Factors, Trailing Zeros and Recurring Decimals

SSC CGL · CHSL · CPO — method by method

4 Part41 worked examples130 practice questions

A factor of a number is a number that divides it exactly: the factors of 12 are 1, 2, 3, 4, 6, 12. A prime number is one with exactly two factors — 1 and itself: 2, 3, 5, 7, 11, 13 … One solid fact: every number can be written as a product of primes in only one way (apart from the order). It is the number’s fingerprint, and the key to this whole chapter.

Once a number is written as $N = 2^{3} \times 3^{2} \times 5$ , three big families of exam questions can be read straight from that form:

Part	Typical question	Solved by
1. Factors	how many factors? their sum? how many pairs? perfect square?	multiply the (power + 1) of every prime
2. Trailing zeros	how many zeros at the end of $n!$ or of a product?	counting pairs of 2 and 5
3. Recurring decimals	fraction ↔ decimal, sums, comparing, the $n$ -th digit	the 2s, 5s and other primes of the denominator; denominators of 9s and 0s

How to studyAfter every method come solved examples and then questions to do yourself — open a solution only after you have tried. The divisibility tests (2, 3, 5, 7, 11) are in the chapter “Remainders and Divisibility”, and the basics of fractions in the chapter “Simple and Decimal Fractions”.

Part 1 — Factors

Count · sum · pairs · squares

From prime factorisation to the count, sum, product, odd/even and pairs of factors.

There are five methods, each resting on the one before. First learn to break a number into primes; everything else is its fruit.

Prime factorisation and the factor tree

N = p^{a} q^{b} r^{c} \dots

Breaking a number into primes, and counting the prime factors.

The idea

Take a number like 924. It is even, so 2 divides it: $924 = 2 \times 462$ . 462 is even too: $462 = 2 \times 231$ . The digit sum of 231 is 6, so 3 divides it: $231 = 3 \times 77$ . And $77 = 7 \times 11$ . Now every piece is prime, so stop. $924 = 2 \times 2 \times 3 \times 7 \times 11 = 2^{2} \times 3 \times 7 \times 11$ . This is the prime factorisation. Always divide by the smallest prime that works; that way no prime is ever missed.

◭ Diagram

The factor tree of 924: at every step the smallest prime (a coloured circle) drops out and the rest goes down; multiplying all the circles gives back 924

∑ Formula

N = p_{1}^{a_{1}} \times p_{2}^{a_{2}} \times p_{3}^{a_{3}} \times \dots

The standard form (every

p

prime, every power

\geq 1

)

$N$: the given number
$p_{1}, p_{2}, \dots$: the different primes, smallest first
$a_{1}, a_{2}, \dots$: the power of each prime (how many times it appears)

Which prime to try	It divides when	Why
2	last digit is even	multiples of 10 are divisible by 2
3	digit sum divisible by 3	$10 \equiv 1$ (mod 3)
5	last digit 0 or 5	multiples of 10 are divisible by 5
7	double the last digit, subtract from the rest; if 7 divides the result, it divides N	$10 a + b \to a - 2 b$
11	difference of alternate digit sums is 0 or a multiple of 11	$10 \equiv - 1$ (mod 11)

Here: $a$ = the number left after cutting off the last digit, $b$ = the last digit (in the row for 7); (mod $k$ ) means “the remainder on division by $k$ ”.

RememberWhere to stop? To break a number

N

, try only primes whose square does not exceed

N

. If no prime up to

N

divides it,

N

itself is prime. (Reason: of any two factors that multiply to

N

, one is always at most

N

Counting the prime factors

“How many prime factors?” has two meanings. In $924 = 2^{2} \times 3 \times 7 \times 11$ there are 4 different primes (2, 3, 7, 11), but if every prime is counted as many times as it appears, there are $2 + 1 + 1 + 1 = 5$ prime factors. The exam usually means the second — the sum of the powers. When a base in the product is not prime (like $1 5^{8}$ ), break it first: $1 5^{8} = 3^{8} \times 5^{8}$ .

∑ Formula

Ω (N) = a_{1} + a_{2} + a_{3} + \dots

prime factors counted with repeats: add all the powers; the bases must be prime

$Ω (N)$: number of prime factors of $N$ (with repeats)
$a_{1}, a_{2}, \dots$: the powers of the primes in standard form

Steps

Divide by the smallest prime (2, then 3, 5, 7 …) as long as it divides; write down each quotient.
Stop when the quotient is 1 (or a prime is left).
Write a prime that repeats as a power.
Multiply back to check that you get the original number.

Example 1 · Basic

Write 756 as a product of primes.

Solution, step by step

1756 is even, so divide by 2: $756 \div 2 = 378$ ; again $378 \div 2 = 189$ . 189 is odd, so 2 will not work any more.
2The digit sum of 189 is $1 + 8 + 9 = 18$ , so 3 divides it: $189 \div 3 = 63$ , $63 \div 3 = 21$ , $21 \div 3 = 7$ . 7 is prime, so stop.
3The divisors were 2, 2, 3, 3, 3 and finally 7, so $756 = 2^{2} \times 3^{3} \times 7$ .
4Check: $4 \times 27 \times 7 = 756$ ✓

Answer $2^{2} \times 3^{3} \times 7$

At each stage check whether the quotient still divides by a small prime; that way no factor is ever missed.

Example 2 · Standard

How many prime factors does the product

4^{6} \times 9^{3} \times 3 5^{2} \times 11

have in all (each prime counted as many times as it appears), and how many different primes are there?

Solution, step by step

1The bases 4, 9 and 35 are not prime, so break them first: $4^{6} = (2^{2})^{6} = 2^{12}$ ; $9^{3} = (3^{2})^{3} = 3^{6}$ ; $3 5^{2} = (5 \times 7)^{2} = 5^{2} \times 7^{2}$ .
2The whole product is $2^{12} \times 3^{6} \times 5^{2} \times 7^{2} \times 11$ .
3Total prime factors = sum of the powers $= 12 + 6 + 2 + 2 + 1 = 23$ .
4Different primes: 2, 3, 5, 7, 11 — that is 5.

Answer 23 (in all), 5 (different)

Power rules:

(a^{m})^{n} = a^{mn}

and

(ab)^{n} = a^{n} b^{n}

Example 3 · Advanced

Is 1517 prime? If not, find its prime factors.

Solution, step by step

1 $3 8^{2} = 1444 < 1517 < 3 9^{2} = 1521$ , so $1517$ is about 38.9. We need to try only primes up to 37: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37.
2The last digit is 7, so not 2 or 5; the digit sum $1 + 5 + 1 + 7 = 14$ , so not 3. $1517 \div 7$ : $7 \times 216 = 1512$ , remainder 5 — no. 11: $(7 + 5) - (1 + 1) = 10$ — no.
313: $13 \times 116 = 1508$ , remainder 9 — no. 17: $17 \times 89 = 1513$ , remainder 4 — no. 19: $19 \times 79 = 1501$ , remainder 16 — no. 23: $23 \times 65 = 1495$ , remainder 22 — no. 29: $29 \times 52 = 1508$ , remainder 9 — no. 31: $31 \times 48 = 1488$ , remainder 29 — no.
437: $37 \times 41 = 1517$ — it divides! And 41 is prime.

Answer No, $1517 = 37 \times 41$

This is also the test for primality: if no prime up to

N

divides

N

, then

N

is prime.

BasicTry it yourself first

Write 1092 as a product of primes.

BasicTry it yourself first

How many prime factors (counting repeats) does

2^{5} \times 3^{4} \times 7

have?

StandardTry it yourself first

What is the sum of all the different prime factors of 3003?

StandardTry it yourself first

How many prime factors (counting repeats) does

6^{5} \times 1 0^{3} \times 1 5^{2}

have?

StandardTry it yourself first

2^{x} \times 3^{2} \times 5 = 1440

, what is the value of

x

AdvancedTry it yourself first

What is the largest prime factor of 8051?

AdvancedTry it yourself first

The product of two prime numbers is 2491. What is their sum?

The number of factors

(a + 1) (b + 1) (c + 1) \dots

Adding 1 to every power and multiplying gives the count of all factors.

The idea

Look at $72 = 2^{3} \times 3^{2}$ . Any factor of 72 has the form $2^{i} \times 3^{j}$ , where the power $i$ of 2 can be only one of $0, 1, 2, 3$ (4 choices) and the power $j$ of 3 only one of $0, 1, 2$ (3 choices). Every pair $(i, j)$ gives a different factor, so there are $4 \times 3 = 12$ factors in all. The figure below is exactly this table: row = power of 2, column = power of 3, every cell one factor.

◭ Diagram

The 12 factors of 72: powers of 2 down the side (blue), powers of 3 across the top (green); total

4 \times 3 = 12

∑ Formula

d (N) = (a_{1} + 1) (a_{2} + 1) (a_{3} + 1) \dots

The total number of factors of

N

(1 and

N

included)

$d (N)$: number of factors of $N$
$a_{1}, a_{2}, \dots$: the powers of the primes in $N = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots$

Why “+1”? Because a power may also be zero — the prime may be absent from that factor. For $2^{3}$ the choices are $2^{0}, 2^{1}, 2^{2}, 2^{3}$ : one more than the power. This also brings in 1 (all powers 0) and $N$ (all powers full).

Form of $N$	Number of factors	Example
$p$ (prime)	2	13: 1, 13
$p^{2}$	3	49: 1, 7, 49
$p^{k}$	$k + 1$	$2^{6} = 64$ : 7
$pq$	4	15: 1, 3, 5, 15
$p^{2} q$	6	12: 1, 2, 3, 4, 6, 12

Here $p, q$ are different primes and $k$ is any power. Two useful facts follow: a number with exactly 3 factors is always the square of a prime; a number with exactly 4 factors is either $p^{3}$ or $pq$ .

TrapTrap: “how many factors” includes both 1 and

N

. “Excluding 1 and

N

” means

d - 2

; “less than

N

” means

d - 1

. Read the wording carefully.

The reverse question: finding the number

Sometimes the number of factors is given and the number or a power is asked. Then split $d$ into a product; each part $=$ (power $+ 1$ ). For the smallest number, give the biggest power to the smallest prime (2), the next to 3, then 5. Try every way of splitting.

Split $d = 12$	Powers	Number
12	11	$2^{11} = 2048$
$6 \times 2$	5, 1	$2^{5} \times 3 = 96$
$4 \times 3$	3, 2	$2^{3} \times 3^{2} = 72$
$3 \times 2 \times 2$	2, 1, 1	$2^{2} \times 3 \times 5 = 60$ ← smallest

Steps

Break the number into primes: $p^{a} q^{b} r^{c}$ .
Add 1 to every power: $(a + 1), (b + 1), (c + 1)$ .
Multiply them all.
Look at the condition (count 1 and $N$ or not) and subtract accordingly.

