Number System

high importance~2 Q in Tier 133 formulas⚡ 18 shortcuts6 subtopics

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Subtopic 2 of 6·← Number types, place value & counting·Remainders & remainder theorem →

Divisibility rules

Divisor	Rule
2, 5, 10	last digit
4	last two digits divisible by 4
8	last three digits divisible by 8
3, 9	digit sum divisible by 3 / 9
11	(sum of digits at odd places) − (sum at even places) = 0 or a multiple of 11
7, 11, 13	alternating sum of 3-digit groups from the right divisible by 7/11/13 (since 1001 = 7 × 11 × 13)
composite	break into co-prime factors: 12 = 3 × 4, 72 = 8 × 9, 88 = 8 × 11, 99 = 9 × 11

Algebraic divisibility (very popular): $a^n - b^n$ is always divisible by $a-b$ ; by $a+b$ too when $n$ is even. $a^n + b^n$ is divisible by $a+b$ when $n$ is odd.

For missing-digit questions, apply the most restrictive rule first (usually 8 or 11), then the digit-sum rule.

📖 Notes 🧩 6 question types ⚡ Formulas 🎯 18 practice Q

Detailed notes

What does "divisible" mean?

A number $N$ is divisible by $d$ if $d$ goes into $N$ exactly, with remainder 0. For example 84 is divisible by 7 because 84 = 7 × 12. Divisibility rules let you check this without long division — a big time saver in exams.

The basic rules

Divisor	Test	Example
2	last digit even	3,578 ✓
3	sum of digits divisible by 3	4,521 → 12 ✓
4	last two digits divisible by 4	7,316 → 16 ✓
5	last digit 0 or 5	4,215 ✓
8	last three digits divisible by 8	91,224 → 224 ✓
9	sum of digits divisible by 9	6,453 → 18 ✓
10	last digit 0	560 ✓
11	(sum of digits at odd places) − (sum at even places) = 0 or a multiple of 11	9,48,475 ✓
7, 11, 13	alternating sum of 3-digit groups from the right	2,47,247 → 247 − 247 = 0 ✓

Rule of 11, step by step. Take 948475. From the right, odd places hold 5, 4, 4 (sum 13) and even places hold 7, 8, 9 (sum 24). Difference 24 − 13 = 11 → divisible by 11.

Composite divisors: split into co-prime parts

To test 72, there is no "rule of 72". Write 72 = 8 × 9 (8 and 9 share no factor) and check both rules. Useful splits: 12 = 3 × 4, 18 = 2 × 9, 24 = 3 × 8, 36 = 4 × 9, 44 = 4 × 11, 72 = 8 × 9, 88 = 8 × 11, 99 = 9 × 11. Do not split 24 as 4 × 6 — they share the factor 2, so a number divisible by 4 and 6 (like 36) need not be divisible by 24.

Missing digits

When digits are hidden (x, y), apply the rule with the fewest choices first. The rule of 8 fixes the last three digits; the rule of 11 gives one equation; the digit-sum rule gives another. Often two or three pairs (x, y) work, but x + y (or x − y) is the same in all of them — that is exactly what the question asks.

Algebraic divisibility (very popular)

$a^n - b^n$ is always divisible by $a - b$ .
$a^n - b^n$ is divisible by $a + b$ when $n$ is even.
$a^n + b^n$ is divisible by $a + b$ when $n$ is odd. Example: $17^{25} + 23^{25}$ has an odd power, so it is divisible by 17 + 23 = 40.

Take out the common power

$3^{25} + 3^{26} + 3^{27} = 3^{25}(1 + 3 + 9) = 3^{25} \times 13$ . So the sum is divisible by 13. Always factor out the smallest power first.

Special number forms

$abcabc = abc \times 1001 = abc \times 7 \times 11 \times 13$ .
$aaa = a \times 111 = a \times 3 \times 37$ , so it is always divisible by 37.
$ab + ba = 11(a + b)$ , always divisible by 11. $ab - ba = 9(a - b)$ .
The product of any 3 consecutive integers is divisible by 6 (= 3!).

