time.

StandardTry it yourself first

Find the greatest number which, on dividing 901, 1637 and 2323, leaves the remainders 5, 13, 27 respectively.

StandardTry it yourself first

Find the smallest number which leaves the remainders 3, 6, 10, 13 when divided by 5, 8, 12 and 15 respectively.

StandardTry it yourself first

What is the least number to add to 16000 so that the sum is exactly divisible by 24, 30 and 40?

StandardTry it yourself first

How many numbers from 300 to 3000 (both included) leave remainder 4 when divided by each of 10, 12 and 15?

StandardTry it yourself first

Find the smallest number which leaves remainder 5 when divided by each of 6, 8 and 10 but is exactly divisible by 7.

StandardTry it yourself first

Find the smallest 3-digit number which leaves remainder 5 when divided by each of 6, 9 and 12.

StandardTry it yourself first

By what smallest number should 2160 be multiplied so that the result is a perfect square?

StandardTry it yourself first

The bells of three temples ring at intervals of 32, 40 and 48 seconds respectively. They start together. How many times do they ring together in 1 hour (not counting the start)?

StandardTry it yourself first

Three runners start together from one point of a circular track 24 km long. Their speeds are 8, 12, 18 km/h. After how many hours will they first be together again at the starting point?

StandardTry it yourself first

Two meshed gear wheels have 15 and 24 teeth. The bigger wheel makes 6 turns per hour. At the start two particular teeth, one of each, are in contact. How many times in 10 hours will the same two teeth meet again after the start?

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150 drummers, 225 flautists, 375 singers take part in a parade. They are to be split into squads; a squad cannot mix two kinds and all squads must be equally strong. What is the least number of squads?

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A hall floor is 14 m 70 cm long and 10 m 50 cm wide. It is to be covered with square tiles of one size, without cutting any tile. What is the least number of tiles needed?

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Rectangular tiles of 24 cm × 40 cm, all laid the same way, are to make the smallest possible solid square. How many tiles will it need in all?

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Find the smallest number which leaves remainder 4 when divided by 11 and remainder 9 when divided by 12.

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Find the smallest perfect square that is exactly divisible by 8, 10 and 12.

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Find the largest 4-digit number which leaves the remainders 8, 11, 14 when divided by 12, 15 and 18 respectively.

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One doctor visits a hospital every 12 days and another every 18 days. Both were there on 14 June (Monday). On what date and weekday will they next be there together?

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One agency cleans a park every 7 days, its first cleaning being on day 3 of the month. Another agency cleans every 10 days, its first cleaning being on day 8. In the first 150 days, on how many days do both clean on the same day?

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A solid wooden block measures 70 cm × 105 cm × 245 cm. It is to be cut into equal cubes with no waste. Cutting the biggest possible cubes, how many cubes are got?

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Solid bricks of 6 cm × 9 cm × 15 cm, all laid the same way, are to make the smallest possible cube. How many bricks will be needed in all?

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A tank can be filled exactly, with no spilling or left-over, by any one of four cups of 6, 8, 10 and 12 litres. What is the least possible capacity of the tank (in litres)?

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Three runners start together from one point of a circular track 600 m long. Their speeds are 48, 80, 100 m/min. By the time they are first together again, how many laps will the fastest runner have run?

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Find the smallest number which becomes exactly divisible by 14, 21 and 35 when 6 is added to it.

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The greatest number which, on dividing 662, 1230 and 1869, leaves the same remainder each time is

N

. Find the sum of the digits of

N

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How many numbers from 100 to 1500 (both included) leave remainder 5 when divided by each of 8, 12 and 18?

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One volunteer helps at a library every 15 days and another every 20 days. Both were there on 2 September (Thursday). On which day of the week will they next be there together?

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Iron rods of lengths 3 m 15 cm, 5 m 25 cm and 8 m 40 cm are to be cut into equal pieces with nothing left over, each piece as long as possible. How many pieces will there be in all?

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N

is the greatest number which, on dividing 959, 2157 and 3274, leaves the remainders 11, 24, 35 respectively. Find the sum of all the quotients.

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The bells of four temples ring at intervals of 30, 42, 54 and 63 seconds respectively. They did so together at 7:15:20 am. When will they next ring together?

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Two meshed gear wheels have 27 and 45 teeth. The bigger wheel makes 4 turns per hour. How many turns does the smaller wheel make in 15 hours?

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A hall floor is 15 m 60 cm long and 11 m 70 cm wide. It is to be covered with square tiles of one size, without cutting any tile. Using the least number of tiles at ₹95 per tile, find the total cost.

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A rectangular field measures 150 m × 120 m. Posts are to be planted along its boundary at equal gaps, with a post at every corner. With the greatest gap, how many posts are needed in all?

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Rectangular tiles of 28 cm × 35 cm, all laid the same way, are to make the smallest possible solid square. How many tiles will it need in all?

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A tank can be filled exactly, with no spilling or left-over, by any one of four cups of 16, 20, 28 and 35 litres. For the tank of least capacity, how many times must the biggest cup be used to fill it?

Revision

Cards · common traps

Cards to memorise formulas and tricks, and the most common exam mistakes.

Mixing the same-remainder and given-remainder questions — the first takes differences, the second subtracts each number’s own remainder.
Taking just the LCM in a smallest-number question — do not forget to add the remainder; and in the shortfall form do not forget to subtract.
Starting from $k = 0$ — $L CM \times 0 + r = r$ is just the remainder, not the answer.
Crossing the limit in $n$ -digit questions — count the digits after adding $r$ or subtracting $g$ .
Counting or skipping the start — in bells read “counting the start?”
Not matching units — seconds and hours, cm and metres.
Taking “meeting anywhere” on a circular track as an LCM problem — the LCM is only for the starting point.
Fewest groups ⟹ biggest size — the HCF is the size, not the number of groups.

BackThe basics behind every method here — the HCF, the LCM, the product rule, fractions and decimals — are in the previous chapter, “HCF and LCM”; go back there if you get stuck.