Example 4 · Basic

How many factors does 504 have?

Solution, step by step

1Break 504: $504 \div 2 = 252$ , $\div 2 = 126$ , $\div 2 = 63$ ; $63 = 3 \times 21 = 3 \times 3 \times 7$ . So $504 = 2^{3} \times 3^{2} \times 7$ .
2The powers are 3, 2, 1; adding 1 to each gives $4, 3, 2$ .
3Total factors $= 4 \times 3 \times 2 = 24$ .

Answer 24

While counting, 1 and 504 itself are already inside this count.

Example 5 · Standard

N = 2^{a} \times 3^{2} \times 7

has exactly 36 factors. Find

a

and

N

Solution, step by step

1Number of factors $= (a + 1) (2 + 1) (1 + 1) = 6 (a + 1)$ .
2 $6 (a + 1) = 36 \Rightarrow a + 1 = 6 \Rightarrow a = 5$ .
3 $N = 2^{5} \times 3^{2} \times 7 = 32 \times 9 \times 7 = 2016$ . Check: $6 \times 3 \times 2 = 36$ ✓

Answer $a = 5, N = 2016$

Running the same formula backwards gives the unknown power.

Example 6 · Advanced

Find the smallest number that has exactly 18 factors.

Solution, step by step

1Split 18 into parts: $18$ ; $9 \times 2$ ; $6 \times 3$ ; $3 \times 3 \times 2$ . Each part is a power $+ 1$ .
2The powers are ${17}$ , ${8, 1}$ , ${5, 2}$ , ${2, 2, 1}$ . Giving the biggest power to the smallest prime: $2^{17} = 131072$ ; $2^{8} \times 3 = 768$ ; $2^{5} \times 3^{2} = 288$ ; $2^{2} \times 3^{2} \times 5 = 180$ .
3The smallest is 180; check: $(2 + 1) (2 + 1) (1 + 1) = 18$ ✓

Answer 180

Using more primes makes the powers smaller and the number often smaller — but check every split.

BasicTry it yourself first

How many factors does

2^{4} \times 3^{2} \times 5

have?

BasicTry it yourself first

The receipt number of a shop is 1001. How many factors does 1001 have?

BasicTry it yourself first

How many factors does

1 0^{4}

have?

StandardTry it yourself first

How many factors does

6^{3} \times 1 0^{2}

have in all?

StandardTry it yourself first

How many factors of

3^{5} \times 7^{2}

remain if 1 and the number itself are left out?

AdvancedTry it yourself first

What is the smallest number with exactly 10 factors?

AdvancedTry it yourself first

3^{a} \times 5^{b}

has exactly 15 factors, and

a > b \geq 1

. Find the number.

Sum and product of factors

\prod \frac{p ^{a + 1} - 1}{p - 1}

N^{d /2}

The sum of all factors, their product and the sum of their reciprocals.

The idea: the sum

In the table of the last method each cell held a factor. Now we want to add them all. Take $100 = 2^{2} \times 5^{2}$ : the cells are $2^{i} \times 5^{j}$ . The sum of the whole table is exactly what you get by multiplying $(1 + 2 + 4) \times (1 + 5 + 25)$ — because every $2^{i}$ pairs with every $5^{j}$ exactly once, which is what expanding a product does. So $σ (100) = 7 \times 31 = 217$ .

◭ Diagram

The 9 factors of 100 and their sum: the sum of the powers of 2 is

7

, of the powers of 5 is

31

, and the product is

217

∑ Formula

σ (N) = \prod \frac{p ^{a + 1} - 1}{p - 1}

The sum of all factors: one fraction for every prime, all multiplied (each fraction equals

1 + p + p^{2} + \dots + p^{a}

)

$σ (N)$: sum of all factors of $N$
$p, a$: a prime of $N$ and its power
$\prod$: multiply over all the primes

TrickInstead of memorising the formula, write the brackets directly: for

2^{4}

it is

(1 + 2 + 4 + 8 + 16) = 31

; for

3^{2}

it is

(1 + 3 + 9) = 13

. For small powers this is faster than the formula, and there is less room for error.

The product, and the sum of reciprocals

Tie the factors in pairs: $f$ and $\frac{N}{f}$ always multiply to $N$ . With $d$ factors there are $\frac{d}{2}$ pairs, each giving $N$ , so the product of all factors is $N^{d /2}$ . (When $N$ is a perfect square the middle factor $N$ stands alone, yet the same formula still holds.)

∑ Formula

\prod f = N^{d /2}, \sum \frac{1}{f} = \frac{σ ( N )}{N}

The product, and the sum of the reciprocals

$\prod f$: product of all the factors
$d$: number of factors
$σ (N)$: sum of the factors

The sum of reciprocals is $\frac{σ ( N )}{N}$ because $\frac{1}{f} = \frac{N / f}{N}$ and $N / f$ is itself a factor; so $\sum \frac{1}{f} = \frac{\sum ( N / f )}{N} = \frac{σ ( N )}{N}$ . The sum of the proper factors (leaving out $N$ itself) is $σ (N) - N$ . If this equals $N$ , the number is a perfect number (6, 28, 496).

Steps

Write the prime factorisation.
For every prime evaluate the bracket $(1 + p + \dots + p^{a})$ .
Multiply the brackets for $σ (N)$ ; for the product find $d$ and use $N^{d /2}$ .

Example 7 · Basic

Find the sum of all the factors of 1400.

Solution, step by step

1 $1400 = 14 \times 100 = 2 \times 7 \times 2^{2} \times 5^{2} = 2^{3} \times 5^{2} \times 7$ .
2Brackets: for $2^{3}$ , $1 + 2 + 4 + 8 = 15$ ; for $5^{2}$ , $1 + 5 + 25 = 31$ ; for $7$ , $1 + 7 = 8$ .
3Sum $= 15 \times 31 \times 8 = 3, 720$ .

Answer 3,720

The sum is much larger than the count — there are only

d = 4 \times 3 \times 2 = 24

factors, but their sum is 3,720.

Example 8 · Standard

Find the product of all the factors of 36 and of all the factors of 30.

Solution, step by step

1 $36 = 2^{2} \times 3^{2}$ , so $d = 3 \times 3 = 9$ . Product $= 3 6^{9/2} = (6^{2})^{9/2} = 6^{9}$ .
2 $6^{9} = 1, 00, 77, 696$ . (Check: in the factors 1, 2, 3, 4, 6, 9, 12, 18, 36, four pairs give 36 each and the middle 6 is left: $3 6^{4} \times 6 = 6^{8} \times 6 = 6^{9}$ ✓)
3 $30 = 2 \times 3 \times 5$ , so $d = 2 \times 2 \times 2 = 8$ and the product $= 3 0^{4} = 8, 10, 000$ .

Answer 1,00,77,696 and 8,10,000

In a perfect square the middle factor

N

stands alone, yet

N^{d /2}

still fits.

Example 9 · Advanced

(a) By comparing 496 with the sum of its proper factors, say whether 496 is a perfect number. (b) Find the sum of the reciprocals of all the factors of 60.

Solution, step by step

1(a) $496 = 16 \times 31 = 2^{4} \times 31$ . $σ = (1 + 2 + 4 + 8 + 16) (1 + 31) = 31 \times 32 = 992$ .
2Sum of proper factors $= 992 - 496 = 496$ , equal to the number — so 496 is a perfect number.
3(b) $60 = 2^{2} \times 3 \times 5$ ; $σ (60) = 7 \times 4 \times 6 = 168$ . Sum of reciprocals $= \frac{σ}{N} = \frac{168}{60} = \frac{14}{5}$ .

Answer (a) yes; (b) $\frac{14}{5}$

The proper factors 1, 2, 3 of 6 add to 6; those of 28 (1, 2, 4, 7, 14) add to 28 — these are perfect numbers too.

BasicTry it yourself first

What is the sum of all the factors of 96?

BasicTry it yourself first

What is the sum of all the factors of

7^{3} \times 11

StandardTry it yourself first

What is the sum of the proper factors of 220 (leaving out 220 itself)?

StandardTry it yourself first

What is the product of all the factors of 24?

StandardTry it yourself first

What is the sum of the reciprocals of all the factors of 12?

AdvancedTry it yourself first

The sum of all the factors of

5^{a}

is 781. What is

a

AdvancedTry it yourself first

The product of all the factors of a number

N

N^{10}

. How many factors does

N

have?

Even, odd and special factors

Power of 2: from 1 to

a

, or only 0

Counting and summing even and odd factors, factors that are multiples of a given number, square and cube factors.

The idea

Every factor of $N = 2^{a} \times 3^{b} \times 7^{c}$ looks like $2^{i} 3^{j} 7^{k}$ . The factor is odd when 2 does not appear at all, i.e. $i = 0$ . It is even when at least one 2 appears, i.e. $i \in {1, 2, \dots, a}$ . The counting rule is the same as before; only the choices for 2 change:

Which factors	Count	Sum
all	$(a + 1) (b + 1) (c + 1)$	$(1 + 2 + \dots + 2^{a}) (1 + 3 + \dots) (1 + 7 + \dots)$
odd ( $i = 0$ )	$(b + 1) (c + 1)$	drop the bracket of 2
even ( $i \geq 1$ )	$a (b + 1) (c + 1)$	$(2 + 4 + \dots + 2^{a}) (1 + 3 + \dots) (1 + 7 + \dots)$
multiples of $m$ ( $m ∣ N$ )	$d (\frac{N}{m})$	$m \times σ (\frac{N}{m})$
perfect squares	$(⌊ \frac{a}{2} ⌋ + 1) (⌊ \frac{b}{2} ⌋ + 1) \dots$	keep only even powers
perfect cubes	$(⌊ \frac{a}{3} ⌋ + 1) (⌊ \frac{b}{3} ⌋ + 1) \dots$	keep only powers that are multiples of 3

Here: $a, b, c$ = the powers of the primes 2, 3, 7; $d (\cdot)$ = number of factors; $σ (\cdot)$ = their sum; $⌊ x ⌋$ = the whole-number part of $x$ (drop the decimals).

The row for “multiples of $m$ ” is the most useful. A factor of $N$ that is a multiple of $m$ can be written $m \times (something)$ , and that something must be a factor of $\frac{N}{m}$ . So $N$ has exactly as many factors that are multiples of $m$ as $\frac{N}{m}$ has factors. Taking $m = 2$ gives the even factors: $d (N /2)$ of them, with sum $2 σ (N /2)$ — both agree with the rows above. Taking $m = 10$ gives the “factors ending in zero”.

CheckEven + odd = all, for the count and for the sum. You can get one from the other by subtracting, and it doubles as your check.