Nearest multiple questions

Least number to add to $N$ to make it divisible by $d$ : $d - (N \bmod d)$ (0 if already divisible).
Least number to subtract: $N \bmod d$ .
Largest $n$ -digit number divisible by $d$ : $99\dots9 - (99\dots9 \bmod d)$ .
Smallest $n$ -digit number divisible by $d$ : $10\dots0 + (d - 10\dots0 \bmod d)$ . Example: 8,357 ÷ 12 leaves 5, so add 12 − 5 = 7 to get 8,364.

Quick revision

2/5/10 → last digit; 4 → last 2 digits; 8 → last 3 digits; 3/9 → digit sum; 11 → alternating sum.
Composite divisor → split into co-prime factors only.
$a^n + b^n$ divisible by $(a+b)$ needs odd $n$ ; $a^n - b^n$ divisible by $(a-b)$ always.
abcabc → 7, 11, 13; aaa → 37.
To reach a multiple: add $d - r$ or subtract $r$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Missing digits for a composite divisor (72, 88, 36, 99)very common3 practice Q

How to spot it:

A number with blanks x and y 'is divisible by 72/88/99'; the options are values of x + y or x − y.

72 = 8 \times 9,\ 88 = 8 \times 11,\ 99 = 9 \times 11

Split the divisor into co-prime factors.
Use the rule with fewer possibilities first (8 → last three digits, 11 → alternating sum).
Put that value into the second rule (digit sum for 9, alternating sum for 11).
List the few cases; the asked quantity is the same in all cases.

Why: a number divisible by two co-prime numbers is divisible by their product, and vice versa.

Example: If the five-digit number 3x5y2 is divisible by 88, find x + y.

$88 = 8 \times 11$ . Rule of 8: $5y2$ divisible by 8 → 512, 552, 592 → $y = 1, 5, 9$ . Rule of 11 (from the right): odd places $2 + 5 + 3 = 10$ , even places $y + x$ . So $10 - (x + y) = 0$ → $x + y = 10$ . Pairs (9, 1), (5, 5), (1, 9) all give x + y = 10.

Type 2: Divisibility by 11 (single missing digit or pick the number)very common2 practice Q

How to spot it:

'If 9x8475 is divisible by 11, find x', or 'which of these numbers is divisible by 11?'.

(\text{sum of odd-place digits}) - (\text{sum of even-place digits}) \in \{0, \pm 11, \pm 22, \dots\}

Number the digits from the right: 1st, 2nd, 3rd …
Add the odd-place digits and the even-place digits separately.
The difference must be 0, 11, 22 … Solve for the missing digit (0 to 9).

Why: $10$ leaves remainder $-1$ with 11, so place values alternate between +1 and −1.

Example: Is 7,29,135 divisible by 11?

From the right: odd places 5, 1, 2 → 8; even places 3, 9, 7 → 19. Difference $= 11$ → yes, divisible.

Type 3: Algebraic divisibility: $a^n \pm b^n$very common3 practice Q

How to spot it:

Big powers with the same exponent added or subtracted, e.g. ' $17^{25} + 23^{25}$ is divisible by'.

(a-b) \mid a^n - b^n;\quad (a+b) \mid a^n - b^n\ (n \text{ even});\quad (a+b) \mid a^n + b^n\ (n \text{ odd})

Note whether it is a sum or a difference and whether $n$ is odd or even.
Compute $a + b$ and $a - b$ and match with the options.
For $a^n - 1$ : any $a^k - 1$ where $k$ divides $n$ is a factor.

Why: $a^n - b^n = (a-b)(a^{n-1} + \dots + b^{n-1})$ ; the sum version works for odd $n$ by replacing $b$ with $-b$ .

Example: Is $29^{16} - 13^{16}$ divisible by 42?

$n = 16$ is even, so $a^n - b^n$ is divisible by $a + b = 29 + 13 = 42$ . Yes.

Type 4: Sum of powers of the same base: factor out the smallest powercommon2 practice Q

How to spot it:

A sum like $3^{25} + 3^{26} + 3^{27}$ or $5^{21} + 5^{22} + 5^{23}$ and 'is divisible by'.

a^m + a^{m+1} + a^{m+2} = a^m(1 + a + a^2)

Take the smallest power outside the bracket.
Simplify the bracket to a small number.
The number is divisible by that bracket value (and by powers of the base).

Why: every term shares the smallest power as a common factor.

