Aptitude · Foundations

Four families, not fifteen rules to memorise

Divisibility rules look like a list to be learned by heart. They are not. They fall into four families, each with one reason behind it, and knowing the reason means you can rebuild any rule you have forgotten — including ones nobody taught you.

Pick a number and a divisor, and step through the right rule →

Do not memorise rules one divisor at a time. Learn why each family works and every rule in it follows — including the ones for 7, 13 and 25.

01 The idea

Why a rule exists at all

Every divisibility rule is the same observation in different clothes: a big number can be broken into pieces, and if you already know each piece is a multiple of your divisor, you can throw it away and look at what is left.

Take 4. One hundred is divisible by 4, so every multiple of a hundred is too. That means in 3564 the “3500” part is automatically a multiple of 4 and tells you nothing. Only the last two digits, 64, can decide the question. That is the whole rule, and it explains why the rule for 4 uses two digits while the rule for 8 uses three.

Take 9. Here nothing about hundreds helps, because 100 is not a multiple of 9. But 100 leaves remainder 1, and so does 10, and 1000, and every power of ten. So the number leaves the same remainder as the sum of its digits — which is why 9 gets a digit-sum rule and 4 does not.

Those two mechanisms — the divisor divides a power of ten, or the divisor leaves a tidy remainder in every power of ten — generate all four families below. Once you see which mechanism a divisor sits under, you can derive its rule instead of recalling it.

A rule is a way of discarding the part of the number that cannot affect the answer. Which part you may discard depends on how your divisor relates to powers of ten.

Tail ruleUsed when the divisor divides a power of ten exactly. Everything above that many digits is already a multiple, so only the tail is tested. 2, 4, 8, 5, 10, 25 all work this way.

Digit-sum ruleUsed when every power of ten leaves remainder 1. The number then leaves the same remainder as its digit sum. Only 3 and 9 (and their multiples) qualify.

Alternating-sum ruleUsed when powers of ten alternate between remainder +1 and −1, which is what happens for 11. The digits inherit the alternation, so signs flip along the number.

02 Worked example

One number, three families: is 3564 divisible by 4, by 9, by 11?

3564 is used for the whole lesson because it happens to pass several rules at once, so the families can be compared on identical material. Test 3564 for divisibility by 4, then by 9, then by 11 — without dividing.

By 4 — keep two digits100 is a multiple of 4, so the 3500 part is irrelevant. Only 64 can decide it.last two digits = 64, and 64 / 4 = 16 exactly ⇒ divisible by 4

By 9 — add the digitsEvery power of ten leaves remainder 1 on division by 9, so 3564 leaves the same remainder as its digit sum.3 + 5 + 6 + 4 = 18, and 18 / 9 = 2 exactly ⇒ divisible by 9

By 3 — the same sum, for freeThe digit sum 18 is also a multiple of 3, so no new work is needed. Anything divisible by 9 is automatically divisible by 3.18 / 3 = 6 ⇒ divisible by 3 as well

By 11 — alternate the signsTen leaves remainder −1 on division by 11, so the digits alternate in sign starting from the units.4 − 6 + 5 − 3 = 0

Read the alternating resultA difference of exactly 0 counts as a multiple of 11. So 3564 clears this family too.0 is a multiple of 11 ⇒ divisible by 11 (3564 / 11 = 324)

Three rules, three different mechanisms, one number, and not a single long division. Note also what fell out for free: 3564 passes 2 and 3, so it passes 6; and it passes 4 and 9, so it passes 36. Combining rules you have already run is cheaper than testing a new divisor from scratch — provided the parts are co-prime, which is the trap in section 03.

03 The method

The four families, and the trap in the fourth

The first three families are safe. The fourth is where students lose marks, because the rule has a condition that is usually taught without emphasis.

Family 1 — tails: 2 keeps 1 digit, 4 keeps 2, 8 keeps 3, 5 and 10 keep 1, 25 keeps 2. Family 2 — digit sum: 3 and 9. Family 3 — alternating sum: 11. Family 4 — split into co-prime factors: 6 = 2×3, 12 = 3×4, 15 = 3×5, 18 = 2×9.

The trap: splitting only works when the two factors are co-prime. 8 = 2×4, but 2 and 4 share a factor, so passing both tests does not prove divisibility by 8 — 12 is divisible by 2 and by 4 and is not divisible by 8. For 8 you must use the tail rule. Likewise 12 must be split as 3×4, never as 2×6.

Divisor	Rule	Family
2	`last digit is 0 2 4 6 8`	Tail, 1 digit
4	`last 2 digits divisible by 4`	Tail, 2 digits
8	`last 3 digits divisible by 8`	Tail, 3 digits
5 and 10	`last digit 0 or 5 / 0`	Tail, 1 digit
3 and 9	`digit sum divisible by 3 / 9`	Digit sum
11	`alternating sum is 0 or a multiple of 11`	Alternating
6	`passes 2 and passes 3`	Co-prime split
15	`passes 3 and passes 5`	Co-prime split
8 via 2 and 4	`invalid — not co-prime`	Use the tail rule

05 Cheat sheet

Every rule worth knowing

Grouped by family so the reasons stay attached to the rules. The last two rows are the ones nobody teaches and both follow from the tail mechanism.