Steps

Write the prime factorisation.
Decide which powers of which prime the question allows (even: power of 2 at least 1; odd: power of 2 equal to 0; multiple of $m$ : take $m$ ’s powers off first; square: even powers).
For every prime write the number of choices (for a count) or the bracket-sum of those choices (for a sum).
Multiply them all.

Example 10 · Basic

How many odd factors and how many even factors does

N = 2352

have?

Solution, step by step

1 $2352 = 16 \times 147 = 2^{4} \times 147$ and $147 = 3 \times 49$ , so $2352 = 2^{4} \times 3 \times 7^{2}$ .
2Odd factors contain no 2: the choices for the other primes are $(1 + 1) (2 + 1) = 2 \times 3 = 6$ .
3Even factors have a power of 2 from $1, 2, 3, 4$ — 4 choices: $4 \times 2 \times 3 = 24$ .
4Check: total $= 5 \times 2 \times 3 = 30 = 6 + 24$ ✓

Answer 6 and 24

The choices of 2 decide the count: only 1 choice (power 0) for odd,

a

choices for even.

Example 11 · Standard

Find the sum of the odd factors and the sum of the even factors of 1960.

Solution, step by step

1 $1960 = 8 \times 245 = 2^{3} \times 5 \times 7^{2}$ .
2Odd: leave out the bracket of 2: $(1 + 5) (1 + 7 + 49) = 6 \times 57 = 342$ .
3Even: the bracket of 2 without the 1: $(2 + 4 + 8) = 14$ , so $14 \times 6 \times 57 = 4788$ .
4Check: total $σ = 15 \times 6 \times 57 = 5130$ and $342 + 4788 = 5130$ ✓

Answer 342 and 4,788

Even factors are not just “doubled” odd ones: the bracket of 2 starts at 2 instead of 1.

Example 12 · Advanced

How many factors of 5400 are (a) multiples of 15, (b) perfect squares, (c) perfect cubes?

Solution, step by step

1 $5400 = 2^{3} \times 3^{3} \times 5^{2}$ (since $5400 = 54 \times 100$ and $54 = 2 \times 27$ ).
2(a) Factors that are multiples of 15 are as many as the factors of $\frac{5400}{15} = 360 = 2^{3} \times 3^{2} \times 5$ : $4 \times 3 \times 2 = 24$ .
3(b) In a perfect square every power must be even: power of 2 from ${0, 2}$ (2 choices), of 3 from ${0, 2}$ (2 choices), of 5 from ${0, 2}$ (2 choices): $2 \times 2 \times 2 = 8$ .
4(c) In a perfect cube every power is a multiple of 3: power of 2 from ${0, 3}$ (2), of 3 from ${0, 3}$ (2), of 5 only ${0}$ (1): $2 \times 2 \times 1 = 4$ .

Answer 24, 8, 4

All three parts use one idea: which powers of which prime are allowed — count them and multiply.

BasicTry it yourself first

How many factors of

2^{5} \times 3^{2} \times 5^{2}

are odd?

BasicTry it yourself first

How many factors of

2^{6} \times 3 \times 5^{2}

are even?

StandardTry it yourself first

What is the sum of all the odd factors of

2^{4} \times 3^{2} \times 7

StandardTry it yourself first

How many factors of 4500 end in 0?

StandardTry it yourself first

How many factors of

2^{6} \times 3^{4} \times 5^{3} \times 7

are perfect squares?

AdvancedTry it yourself first

How many factors of

2^{3} \times 3^{2} \times 5 \times 7

are multiples of 6 but not divisible by 5?

AdvancedTry it yourself first

How many factors of

2^{3} \times 5^{2} \times 7

have 5 as the last digit?

Perfect squares, pairs and ways of writing a product

d

odd

⟺

perfect square

Pairs of factors, spotting a perfect square, ways of writing a number as a product of two factors, and making a perfect square or cube.

The idea

Every factor $f$ of $N$ has a partner $\frac{N}{f}$ , and the two multiply to $N$ . For 30: $1 \times 30$ , $2 \times 15$ , $3 \times 10$ , $5 \times 6$ — 8 factors, 4 pairs. But for 36: $1 \times 36$ , $2 \times 18$ , $3 \times 12$ , $4 \times 9$ and $6 \times 6$ . Here 6 is its own partner, so it is counted once and the total is $4 \times 2 + 1 = 9$ — an odd number.

◭ Diagram

30: in all four pairs the partners are different, so the number of factors is even (8)

◭ Diagram

36: in the pair

6 \times 6

the factor is its own partner (rose), so the number of factors is odd (9)

∑ Formula

N = k^{2} ⟺ d (N) odd

N

is a perfect square ⟺ its number of factors is odd ⟺ every power in

N

is even

$d (N)$: number of factors of $N$

Reason: $d = (a + 1) (b + 1) \dots$ is odd only when every bracket is odd, i.e. every power is even — and when all powers are even the number is a square anyway. (The locker puzzle: 100 lockers are shut; student $k$ flips every $k$ -th locker, for $k = 1$ to $100$ . Locker $N$ is flipped once for each of its factors; it ends up open only if the number of factors is odd — that is lockers 1, 4, 9, …, 100.)

Questions about pairs

Question	Answer	When
ways of writing $N = f \times g$ ( $f \leq g$ )	$\frac{d}{2}$	$N$ not a perfect square
	$\frac{d + 1}{2}$	$N$ a perfect square ( $f = g$ counted)
ways with two different factors	$\frac{d - 1}{2}$	$N$ a perfect square
counting the order ( $f$ first, $g$ second)	$d$	always
co-prime pairs ( $g cd (f, g) = 1$ , $f \leq g$ )	$2^{k - 1}$	$k$ different primes, $N > 1$

Here: $d$ = number of factors of $N$ ; $k$ = number of different primes in $N$ . In a co-prime pair the whole power of each prime goes either to $f$ or to $g$ (it cannot be split), so there are $2^{k}$ ordered ways and $2^{k - 1}$ unordered.

Making a perfect square or cube

For every prime whose power is not even, multiply by it once more (or divide by it once); for a cube make the power a multiple of 3. The smallest multiplier or divisor is the product of exactly the primes whose power is “incomplete”.

Power $e$	Multiply (square)	Divide (square)	Multiply (cube)	Divide (cube)
1	p	p	$p^{2}$	p
2	—	—	p	$p^{2}$
3	p	p	—	—
4	—	—	$p^{2}$	p
5	p	p	p	$p^{2}$

Here: $e$ = the power of that prime $p$ in the number; “—” means do nothing.

Steps

Find the prime factorisation and $d$ .
Perfect square test: all powers even (or $d$ odd).
For pair questions choose the right row of the table (square or not, order counted or not).
To make a square or cube, look at each prime’s power and build the multiplier or divisor.

Example 13 · Basic

Is 7056 a perfect square? In how many ways can it be written as a product of two factors (order not counted, equal factors allowed)?

Solution, step by step

1 $7056 = 16 \times 441 = 2^{4} \times 2 1^{2} = 2^{4} \times 3^{2} \times 7^{2}$ .
2All the powers (4, 2, 2) are even, so it is a perfect square: $7056 = 8 4^{2}$ .
3The number of factors is $d = (4 + 1) (2 + 1) (2 + 1) = 45$ (odd, as it should be).
4It is a perfect square, so the ways are $\frac{d + 1}{2} = \frac{46}{2} = 23$ (this includes $84 \times 84$ ).

Answer a perfect square; 23 ways

Two ways to test: all powers even, or

d

odd. Both say the same thing.

Example 14 · Standard

By what smallest number must 1350 be multiplied to get (a) a perfect square, (b) a perfect cube?

Solution, step by step

1 $1350 = 135 \times 10 = 2 \times 3^{3} \times 5^{2}$ .
2(a) For a square all powers must be even. The power 1 of 2 and the power 3 of 3 are odd; the power 2 of 5 is fine. Multiplier $= 2 \times 3 = 6$ . Result $1350 \times 6 = 8100 = 9 0^{2}$ ✓
3(b) For a cube every power must be a multiple of 3. The power 1 of 2 needs $2^{2}$ more; the power 3 of 3 is already fine; the power 2 of 5 needs one more 5. Multiplier $= 2^{2} \times 5 = 20$ . Result $1350 \times 20 = 27000 = 3 0^{3}$ ✓

Answer 6 and 20

The smallest multiplier contains only the primes whose power is incomplete — the others are left alone.

Example 15 · Advanced

In how many ways can

N = 2^{3} \times 3 \times 5^{2} \times 7

be written as a product of two co-prime factors (order not counted)? And when order is counted?

Solution, step by step

1There are four different primes: 2, 3, 5, 7, so $k = 4$ .
2For the factors to be co-prime the whole power of every prime must go into one factor: $2^{3}$ into the first or the second, $3$ into the first or the second, … — 2 choices for each.
3With order counted there are $2^{4} = 16$ ways. But $(f, g)$ and $(g, f)$ are the same pair, so without order $\frac{16}{2} = 2^{3} = 8$ .

Answer 8 and 16

In a co-prime pair a prime’s power cannot be split — so each prime gives only 2 choices, not

(a + 1)

BasicTry it yourself first

In how many ways can 180 be written as a product of two factors (order not counted)?

BasicTry it yourself first

In how many ways can 2025 be written as a product of two different factors (order not counted)?

StandardTry it yourself first

By what smallest number must 2450 be multiplied to make a perfect square?

StandardTry it yourself first

By what smallest number must 2160 be divided to get a perfect square?

StandardTry it yourself first

By what smallest number must 1125 be multiplied to make a perfect cube?

AdvancedTry it yourself first

In how many ways can 2310 be written as a product of two co-prime factors (order not counted)?

AdvancedTry it yourself first

Which is the largest number below 100 that has exactly 3 factors?

AdvancedTry it yourself first

Which is the smallest perfect square divisible by each of 18, 20 and 30?

Part 2 — Trailing zeros

Pairs of 2 × 5 ·

n!

· products

How many zeros end a product or $n!$ , and the reverse question.

“With how many zeros does this product end?” is a sure exam question, and you never need to multiply it out. Four methods: the pairing idea, the formula for $n!$ , the reverse question and different kinds of series.

Where zeros come from: a pair of 2 and 5

zeros

= min (power of 2, power of 5)

Breaking a product into primes and counting the pairs of 2 and 5.

The idea

One zero at the end means the number is divisible by 10; two zeros mean 100, i.e. $1 0^{2}$ ; $k$ zeros mean $1 0^{k}$ . But $10 = 2 \times 5$ , so $1 0^{k} = 2^{k} \times 5^{k}$ . So the number must contain $k$ twos and $k$ fives, and a pair needs both. An example: $2^{7} \times 5^{4} \times 3 = 128 \times 625 \times 3 = 240000$ — the four 5s found four 2s, so 4 zeros; the remaining three 2s and the 3 had no partner.