Example: Find a factor of $4^{61} + 4^{62} + 4^{63}$ other than a power of 2.

$4^{61}(1 + 4 + 16) = 4^{61} \times 21$ . So it is divisible by 21 (and by 3 and 7).

Type 5: Least number to add/subtract; largest or smallest n-digit multiplevery common3 practice Q

How to spot it:

'What least number must be added to/subtracted from N so that it is divisible by d?', 'largest five-digit number divisible by 47'.

\text{add} = d - (N \bmod d),\quad \text{subtract} = N \bmod d

Divide $N$ by $d$ and find the remainder $r$ .
Subtract $r$ (goes down to the lower multiple) or add $d - r$ (goes up to the next multiple).
Largest $n$ -digit multiple: $99\dots9 - r$ . Smallest $n$ -digit multiple: $10\dots0 + (d - r)$ .

Why: multiples of $d$ are $d$ apart, and the remainder tells how far $N$ is above the lower one.

Example: What least number must be subtracted from 4,527 to make it divisible by 16?

$4527 = 16 \times 282 + 15$ . Subtract the remainder 15. Check: $4512 = 16 \times 282$ .

Type 6: Special forms: abcabc, aaa, ab + bacommon2 practice Q

How to spot it:

'A six-digit number formed by repeating a three-digit number is always divisible by …', 'a three-digit number with all digits equal'.

abcabc = 1001 \times abc = 7 \times 11 \times 13 \times abc,\quad aaa = 3 \times 37 \times a

Write the number in expanded form.
Take out the fixed factor (1001, 111, 11, 9).
Choose the option that divides this fixed factor.

Why: the pattern of the digits builds in a fixed multiplier, whichever digits are used.

Example: Is 4,84,484 divisible by 13? (It is 484 written twice.)

$484484 = 484 \times 1001$ and $1001 = 7 \times 11 \times 13$ . So yes, it is divisible by 13.

Formulas

Divisibility by 11

\left(\sum \text{odd-place digits}\right) - \left(\sum \text{even-place digits}\right) \in \{0, \pm 11, \pm 22, \dots\}

Composite divisor

pq \mid N \iff p \mid N \text{ and } q \mid N, \quad \gcd(p,q)=1

Difference of powers

(a-b) \mid (a^n - b^n) \text{ for all } n

Difference of even powers

(a+b) \mid (a^n - b^n) \text{ when } n \text{ is even}

Sum of odd powers

(a+b) \mid (a^n + b^n) \text{ when } n \text{ is odd}

abcabc form

\overline{abcabc} = \overline{abc} \times 1001 = \overline{abc} \times 7 \times 11 \times 13

Shortcut tricks

⚡ Split into co-prime factors

Never split 12 as 2 × 6 or 72 as 4 × 18 — the factors must be co-prime. Test each factor's rule separately.

Example: If the five-digit number 37x84 is divisible by 12, find the smallest value of x.

12 = 3 × 4. By 4: last two digits 84 ✓ (any x). By 3: $3+7+x+8+4 = 22+x$ must be a multiple of 3 ⇒ $x \in \{2, 5, 8\}$ . Smallest $x = 2$ .

⚡ The 1001 family

Any six-digit number of the form $abcabc$ equals $abc \times 1001$ , so it is always divisible by 7, 11, 13 (and 77, 91, 143, 1001).

Example: Which of 7, 11 and 13 divide 345345?

$345345 = 345 \times 1001 = 345 \times 7 \times 11 \times 13$ — all three divide it.

⚡ Sum of odd powers

$a^n + b^n$ with odd $n$ is divisible by $a+b$ — look at the options for $a+b$ first.

Example: Is $17^{15} + 23^{15}$ divisible by 40?

Exponent 15 is odd, so $17 + 23 = 40$ divides $17^{15} + 23^{15}$ . Yes.

Where students lose marks

Splitting a composite divisor into non-co-prime factors (e.g. 18 is divisible by 2 and 6 but not by 12).
Checking only the last two digits for 8 (need last three).
For 11, forgetting that the difference can be 0 or a negative multiple of 11.
Applying the $(a+b)$ rule to $a^n + b^n$ when $n$ is even — it fails.

Practice sets — 18 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.