Divisor	Test	Reason
2, 5, 10	`last 1 digit`	10 is a multiple of 2, 5 and 10
4, 25	`last 2 digits`	100 is a multiple of 4 and of 25
8, 125	`last 3 digits`	1000 is a multiple of 8 and of 125
3, 9	`digit sum`	Every power of 10 leaves remainder 1
11	`alternating sum`	Powers of 10 alternate +1 and −1
6, 12, 15, 18	`two co-prime rules`	Only valid when the factors are co-prime
7, 13	`no cheap rule — just divide`	Neither divides a power of ten nor leaves a tidy remainder

Co-prime or the split is invalid8 is not 2×4 for this purpose — 12 passes both and fails 8. Split 12 as 3×4, never 2×6, and use the tail rule for 8.

Divisible by 9 means divisible by 3The digit sum does both at once, so a single addition settles two divisors. The reverse does not hold: 12 passes 3 and fails 9.

A difference of 0 counts for 11Zero is a multiple of every number. Exam setters like alternating sums that come to exactly 0, and students often read that as a failure.

06 Where & why

Where these actually earn marks

Divisibility rules are rarely the question. They are how you get through the question fast enough to reach the next one.

Ratio and fraction simplification

Cancelling without trial and error

Every ratio question needs common factors spotted quickly. The digit-sum rule finds a 3 or a 9 in a five-digit number instantly.

TCS NQT · number-system questions

“Find the missing digit”

“If 34_86 is divisible by 9, find the digit.” Set the digit sum to the nearest multiple of 9 and solve — a one-line answer to a question that looks like a search.

HCF and LCM

Factorising before you start

Both operations need prime factorisations, and the family rules are the fastest way to strip out the small primes before anything harder is attempted.

Remainder questions

The rules give remainders, not just yes or no

The digit sum of a number leaves the same remainder on 9 as the number does, so the same work answers “what is the remainder” as well as “is it divisible”.

Note what the last card means: these rules are stronger than they are usually taught. The digit sum does not merely test divisibility by 9 — it hands you the exact remainder, which turns a whole class of remainder questions into one addition.

07 Interview questions

What gets asked about divisibility

Ten questions, and two of them are the ones that separate memorising from understanding.

Why does the rule for 4 use two digits rather than one?

Because 100 is divisible by 4 but 10 is not. Every hundred is automatically a multiple of 4, so everything above the last two digits contributes nothing, and the last two digits carry the whole answer. For 8 you need three digits, because 1000 is the first power of ten divisible by 8.

Why does adding the digits test for 9?

Because every power of ten leaves remainder 1 when divided by 9 — 10, 100 and 1000 are all one more than a multiple of 9. So each digit contributes just itself to the remainder, and the number leaves the same remainder as the sum of its digits.

Why does 11 use an alternating sum instead of a plain one?

Because 10 leaves remainder −1 on division by 11, not +1. So 100 leaves +1, 1000 leaves −1, and so on alternately. Each digit inherits the sign of its place value, which is exactly the alternating sum.

Is 3564 divisible by 11?

Yes. Alternating from the units digit: 4 − 6 + 5 − 3 = 0, and 0 is a multiple of 11. It checks out: 3564 / 11 = 324. A difference of exactly 0 is the commonest case in exam numbers and is often misread as a failure.

Can you test for 8 by checking 2 and then 4?

No, and this is the most important trap in the topic. 2 and 4 are not co-prime, so passing both does not imply divisibility by 8 — 12 is divisible by 2 and by 4 but not by 8. Splitting a divisor only works when the two factors share no common factor. For 8, use the last-three-digits rule.

So how do you test for 12?

Split it as 3 × 4, because 3 and 4 are co-prime. Test the digit sum for 3 and the last two digits for 4, and both passing does prove divisibility by 12. Splitting it as 2 × 6 is invalid, since 2 and 6 share a factor.

If a number is divisible by 9, is it divisible by 3?

Yes, always, because 3 divides 9. So one digit sum answers both. The converse fails: 12 has digit sum 3, which is divisible by 3 but not by 9, so 12 passes 3 and fails 9.

If 34_86 is divisible by 9, what is the missing digit?

The known digits sum to 3 + 4 + 8 + 6 = 21, so the total must reach the next multiple of 9, which is 27. That needs 6. This is the standard exam use of the digit-sum rule — setting the sum to a target rather than testing a given number.

Is there a quick rule for 7?

Hard

(a) Using the tail mechanism, work out and state the divisibility rule for 25, and for 125. (b) Explain why no rule of that shape can exist for 7. (c) A student claims that because 3564 is divisible by 4 and by 9, it must be divisible by 36, and separately that because it is divisible by 2 and by 6 it must be divisible by 12. One claim is sound and one is not. Say which, and give the general condition that decides it.

Follow-up

This is the lesson's real test: part (a) asks you to generate rules from the mechanism rather than recall them, and part (c) puts a valid split and an invalid one side by side so the co-primality condition has to be stated precisely rather than gestured at.

Show the hint

For (a) ask which power of ten each divisor divides. For (c) check the gcd of each pair before trusting either claim.