◭ Diagram

2^{7} \times 5^{4}

: four pairs (rose line) = four tens = 4 zeros; three 2s left without a partner

∑ Formula

Z = min (x, y)

where

2^{x}

and

5^{y}

are the total powers of 2 and 5 in the prime factorisation of the product

$Z$: number of trailing zeros
$x$: total power of 2 in the product
$y$: total power of 5 in the product

Break every number of the product separately and add up the total powers. The reason adding works: $2^{a} \times 2^{b} = 2^{a + b}$ . There is no need to multiply out.

RememberTwo firm rules: (1) a product of odd numbers only never ends in 0, because it has not a single 2. (2) As soon as there is a multiple of 5 and an even number, zeros appear — but how many is told by the powers.

Steps

Break every number (or every power) of the product into primes; multiply the exponents by the outer power: $(2^{3} \cdot 5)^{4} = 2^{12} \cdot 5^{4}$ .
Add the powers of 2 in all of them → $x$ ; add the powers of 5 → $y$ .
Number of zeros $= min (x, y)$ .

Example 16 · Basic

How many zeros are at the end of the product

24 \times 35 \times 50 \times 64 \times 125

Solution, step by step

1Break each number: $24 = 2^{3} \times 3$ ; $35 = 5 \times 7$ ; $50 = 2 \times 5^{2}$ ; $64 = 2^{6}$ ; $125 = 5^{3}$ .
2Total power of 2: $x = 3 + 1 + 6 = 10$ .
3Total power of 5: $y = 1 + 2 + 3 = 6$ .
4Zeros $= min (10, 6) = 6$ . (Of the ten 2s only six find a 5 to pair with.)

Answer 6

Always add both powers, then pick the smaller — here the power of 5 was smaller.

Example 17 · Standard

How many zeros are at the end of

1 2^{7} \times 1 5^{6} \times 2 0^{3}

Solution, step by step

1 $1 2^{7} = (2^{2} \times 3)^{7} = 2^{14} \times 3^{7}$ ; $1 5^{6} = (3 \times 5)^{6} = 3^{6} \times 5^{6}$ ; $2 0^{3} = (2^{2} \times 5)^{3} = 2^{6} \times 5^{3}$ .
2Power of 2: $x = 14 + 6 = 20$ and power of 5: $y = 6 + 3 = 9$ .
3Zeros $= min (20, 9) = 9$ .

Answer 9

Breaking the bases into primes and multiplying by the outer power — that is the whole job.

Example 18 · Standard

How many zeros are at the end of

2 5^{4} \times 6^{3} \times 7^{5}

Solution, step by step

1 $2 5^{4} = (5^{2})^{4} = 5^{8}$ ; $6^{3} = 2^{3} \times 3^{3}$ ; $7^{5}$ has neither a 2 nor a 5.
2Power of 2: $x = 3$ ; power of 5: $y = 8$ .
3This time the 2s run short: zeros $= min (3, 8) = 3$ . (Five 5s are left without a partner.)

Answer 3

Assuming “the 5s will always be fewer” is a trap. Here the 2s were fewer. Count both.

BasicTry it yourself first

How many zeros are at the end of

4^{5} \times 2 5^{3}

BasicTry it yourself first

How many zeros are at the end of

125 \times 32 \times 7

StandardTry it yourself first

How many zeros are at the end of

3 5^{6} \times 1 8^{4} \times 4

StandardTry it yourself first

How many zeros are at the end of

15 \times 25 \times 35 \times 45 \times 55 \times 8

StandardTry it yourself first

How many zeros are at the end of

6^{8} \times 5^{12}

StandardTry it yourself first

How many zeros are at the end of

1 2^{15} \times 2 5^{9}

AdvancedTry it yourself first

Find the smallest natural number

n

for which

14 \times 15 \times 16 \times n

ends in exactly 3 zeros.

AdvancedTry it yourself first

How many zeros are at the end of

2^{3} \times 3^{4} \times 5^{5} \times 7^{6} \times 11

Zeros of $n!$ and Legendre’s formula

⌊ \frac{n}{5} ⌋ + ⌊ \frac{n}{25} ⌋ + \dots

The highest power of a prime in $n! = 1 \times 2 \times \dots \times n$ , and from it the trailing zeros.

The idea

To count zeros in $n!$ the earlier method still applies: count 2s and 5s. But here there is a convenience — the 2s always outnumber the 5s (every second number is even, but only every fifth is divisible by 5). So zeros $=$ the total power of 5. Now count the 5s: every number from $1$ to $n$ divisible by 5 gives at least one 5 — there are $⌊ \frac{n}{5} ⌋$ such numbers. But 25, 50, 75, … contain two 5s ( $25 = 5 \times 5$ ); the first is already counted above, the second is still due — there are $⌊ \frac{n}{25} ⌋$ such numbers. Multiples of 125 hold a third 5, and so on.

◭ Diagram

The power of 5 in

130!

: 26 numbers are divisible by 5, 5 of them hold a second 5 and 1 (the number 125) holds a third — 32 zeros in all

∑ Formula

v_{p} (n!) = k \geq 1 \sum ⌊ \frac{n}{p ^{k}} ⌋

The highest power of the prime

p

n!

(Legendre’s formula); keep adding until the quotient becomes 0

$v_{p} (n!)$: the largest $k$ such that $p^{k}$ divides $n!$
$p$: a prime (for zeros $p = 5$ )
$⌊ x ⌋$: the whole-number part of $x$ (divide and drop the remainder)

CheckA check: write

n

in base

p

and add its digits (

s

); then

v_{p} (n!) = \frac{n - s}{p - 1}

. For example

130

in base 5 is

101 0_{5}

(

125 + 5

), digit sum 2, so

\frac{130 - 2}{4} = 32

✓.

The highest power of a composite number

If asked for the largest $k$ such that $1 2^{k}$ divides $n!$ , break $12 = 2^{2} \times 3$ . For $1 2^{k} = 2^{2 k} \times 3^{k}$ the number $n!$ needs a power of 2 of at least $2 k$ and a power of 3 of at least $k$ . So $k = min (⌊ \frac{v _{2}}{2} ⌋, v_{3})$ — divide each prime’s power by its exponent in the composite and take the smallest. This is also the question “how many zeros end $n!$ when written in base 12?”.

Steps

Pick the prime $p$ (for zeros, 5).
Divide $n$ by $p$ (whole-number part), then that quotient by $p$ again, and so on until the quotient is less than $p$ .
Add all the quotients.
For a composite do this for every prime, and take the minimum of power ÷ exponent.

Example 19 · Basic

How many zeros are at the end of

90!

Solution, step by step

1Zeros $=$ power of 5 (the 2s are in surplus).
2 $⌊ 90 \div 5 ⌋ = 18$ (5, 10, …, 90 — each has at least one 5).
3 $⌊ 90 \div 25 ⌋ = 3$ (25, 50, 75 — one extra 5 each). $⌊ 90 \div 125 ⌋ = 0$ , so stop here.
4Total $= 18 + 3 = 21$ zeros.

Answer 21

Add only until the quotient reaches 0; all later terms are 0.

Example 20 · Standard

What is the highest power of 7 in

300!

, i.e. the largest

k

for which

7^{k}

divides

300!

Solution, step by step

1Legendre’s formula for $p = 7$ : first $⌊ 300 \div 7 ⌋ = 42$ (the multiples of 7).
2Now multiples of $7^{2} = 49$ : $⌊ 300 \div 49 ⌋ = 6$ .
3 $7^{3} = 343 > 300$ , so the rest are 0.
4 $k = 42 + 6 = 48$ .

Answer 48

The formula works for every prime, not only 5.

Example 21 · Advanced

Find the largest

k

for which

1 2^{k}

divides

50!

exactly.

Solution, step by step

1 $12 = 2^{2} \times 3$ , so $1 2^{k} = 2^{2 k} \times 3^{k}$ .
2Power of 2 in $50!$ : $25 + 12 + 6 + 3 + 1 = 47$ .
3Power of 3 in $50!$ : $16 + 5 + 1 = 22$ .
4From the 2s: $2 k \leq 47 \Rightarrow k \leq 23$ ; from the 3s: $k \leq 22$ .
5Both conditions must hold: $k = min (23, 22) = 22$ .

Answer 22

Here the 3s set the limit, though the 2s alone could go up to 23 — always take the weakest link.

BasicTry it yourself first

How many zeros are at the end of

60!

BasicTry it yourself first

How many zeros are at the end of

205!

StandardTry it yourself first

What is the highest power of 3 in

100!

StandardTry it yourself first

What is the highest power of 2 in

64!

StandardTry it yourself first

Find the largest

k

for which

1 5^{k}

divides

80!

AdvancedTry it yourself first

Find the largest

k

for which

2 0^{k}

divides

100!

AdvancedTry it yourself first

How many zeros end

150!

when it is written in base 12?

The reverse question: zeros given, find $n$

n \approx 4 \times

zeros, then adjust

The smallest or largest $n$ with given zeros, how many $n$ are possible, and which zero-counts are impossible.

The idea

Now $n$ is unknown and the count of zeros is known. Some things to understand. The zeros of $n!$ grow with $n$ like a staircase: until $n$ reaches the next multiple of 5 the zeros stay the same (before $25!$ , each of $20!$ to $24!$ has 4 zeros). Then at $n = 5, 10, 15, \dots$ one zero is added, but at $n = 25, 50, 75, 100$ two are added and at $n = 125$ three. So some zero-counts never occur at all.

◭ Diagram

The staircase of the zeros of

n!

: a jump of two at

25!

(the count 5 never occurs) and again at

50!

(the count 11 never occurs)

Fact	Why
For a zero-count $k$ the possible $n$ are either exactly 5 consecutive numbers ( $5 m, 5 m + 1, \dots, 5 m + 4$ ) or none	zeros rise only at multiples of 5
The smallest such $n$ is always a multiple of 5	the step begins there
$n$ lies roughly between $4 k$ and $5 k$	the power of 5 in $n!$ is a little under $\frac{n}{4}$
counts hidden by a jump: $25! \to 6$ , $50! \to 12$ , $75! \to 18$ , $100! \to 24$	two zeros were added at each

Steps

Estimate from the zeros $k$ : $n \approx 4 k$ , and take a multiple of 5 near it.
Find the zeros of that $n$ with Legendre.
If the zeros are too few move $n$ to the next multiple of 5; if too many, to the previous one. Repeat until exactly $k$ .
The $n$ where $k$ first appears is the smallest, and 4 more than it is the largest.

Example 22 · Basic

Find the smallest

n

for which

n!

has at least 31 trailing zeros.

Solution, step by step

1Estimate: $n \approx 4 \times 31 = 124$ . Check: $124!$ has $24 + 4 + 0 = 28$ zeros — fewer than 31.
2Zeros rise only at multiples of 5, so try the next multiple of 5, which is 125: $125!$ has $25 + 5 + 1 = 31$ zeros.
3The previous multiple of 5 is 120: $120!$ has $24 + 4 = 28$ zeros (fewer than 31). So the smallest $n$ is 125.

Answer 125

When

n

is not a multiple of 5 the zeros equal those of the previous multiple of 5 — so test multiples of 5 directly.

Example 23 · Standard

Is there any

n

for which

n!

ends with exactly 5 zeros? Explain. Also name the next three zero-counts after 5 that never occur.

Solution, step by step

1 $24!$ has $4$ zeros ( $⌊ 24 \div 5 ⌋ = 4$ , $24 < 25$ ).
2 $25!$ has $5 + 1 = 6$ zeros, because 25 holds two 5s and so two zeros were added at once. From 4 straight to 6 — the 5 was skipped.
3So no $n$ has exactly 5 zeros.
4The same happens at $50!$ ( $49! \to 10$ , $50! \to 12$ : 11 skipped), at $75!$ ( $74! \to 16$ , $75! \to 18$ : 17 skipped) and at $100!$ ( $99! \to 22$ , $100! \to 24$ : 23 skipped). The next three: 11, 17, 23.

Answer No; 11, 17, 23

The staircase figure shows exactly this: double jumps at 25, 50, 75, 100.

Example 24 · Advanced

n!

ends with exactly 12 zeros. Find the smallest and the largest value of

n

and the number of possible values of

n

Solution, step by step

1Estimate $4 \times 12 = 48$ . $49!$ has $9 + 1 = 10$ zeros (too few); $50!$ has $10 + 2 = 12$ — exactly 12 for the first time here.
2The smallest $n$ is 50.
3 $51!, 52!, 53!, 54!$ also have $⌊ n \div 5 ⌋ = 10$ and $⌊ n \div 25 ⌋ = 2$ , i.e. 12. At $55!$ it becomes $11 + 2 = 13$ .
4The largest $n$ is 54 and the possible values are $50, 51, 52, 53, 54$ — 5 in all.

Answer 50, 54 and 5

For any possible zero-count there are exactly five consecutive values of

n

BasicTry it yourself first

Find the smallest

n

for which

n!

has at least 10 trailing zeros.

BasicTry it yourself first

n!

ends with exactly 7 zeros, what is the smallest value of

n

StandardTry it yourself first

n!

ends with exactly 14 zeros. What is the largest possible value of

n

StandardTry it yourself first

Which of these can never be the number of trailing zeros of any

n!

AdvancedTry it yourself first

For how many natural numbers

n

does

n!

end with exactly 29 zeros?

AdvancedTry it yourself first

Find the smallest

n

for which

n!

is exactly divisible by

1 0^{50}

AdvancedTry it yourself first

Find the smallest

n

for which

7^{10}

divides

n!

Products of series

Term by term: powers of 5 and 2

Zeros in the product of a range of numbers, of multiples of a number, of odd numbers only, of terms with powers, and of factorials.

The idea

In the exam the product comes in many shapes, but the job is always the same: find the total power $y$ of 5 and $x$ of 2 in the whole product, then $min (x, y)$ . All the effort goes into counting $x$ and $y$ quickly. Each shape has a small trick:

Shape of the product	How to count the power of 5 (or 2)
$a \times (a + 1) \times \dots \times b$ (consecutive)	subtract: $v (b!) - v ((a - 1)!)$
$k \times 2 k \times 3 k \times \dots \times mk$	$= k^{m} \times m!$ ; add the powers of $k^{m}$ and of $m!$ separately
odd numbers only	power of 2 is 0, so 0 zeros; if a power of 2 is multiplied in, count the 5s and take $min$
terms with powers: $1^{e_{1}} \times 2^{e_{2}} \times \dots$	term $j^{e}$ gives a power of 5 equal to $e \times v_{5} (j)$ ; look only at the multiples of 5 and make a table
$(n!)^{r}$ or $1! \times 2! \times \dots \times n!$	$(n!)^{r}$ : $r \times v (n!)$ ; for the product of factorials add each $v (j!)$

Here: $v (\cdot)$ = the power of that prime (5 or 2) in the term; $v_{5} (j)$ = the power of 5 in $j$ (e.g. $v_{5} (50) = 2$ ); $a, b$ = first and last number of a consecutive product; $k, m$ = the multiple and the number of terms.

TrapThe weakest link: in

n!

it is always the 5s that run short, but in a product like

k \times 2 k \times \dots

with many 5s inside

k

(such as

k = 15

or 25), the 5s are plentiful and the 2s may run short. Count both here.

Steps

Identify the shape (consecutive, multiples, odd, with powers, factorial).
With its trick get the total power $y$ of 5.
Get the total power $x$ of 2 (or argue that it is far larger than $y$ ).
The answer is $min (x, y)$ .

Example 25 · Basic

How many zeros are at the end of the product

61 \times 62 \times 63 \times \dots \times 140

Solution, step by step

1This is the part of $140!$ left after $60!$ : $61 \times \dots \times 140 = \frac{140 !}{60 !}$ .
2Power of 5 in $140!$ : $28 + 5 + 1 = 34$ ; in $60!$ : $12 + 2 = 14$ .
3On division the powers subtract: $y = 34 - 14 = 20$ .
4Power of 2: $x = 137 - 56 = 81$ , far above 20 (of 80 numbers, 40 are even), so the 2s will not run short.
5Zeros $= min (81, 20) = 20$ .

Answer 20

A product “from

a

b

” is always a quotient of two factorials.

Example 26 · Standard

How many zeros are at the end of the product

15 \times 30 \times 45 \times \dots \times 600

(the first 40 multiples of 15)?

Solution, step by step

1 $600 = 15 \times 40$ , so the product $= 15 \times (15 \cdot 2) \times (15 \cdot 3) \times \dots \times (15 \cdot 40) = 1 5^{40} \times 40!$ .
2Power of 5: $1 5^{40} = 3^{40} \times 5^{40}$ gives 40, and $40!$ gives $8 + 1 = 9$ ; total $y = 49$ .
3Power of 2: $1 5^{40}$ has no 2; $40!$ gives $20 + 10 + 5 + 2 + 1 = 38$ ; total $x = 38$ .
4This time the 2s are fewer: zeros $= min (38, 49) = 38$ .

Answer 38

First pull out the multiple to get

k^{m} \times m!

; then it is easy to add the powers of 2 and 5 of the two parts separately.

Example 27 · Standard

How many zeros are at the end of

1^{1} \times 2^{2} \times 3^{3} \times \dots \times 3 0^{30}

Solution, step by step

1The term $j^{j}$ gives a power of 5 equal to $j \times v_{5} (j)$ , so look only at the multiples of 5: 5, 10, 15, 20, 25, 30.
2 $5^{5} \to 5 \times 1 = 5$ ; $1 0^{10} \to 10$ ; $1 5^{15} \to 15$ ; $2 0^{20} \to 20$ ; $2 5^{25} \to 25 \times 2 = 50$ (since $25 = 5^{2}$ ); $3 0^{30} \to 30$ .
3Total $y = 5 + 10 + 15 + 20 + 50 + 30 = 130$ .
4Power of 2: from the even terms the total is $x = 416$ , far above $y$ , so zeros $= 130$ .

Answer 130

With powers on the terms, make a table of exponent ×

v_{5} (j)

; do not forget two 5s at a term like 25.

Example 28 · Standard

How many zeros are at the end of

1 \times 3 \times 5 \times 7 \times \dots \times 49 \times 16

(all odd numbers up to 49, times 16)?

Solution, step by step

1The odd numbers carry no 2, so all the 2s come from 16: $16 = 2^{4}$ , $x = 4$ .
2The odd numbers with a 5: 5, 15, 25, 35, 45 — with powers of 5 equal to $1, 1, 2, 1, 1$ ; total $y = 6$ .
3Zeros $= min (4, 6) = 4$ . (Without the 16 the answer would be 0.)

Answer 4

With odd numbers only the 2s can come from outside, so the answer is usually just the power of 2.

BasicTry it yourself first

How many zeros are at the end of

21 \times 22 \times 23 \times \dots \times 60

BasicTry it yourself first

How many zeros are at the end of

25 \times 50 \times 75 \times \dots \times 400

(the first 16 multiples of 25)?

StandardTry it yourself first

How many zeros are at the end of

1 \times 3 \times 5 \times \dots \times 39 \times 8

(all odd numbers up to 39, times 8)?

StandardTry it yourself first

How many zeros are at the end of

(30!)^{3}

StandardTry it yourself first

How many zeros are at the end of

2 0^{1} \times 1 9^{2} \times 1 8^{3} \times \dots \times 2^{19} \times 1^{20}

AdvancedTry it yourself first

How many zeros are at the end of

1! \times 2! \times 3! \times \dots \times 15!

AdvancedTry it yourself first

How many zeros are at the end of

6 \times 12 \times 18 \times \dots \times 240

(the first 40 multiples of 6)?

AdvancedTry it yourself first

How many zeros are at the end of

51 \times 52 \times 53 \times \dots \times 100

Part 3 — Recurring decimals

Fraction ↔ decimal · sums · comparing

When a fraction gives a terminating or a recurring decimal, recurring decimal to fraction, operations on them and comparing them.

A decimal can be of three kinds: terminating (it ends, like 0.375), pure recurring (it repeats from the very first digit, like $0. \overline{27}$ ) and mixed recurring (it repeats after a few digits, like $0.1 \overline{6}$ ). The bar on top marks the digits that repeat forever. Four methods: telling the kind, decimal to fraction, operations, and comparing/the $n$ -th digit.

Fraction to decimal: when it ends, when it repeats

only 2s and 5s in the denominator ⟹ terminating

The remainder cycle of long division, the test for terminating, counting decimal places and the length of the repeating block.

The idea

Do $\frac{1}{7}$ by long division. At every step we multiply the remainder by 10 and divide by 7; the remainders come as 1, 3, 2, 6, 4, 5 and then 1 returns. The moment a remainder repeats, all the work after it repeats too — and so do the digits: $\frac{1}{7} = 0. \overline{142857}$ . A remainder is always less than 7 (one of $1$ to $6$ , six possibilities, or $0$ ), so within a few steps either a repeat or an end is certain. If the remainder 0 appears, the division stops and the decimal is terminating.

◭ Diagram

Long division of

\frac{1}{7}

: the remainders (blue circles) go round a cycle and each gives one digit (rose); after six steps the 1 returns, so the repeating block has 6 digits

When does it end? Bring $\frac{p}{q}$ to lowest terms. The decimal stops when some $1 0^{k}$ (that is $2^{k} 5^{k}$ ) is a multiple of the denominator $q$ , so the fraction can be rewritten with denominator $1 0^{k}$ . That happens only when $q$ has no prime other than 2 and 5.

∑ Formula

q = 2^{a} 5^{b} \Rightarrow max (a, b) places

For the denominator of the fraction

\frac{p}{q}

in lowest terms

$a, b$: the powers of 2 and 5 in $q$
$max (a, b)$: the larger of the two = number of places after the point

Reason: $\frac{7}{40} = \frac{7}{2 ^{3} \times 5}$ . To make the 5 appear three times multiply numerator and denominator by $5^{2}$ : $\frac{7 \times 25}{2 ^{3} \times 5 ^{3}} = \frac{175}{1000} = 0.175$ — three places, which is the bigger power 3.

The kind of recurring decimal and the length of the block

Now take $q = 2^{a} \times 5^{b} \times m$ where $m > 1$ has no 2 or 5. Then the decimal is recurring, and: before the repeat starts there are $max (a, b)$ digits (if $a = b = 0$ , i.e. the denominator is only $m$ , it is pure recurring); the length of the block is decided by $m$ alone — it is the smallest $k$ for which $m$ divides $99 \dots 9$ ( $k$ nines).

$m$	3	9	11	27	37	101	7	13
Block length	1	1	2	3	3	4	6	6
because it divides	9	9	99	999	999	9999	999999	999999

Here $m$ = the part of the denominator that has no 2 or 5. (For example $\frac{1}{3} = 0. \overline{3}$ , $\frac{1}{11} = 0. \overline{09}$ , $\frac{1}{37} = 0. \overline{027}$ , $\frac{1}{13} = 0. \overline{076923}$ .) And a neat fact: rotating the digits of $\frac{1}{7}$ gives the other sevenths — $\frac{2}{7} = 0. \overline{285714}$ , $\frac{3}{7} = 0. \overline{428571}$ , because the remainder cycle is the same, only the starting point differs.

TrapTrap: reduce the fraction first! The denominator of

\frac{33}{110}

110 = 2 \times 5 \times 11

, yet

\frac{33}{110} = \frac{3}{10} = 0.3

is terminating — the 11 cancelled.

Steps

Bring the fraction to lowest terms.
Break the denominator into primes: $2^{a} \times 5^{b} \times m$ .
If $m = 1$ it terminates, with $max (a, b)$ places. Otherwise recurring: $max (a, b)$ digits before the repeat, and block length from $m$ .
If the digits are needed do the long division and stop when a remainder repeats.

Example 29 · Basic

Which of

\frac{21}{80}, \frac{17}{90}, \frac{33}{110}, \frac{7}{48}, \frac{13}{125}

give terminating decimals?

Solution, step by step

1First reduce and break the denominators into primes. $\frac{21}{80}$ : $g cd (21, 80) = 1$ , denominator $80 = 2^{4} \times 5$ — only 2 and 5 ⟹ terminating.
2 $\frac{17}{90}$ : 17 is prime and does not cancel with 90; $90 = 2 \times 3^{2} \times 5$ contains 3 ⟹ recurring.
3 $\frac{33}{110}$ : $g cd = 11$ , lowest form $\frac{3}{10}$ — denominator $2 \times 5$ ⟹ terminating.
4 $\frac{7}{48}$ : $48 = 2^{4} \times 3$ , a 3 is left ⟹ recurring. $\frac{13}{125}$ : $125 = 5^{3}$ ⟹ terminating.

Answer $\frac{21}{80}, \frac{33}{110}, \frac{13}{125}$

Look at the denominator only after reducing.

Example 30 · Standard

Convert

\frac{5}{13}

to a decimal by long division and state the length of the repeating block.

Solution, step by step

113 has no 2 or 5, so the decimal will be pure recurring. Start with remainder 5: $50 \div 13 = 3$ remainder 11 ⟹ digit 3.
2 $110 \div 13 = 8$ remainder 6 ⟹ digit 8; $60 \div 13 = 4$ remainder 8 ⟹ digit 4; $80 \div 13 = 6$ remainder 2 ⟹ digit 6.
3 $20 \div 13 = 1$ remainder 7 ⟹ digit 1; $70 \div 13 = 5$ remainder 5 ⟹ digit 5.
4The remainder 5 has returned (remainders: 5, 11, 6, 8, 2, 7, then 5), so the repeat begins. $\frac{5}{13} = 0. \overline{384615}$ ; block length 6.

Answer $0. \overline{384615}$ , block length 6

The moment a remainder repeats the digits repeat — the cycle is complete when we return to the first remainder.

Example 31 · Advanced

(a) Without dividing, say how many digits come before the repeat in

\frac{13}{220}

and how long the block is; then write the decimal. (b) For how many integers

n

from

1

60

\frac{n}{84}

terminating?

Solution, step by step

1(a) 13 is prime, so the fraction is in lowest terms. $220 = 2^{2} \times 5 \times 11$ : $a = 2$ , $b = 1$ , $m = 11$ .
2Before the repeat there are $max (a, b) = 2$ digits; the block length comes from 11: $11 ∣ 99$ , so 2 digits.
3Long division gives $\frac{13}{220} = 0.05 \overline{90}$ (the digits 0, 5 first, then 9, 0 repeat).
4(b) $84 = 2^{2} \times 3 \times 7$ . $\frac{n}{84}$ terminates only if $n$ holds both the 3 and the 7 so that only 2s remain in the denominator, i.e. $n$ is a multiple of 21.
5Multiples of 21 up to 60: 21, 42 — 2 in all.

Answer (a) 2 digits, block 2; (b) 2

In (a) the 11 of

220

decided the block, and the powers of 2 and 5 decided the “digits before”.

BasicTry it yourself first

Write

\frac{3}{16}

as a decimal.

BasicTry it yourself first

How many digits repeat in the decimal of

\frac{1}{13}

StandardTry it yourself first

Which of these fractions gives a terminating decimal?

StandardTry it yourself first

How many digits will the decimal of

\frac{13}{2 ^{3} \times 5 ^{6}}

have after the decimal point?

StandardTry it yourself first

Write

\frac{9}{125}

as a decimal.

AdvancedTry it yourself first

For how many integers

n

from

1

100

\frac{n}{70}

a terminating decimal?

AdvancedTry it yourself first

Write

\frac{17}{60}

as a decimal.

AdvancedTry it yourself first

Given

\frac{1}{7} = 0. \overline{142857}

, what is the decimal of

\frac{5}{7}

without dividing?

Recurring decimal to fraction

digits ÷ (nines for the block, then zeros)

Turning pure and mixed recurring decimals into fractions, and why the method works.

The idea: shift and subtract

Take $x = 0. \overline{47} = 0.474747 \dots$ The block has 2 digits, so multiply $x$ by 100: $100 x = 47.4747 \dots$ The digits after the point in the two lines are identical, so they cancel on subtraction: $100 x - x = 47$ , i.e. $99 x = 47$ and $x = \frac{47}{99}$ . An endless repeat leaves nothing behind — that is the heart of the method.

◭ Diagram

The tails of

x

and

100 x

are the same; subtracting gives

99 x = 47

∑ Formula

0. \overline{a} = \frac{a}{r 9 \dots 9}

Pure recurring: the number made by the block of digits ÷ as many 9s

$a$: the digits of the block (read as one number, like 47)
$r$: number of digits in the block

∑ Formula

0. p \overline{a} = \frac{p a - p}{r 9 \dots 9 k 0 \dots 0}

Mixed recurring: (all the digits) − (the digits before the repeat); in the denominator

r

nines followed by

k

zeros

$p$: the $k$ digits before the repeat (as one number)
$a$: the $r$ digits of the block (as one number)
$\overline{p a}$: the digits of $p$ followed by the digits of $a$ : one number

The reason for the mixed case is the same shifting. Let $y = 0.1 \overline{6} = 0.1666 \dots$ : $10 y = 1.666 \dots$ and $100 y = 16.666 \dots$ The tails of these two are the same, so subtract: $90 y = 16 - 1 = 15$ , $y = \frac{15}{90} = \frac{1}{6}$ . Here the “nine” in the denominator is for the one digit of the block and the “zero” for the one digit before the repeat. Keep the whole-number part aside and add it at the end: $3.2 \overline{7} = 3 + \frac{27 - 2}{90} = 3 + \frac{5}{18} = \frac{59}{18}$ .

◭ Diagram

0.3, 0.33, 0.333, 0.3333

get closer to

\frac{1}{3}

each time; infinitely many 3s are exactly

\frac{1}{3}

RememberSurprising but true:

0. \overline{9} = \frac{9}{9} = 1

. In the same way

0.4 \overline{9} = \frac{49 - 4}{90} = \frac{45}{90} = 0.5

. If the decimal ends in endless 9s, it is just the next “round” number.

Steps

Keep the whole-number part aside.
Count the digits before the repeat ( $k$ ) and the digits of the block ( $r$ ).
Numerator = (all digits up to the end of the first block, read as a number) − (the digits before the repeat); denominator = $r$ nines followed by $k$ zeros.
Simplify the fraction and add the whole part.

Example 32 · Basic

Convert

0. \overline{135}

into a fraction in lowest terms.

Solution, step by step

1The block has 3 digits (135) and nothing before it, so the denominator has three 9s: $999$ .
2 $0. \overline{135} = \frac{135}{999}$ .
3Both divide by 27 ( $135 = 27 \times 5$ , $999 = 27 \times 37$ ).
4Lowest terms: $\frac{5}{37}$ .

Answer $\frac{5}{37}$

First see whether there are digits before the repeat — none here, so only nines.

Example 33 · Standard

Convert

0.2 \overline{18}

into a fraction in lowest terms.

Solution, step by step

1One digit before the repeat (2), a block of 2 digits (18). So the denominator: two 9s and one 0 ⟹ $990$ .
2Numerator: all digits (218) − the digits before the repeat (2) $= 216$ .
3 $\frac{216}{990}$ ; both divide by 18 ( $216 = 18 \times 12$ , $990 = 18 \times 55$ ).
4Lowest terms: $\frac{12}{55}$ . Check: $12 \div 55 = 0.21818 \dots$ ✓

Answer $\frac{12}{55}$

The nines in the denominator are for the block, the zeros for the digits before the repeat.

Example 34 · Standard

Write

4.0 \overline{36}

as an improper fraction

\frac{p}{q}

Solution, step by step

1Keep the whole part 4 aside; what remains is $0.0 \overline{36}$ — one digit before the repeat (0) and a block of 2 digits (36).
2Numerator: all digits (036) − the digit before (0) $= 36$ ; denominator: two 9s, one 0 ⟹ $990$ .
3 $\frac{36}{990} = \frac{2}{55}$ (both divide by 18).
4 $4 + \frac{2}{55} = \frac{220 + 2}{55} = \frac{222}{55}$ .

Answer $\frac{222}{55}$

A “0” before the repeat still counts: the denominator gets a zero.

Example 35 · Advanced

Show that

0.3 \overline{9} = 0.4

and

2. \overline{9} = 3

, and explain what “the difference is exactly zero” means.

Solution, step by step

1 $0.3 \overline{9}$ : one digit before the repeat (3), a block of one digit (9). Numerator $= 39 - 3 = 36$ , denominator $= 90$ .
2 $\frac{36}{90} = \frac{2}{5} = 0.4$ .
3 $2. \overline{9} = 2 + \frac{9}{9} = 2 + 1 = 3$ .
4 $0.4 - 0.3999 \dots = 0.0000 \dots$ : there is no place anywhere where a difference could show (every place is 0), so they are one and the same number. Between two different numbers there is always another number; here there is none — which proves they are equal.

Answer $0.3 \overline{9} = \frac{2}{5} = 0.4$ and $2. \overline{9} = 3$

A decimal ending in endless 9s can always be rewritten as a terminating decimal.

BasicTry it yourself first

Convert

0. \overline{8}

into a fraction.

BasicTry it yourself first

Convert

0. \overline{27}

into a fraction in lowest terms.

StandardTry it yourself first

Convert

0. \overline{405}

into a fraction in lowest terms.

StandardTry it yourself first

Convert

0.1 \overline{6}

into a fraction in lowest terms.

StandardTry it yourself first

Write

2. \overline{18}

as an improper fraction.

AdvancedTry it yourself first

Convert

0.12 \overline{345}

into a fraction in lowest terms.

AdvancedTry it yourself first

0. \overline{ab} = \frac{8}{33}

(where

ab

is a two-digit number), what is

ab

AdvancedTry it yourself first

What is the value of

0.4 \overline{9}

Sums, differences, products, quotients and roots of recurring decimals

Make fractions, or add the blocks

Two ways: turn everything into fractions, and for sums stretch the blocks to equal length and add them; plus a few quick patterns.

The idea

Recurring decimals are really fractions, so the safest way is: turn every decimal into a fraction, do the operation, and turn back into a decimal if asked. This works for every operation (+, −, ×, ÷, square root). For sums and differences there is also a quick way, learned below.

The quick way: add the blocks

Let $P$ be the largest number of digits that come before the repeat in any of the decimals, and $L$ the LCM of all the block lengths. Cut every decimal in two: the first $P$ digits (the finite part) and the repeating block after it, stretched to $L$ digits (stretch the block 7 to two digits and it becomes 77). Adding all the finite parts is an ordinary decimal sum, and adding all the stretched blocks as numbers gives $S$ . The blocks together are worth $\frac{S}{99 \dots 9}$ ( $L$ nines). If $S$ reaches the number made of all those nines, the whole-number part of the quotient is added at the $P$ -th place and the remainder becomes the new block.

∑ Formula

S = q (1 0^{L} - 1) + r

Divide

S

99 \dots 9

(

L

nines): quotient

q

and remainder

r

$S$: the sum of the stretched blocks (as numbers)
$L$: the LCM of the block lengths
$q$: whole units: add them at the $P$ -th decimal place
$r$: the remainder: this is the new block (written with $L$ digits, zeros in front)

◭ Diagram

Adding three decimals (the first example below): the finite parts (blue) add as usual, the blocks (saffron) add as numbers to give

S = 195

;

195 = 99 \times 1 + 96

Quick patterns (useful every time): $0. \overline{a} + 0. \overline{b} = \frac{a + b}{9}$ ; $0. \overline{ab} + 0. \overline{ba} = \frac{11 ( a + b )}{99} = \frac{a + b}{9}$ (swapping the digits $a, b$ does not change the sum); $0. \overline{a} \times 9 = a$ ; $0. \overline{ab} \times 99 = ab$ . And when $a + b \geq 9$ the value $\frac{a + b}{9}$ reaches 1 or more, such as $\frac{10}{9} = 1. \overline{1}$ .

CheckTry both ways: if the answers differ there is a mistake somewhere. In a multiple-choice question whose options are decimals the block method is quicker; for multiplication and division make fractions.

Steps (block method for sums)

In each decimal note the digits before the repeat and the block; find $P$ (most digits before) and $L$ (LCM of block lengths).
Write every decimal as its first $P$ digits (finite part) followed by a block of $L$ digits.
Add the whole parts and the finite parts; add the $L$ -digit blocks as numbers to get $S$ .
See how many times $1 0^{L} - 1$ goes into $S$ ( $q$ ) and add it at the $P$ -th place; the remainder is the new block. (If the remainder is 0 the answer terminates.)

Example 36 · Standard

Find

1.2 \overline{36} + 0. \overline{7} + 2.04 \overline{5}

as a recurring decimal.

Solution, step by step

1Digits before the repeat: 1, 0, 2 ⟹ $P = 2$ . Block lengths 2, 1, 1 ⟹ $L = LCM = 2$ .
2Write each decimal as “ $P$ digits | block of $L$ digits”: $1. 23 ∣ 63$ ; $0. 77 ∣ 77$ ; $2. 04 ∣ 55$ (the block 7 stretched to 77, 5 to 55).
3Finite parts: $1.23 + 0.77 + 2.04 = 4.04$ . Blocks as numbers: $S = 63 + 77 + 55 = 195$ .
4 $99$ goes into $S = 195$ once ( $q = 1$ ), with remainder $195 - 99 = 96$ . Add $q = 1$ at the second decimal place: $4.04 + 0.01 = 4.05$ .
5Answer: $4.05 \overline{96}$ . Check (with fractions): the sum $= \frac{13397}{3300} = 4.05 \overline{96}$ ✓

Answer $4.05 \overline{96}$

The “overflow”

q

of the blocks is added to the finite part afterwards — this is where mistakes usually happen.

Example 37 · Basic

Find the value of

0. \overline{37} + 0. \overline{73}

Solution, step by step

1Both blocks have 2 digits and nothing comes before them ⟹ $P = 0$ , $L = 2$ .
2Add the blocks: $S = 37 + 73 = 110$ .
3 $99$ goes in once ( $q = 1$ ) with remainder $110 - 99 = 11$ . This $q = 1$ is added at the whole-number place ( $P = 0$ , i.e. before the point).
4Sum $= 1. \overline{11} = 1. \overline{1}$ . With fractions: $\frac{37}{99} + \frac{73}{99} = \frac{110}{99} = \frac{10}{9} = 1. \overline{1}$ ✓ (pattern: $\frac{a + b}{9} = \frac{3 + 7}{9}$ ).

Answer $1. \overline{1}$

When the block sum is

\geq

99 the whole-number part of the answer rises.

Example 38 · Standard

Find (a)

0. \overline{12} \times 0. \overline{33}

and (b)

0. \overline{7} \div 0. \overline{14}

Solution, step by step

1For multiplication and division making fractions is the right way. $0. \overline{12} = \frac{12}{99} = \frac{4}{33}$ and $0. \overline{33} = \frac{33}{99} = \frac{1}{3}$ .
2(a) $\frac{4}{33} \times \frac{1}{3} = \frac{4}{99} = 0. \overline{04}$ (because $\frac{4}{99} = \frac{04}{99}$ ).
3(b) $0. \overline{7} = \frac{7}{9}$ and $0. \overline{14} = \frac{14}{99}$ . Dividing means multiplying by the reciprocal: $\frac{7}{9} \times \frac{99}{14}$ .
4Simplify: $\frac{7}{9} \times \frac{99}{14} = \frac{1}{1} \times \frac{11}{2} = \frac{11}{2} = 5.5$ .

Answer (a) $0. \overline{04}$ ; (b) 5.5

The block trick does not work for multiplication and division — make fractions.

BasicTry it yourself first

What is the value of

0. \overline{5} + 0. \overline{7}

BasicTry it yourself first

What is the value of

0. \overline{58} + 0. \overline{85}

StandardTry it yourself first

What is the value of

5. \overline{27} - 2. \overline{54}

StandardTry it yourself first

What is the value of

0. \overline{18} \times 0. \overline{55}

StandardTry it yourself first

Find the cube root of

0. \overline{296}

StandardTry it yourself first

x + 0. \overline{36} = 1. \overline{81}

, what is the value of

x

AdvancedTry it yourself first

Find the value of

2. \overline{3} + 1. \overline{14} + 0.2 \overline{7}

AdvancedTry it yourself first

What is the value of

0. \overline{4} \times 0. \overline{3}

Comparing, ordering and the $n$ -th digit

the first difference decides ·

n mod r

Comparing recurring decimals, the digit at the $n$ -th place and the sum of the first $n$ digits.

Comparing

Comparing two decimals is just like ordering words in a dictionary: first the whole part, then go place by place from the first decimal place and find the first place where the digits differ — that settles it. But for recurring decimals the digits have to be written out, so write both out to enough places for the difference to show. Example: $0. \overline{63} = 0.636363 \dots$ and $0.63 \overline{6} = 0.636666 \dots$ : the first three digits agree and the fourth is 3 against 6 — so $0.63 \overline{6}$ is bigger.

◭ Diagram

The digits of

A = 0. \overline{63}

and

B = 0.63 \overline{6}

side by side: the first difference (rose) is at the fourth place — 3 against 6 there, so

B

is bigger

TrickA direct rule: a recurring decimal (like

0. \overline{3}

) is always bigger than its chopped finite form (

0.33

0.333

), because more is still to be added after the cut. To compare with a fraction, write the fraction as a decimal and look at a few digits, or bring both to a common denominator.

The $n$ -th digit and the sum of digits

Say the decimal has $k$ digits before the repeat and a block of $r$ digits. If $n \leq k$ the $n$ -th digit comes straight from the first digits. Otherwise look into the block: drop the digits before the repeat (leaving $n - k$ ), divide $(n - k)$ by $r$ and take the remainder. If the remainder $s \neq = 0$ it is the $s$ -th digit of the block; if the remainder is 0 it is the last ( $r$ -th) digit of the block.

∑ Formula

d_{n} = block [(n - k) mod r]

for

n > k

; if the remainder is 0, the last (

r

-th) digit of the block

$n$: place after the decimal point
$k$: number of digits before the repeat
$r$: number of digits in the block

The sum of the first $n$ digits: the sum of the first $k$ digits $+$ (number of complete blocks $\times$ the digit-sum of one block) $+$ (the sum of the leftover digits, which are the opening digits of the block). The number of complete blocks is $⌊ \frac{n - k}{r} ⌋$ .

Steps

Write the decimal and identify $k$ (before the repeat) and $r$ (the block).
For comparing: write both digit strings side by side and find the first different digit.
For the $n$ -th digit: $(n - k) mod r$ ; for the sum: count the complete blocks and the leftover digits separately.

Example 39 · Standard

Arrange

0. \overline{63}, 0.6 \overline{3}, 0. \overline{6}, 0.63 \overline{6}

in descending order.

Solution, step by step

1Write the first 6 decimal digits of all four: $0.636363$ , $0.633333$ , $0.666666$ , $0.636666$ .
2The first digit is 6 in all. The second digit: 3, 3, 6, 3 — so $0. \overline{6}$ (second digit 6) is the biggest.
3Among the other three the third digit: $0. \overline{63} \to 6$ , $0.6 \overline{3} \to 3$ , $0.63 \overline{6} \to 6$ — $0.6 \overline{3}$ is the smallest.
4Between $0. \overline{63}$ and $0.63 \overline{6}$ the fourth digit: 3 against 6 — $0.63 \overline{6}$ is bigger.
5Descending order: $0. \overline{6} > 0.63 \overline{6} > 0. \overline{63} > 0.6 \overline{3}$ .

Answer $0. \overline{6} > 0.63 \overline{6} > 0. \overline{63} > 0.6 \overline{3}$

In a digit-by-digit comparison the first differing digit decides everything; later digits are never looked at.

Example 40 · Basic

For

\frac{5}{13} = 0. \overline{384615}

find (a) the 100th digit after the point and (b) the sum of the first 50 digits.

Solution, step by step

1No digit before the repeat ( $k = 0$ ) and the block $384615$ has $r = 6$ digits.
2(a) $100 \div 6 = 16$ remainder 4. The remainder is 4, so the 100th digit is the fourth digit of the block: $3, 8, 4, 6, 1, 5$ ⟹ 6.
3(b) The digit-sum of one block is $3 + 8 + 4 + 6 + 1 + 5 = 27$ .
4 $50 \div 6 = 8$ remainder 2: 8 complete blocks and 2 leftover digits (3 and 8). Sum $= 8 \times 27 + (3 + 8) = 216 + 11 = 227$ .

Answer (a) 6; (b) 227

If the remainder is 0 take the last digit of the block, not the first.

Example 41 · Advanced

For

\frac{17}{60} = 0.28 \overline{3}

find (a) the 75th digit and (b) the sum of the first 40 digits.

Solution, step by step

1Two digits before the repeat (2, 8): $k = 2$ ; the block is a single digit (3): $r = 1$ .
2(a) $75 > 2$ , so in the block: $(75 - 2) mod 1 = 0$ — remainder 0 means the last digit of the block, which is 3. The 75th digit is 3.
3(b) The first 2 digits add to $2 + 8 = 10$ . The remaining $40 - 2 = 38$ places are all 3: $38 \times 3 = 114$ .
4Sum $= 10 + 114 = 124$ .

Answer (a) 3; (b) 124

In a mixed recurring decimal count the first

k

digits separately, then do the block arithmetic.

BasicTry it yourself first

Which is the largest of

0. \overline{5}, 0.5 \overline{4}, 0. \overline{54}, 0.54

BasicTry it yourself first

Which is the smallest of

0. \overline{27}, 0.2 \overline{7}, 0.27, 0. \overline{2}

StandardTry it yourself first

What is the 50th digit of

\frac{2}{7} = 0. \overline{285714}

StandardTry it yourself first

What is the sum of the first 30 digits of

\frac{1}{7} = 0. \overline{142857}

StandardTry it yourself first

What is the 41st digit of the decimal of

\frac{1}{12}

AdvancedTry it yourself first

How many times does the digit 8 appear among the first 75 digits of

\frac{3}{37} = 0. \overline{081}

AdvancedTry it yourself first

Arrange

\frac{4}{7}, 0. \overline{57}, 0.5 \overline{7}

in ascending order.

AdvancedTry it yourself first

What is the sum of the first 100 digits of

\frac{1}{13} = 0. \overline{076923}

Summary, mixed practice, revision

All methods in one place

One table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

Which method when

One page to pick the method as soon as you read a question.

Method	Use when	Key formula / move
Prime factorisation	the first move of any factor question	divide by the smallest prime; try up to $N$ ; sum of powers = prime factors
Number of factors	“how many factors”	$(a + 1) (b + 1) (c + 1) \dots$
Sum, product, sum of reciprocals	“of all the factors”	$\prod \frac{p ^{a + 1} - 1}{p - 1}$ ; $N^{d /2}$ ; $\frac{σ}{N}$
Even, odd, multiples, square factors	any condition on the factors	count the allowed powers of each prime; multiples of $m$ : $d (N / m)$
Perfect squares and pairs	“in how many ways as a product”, “make a square/cube”	$\frac{d}{2}$ , $\frac{d + 1}{2}$ , $2^{k - 1}$ ; complete the incomplete powers
Zeros: pairs of 2 and 5	zeros at the end of a product	$min (x, y)$
$n!$ and Legendre	power of a prime or a composite in $n!$	$⌊ \frac{n}{p} ⌋ + ⌊ \frac{n}{p ^{2}} ⌋ + \dots$
Reverse question	zeros given, $n$ asked	$n \approx 4 k$ , test multiples of 5; 5, 11, 17, 23 never occur
Series products	consecutive, multiples, odd, with powers, factorials	$\frac{b !}{( a - 1 )!}$ ; $k^{m} m!$ ; count both 2s and 5s
Terminating or recurring	the decimal form of a fraction	only 2, 5 in the denominator ⟹ terminating ( $max (a, b)$ places)
Decimal → fraction	a recurring decimal is given	(all digits − digits before) ÷ ( $r$ nines, $k$ zeros)
Operations on recurring decimals	sums, products, roots	make fractions; for sums stretch blocks to $L$ and add
Comparing and the $n$ -th digit	order, “the $n$ -th digit”, sum of digits	first differing digit; $(n - k) mod r$

Letters in the table: $N$ = the given number; $a, b, c$ = powers of primes; $d$ = number of factors; $σ$ = their sum; $k$ = number of different primes (in pairs) or the digits before the repeat (in decimals); $x, y$ = total powers of 2 and 5; $n$ = the $n$ of $n!$ , $p$ = a prime; $r$ = digits in the block; $L$ = LCM of the block lengths; $m$ = a factor whose multiples are to be counted.

Spot itAs soon as you read a question ask: (1) is it factors, zeros or decimals? (2) for factors — write the prime factorisation. (3) for zeros — the powers of 2 and 5. (4) for decimals — which prime other than 2, 5 is in the denominator?

Mixed practice — set 1

Basic to standard

Mixed questions from all three parts; spot the method first, then solve.

These questions are not in method order — just like the exam. For each, first decide which method fits.

BasicTry it yourself first

How many factors does 1584 have?

BasicTry it yourself first

What is the sum of all the factors of

2^{2} \times 3 \times 7

BasicTry it yourself first

Which of 1296, 1350, 1458, 1500 has an odd number of factors?

BasicTry it yourself first

How many zeros are at the end of

40 \times 125 \times 18 \times 7

BasicTry it yourself first

How many zeros are at the end of

85!

BasicTry it yourself first

What is the highest power of 5 in

400!

BasicTry it yourself first

What is the simplest fraction for

0. \overline{36}

BasicTry it yourself first

What is the simplest fraction for

0.2 \overline{3}

StandardTry it yourself first

Which of these fractions is a terminating decimal?

StandardTry it yourself first

What is the value of

0. \overline{2} + 0. \overline{9}

StandardTry it yourself first

Which is the largest of

0. \overline{12}, 0.1 \overline{2}, 0.12 \overline{1}

StandardTry it yourself first

What is the 40th digit of

\frac{3}{7} = 0. \overline{428571}

StandardTry it yourself first

How many prime factors (counting repeats) does

8^{4} \times 9^{3} \times 1 0^{2}

have?

StandardTry it yourself first

How many factors of

2^{3} \times 5^{2} \times 11

are even?

StandardTry it yourself first

In how many ways can 1764 be written as a product of two factors (order not counted, equal factors allowed)?

StandardTry it yourself first

Find the smallest

n

for which

n!

has at least 8 trailing zeros.

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

Several questions here need two methods together. Do not look for a method name; decide from how the question looks.

StandardTry it yourself first

What is the smallest number with exactly 14 factors?

StandardTry it yourself first

What is the sum of all the odd factors of

2^{5} \times 3^{3} \times 5^{2} \times 7

StandardTry it yourself first

How many factors of

6^{4} \times 1 5^{2}

are perfect cubes?

AdvancedTry it yourself first

In how many ways can 30030 be written as a product of two co-prime factors (order not counted)?

AdvancedTry it yourself first

By what smallest number must 2058 be multiplied to make a perfect cube?

StandardTry it yourself first

How many zeros are at the end of

31 \times 32 \times 33 \times \dots \times 150

StandardTry it yourself first

How many zeros are at the end of

25! \times 40!

AdvancedTry it yourself first

Find the largest

k

for which

1 8^{k}

divides

100!

AdvancedTry it yourself first

How many zeros are at the end of

35 \times 70 \times 105 \times \dots \times 700

(the first 20 multiples of 35)?

StandardTry it yourself first

What is the value of

0. \overline{81} + 0. \overline{18}

StandardTry it yourself first

Write

2. \overline{45}

as an improper fraction.

StandardTry it yourself first

Convert

0.3 \overline{18}

into a fraction in lowest terms.

AdvancedTry it yourself first

Find

1. \overline{6} + 0. \overline{45} + 2.0 \overline{3}

as a recurring decimal.

AdvancedTry it yourself first

What is the 87th digit of

\frac{7}{13} = 0. \overline{538461}

AdvancedTry it yourself first

For how many integers

n

from

1

200

\frac{n}{150}

a terminating decimal?

AdvancedTry it yourself first

What is the sum of the first 61 digits of

\frac{1}{7} = 0. \overline{142857}

Revision — cards and traps

Flip and recall

Read the front, think, then flip.

TrapsMost common traps: (1) not adding 1 to the powers when counting factors. (2) misjudging

0. \overline{9} = 1

. (3) forgetting the two 5s of numbers like 25 in

n!

. (4) deciding terminating/recurring without first reducing. (5) in a product of a series, counting only the 5s and forgetting to look at the 2s.

Factors, Trailing Zeros and Recurring Decimals

The idea

Counting the prime factors

Steps

The idea

The reverse question: finding the number

Steps

The idea: the sum

The product, and the sum of reciprocals

Steps

The idea

Steps

The idea

Questions about pairs

Making a perfect square or cube

Steps

The idea

Steps

The idea

The highest power of a composite number

Steps

The idea

Steps

The idea

Steps

The idea

The kind of recurring decimal and the length of the block

Steps

The idea: shift and subtract

Steps

The idea

The quick way: add the blocks

Steps (block method for sums)

Comparing

The n-th digit and the sum of digits

Steps

The $n$ -th digit and the sum of digits