Divisibility test rules

Ask any student whether 4,572 is divisible by 3, and most will reach for long division. Ask a number-sense expert, and they will glance at the digits, add them up, and answer in three seconds — no division required. That speed is not a trick to be memorised; it is a direct consequence of how place value works. Every divisibility test — for 2, 3, 4, 5, 6, 9, 10 or 11 — comes from the same idea: write the number in its generalized place-value form, such as $\overline{abc}=100a+10b+c$ , and split it into a piece that is automatically a multiple of the divisor, plus a small leftover that depends on only a few digits. Once you see that split, the "rules" stop being arbitrary and become short proofs you could reconstruct on your own. This article builds that reasoning for every test in the ICSE Class 8 syllabus, gives one fully worked example that applies several tests together, and lists the traps that most often cost marks.

The core insight

Splitting a number's place-value expansion into a term that is always a multiple of the divisor plus a small leftover (last digit, last two digits, digit sum, or alternating digit sum) turns every divisibility rule into a short, provable fact.

The core idea

Take a 3-digit number $\overline{abc} = 100a+10b+c$ , where $a$ , $b$ , $c$ are its hundreds, tens and units digits. The whole strategy is to rewrite $100a+10b+c$ as (a guaranteed multiple of the divisor) $+$ (a small leftover), because a sum is divisible by a number exactly when the leftover is.

Tests that depend only on the last digit. Since $10(10a+b)$ is always a multiple of 10, of 5, and of 2, we get:

100a+10b+c = 10(10a+b)+c

where: $a,b,c$ are the digits of the number, and $10(10a+b)$ is the part built purely from tens and hundreds. Divisibility by 10 needs $c=0$ ; by 5 needs $c\in\{0,5\}$ ; by 2 needs $c$ even.

A test that depends on the last two digits. Since $100=4\times25$ , the term $100a$ is always a multiple of 4, so $100a+10b+c$ is divisible by 4 exactly when the two-digit number $\overline{bc}=10b+c$ is.

Tests that depend on the digit sum. Writing $100=99+1$ and $10=9+1$ separates out multiples of 9 (hence also of 3):

100a+10b+c = 99a+9b+(a+b+c)

where: $99a$ and $9b$ are always multiples of 9 (and of 3), so the number is divisible by 3, or by 9, exactly when the digit sum $a+b+c$ is divisible by 3, or by 9, respectively.

A combined test. Since $6=2\times3$ , a number is divisible by 6 exactly when it passes both the even-units-digit test and the digit-sum-divisible-by-3 test — there is no separate "trick" for 6.

A test that alternates sign. Writing $100=99+1$ and $10=11-1$ isolates multiples of 11:

100a+10b+c = (99a+11b)+(a-b+c)

where: $99a$ and $11b$ are always multiples of 11, so the number is divisible by 11 exactly when the alternating digit sum $(a+c)-b$ is 0 or a multiple of 11.

Figure

A place value number split into a guaranteed multiple block and a small leftover block feeding the divisibility decision

Derivation Athe digit-sum test for 3 and 9

Prove that a whole number is divisible by 3 (or by 9) if and only if the sum of its digits is divisible by 3 (or by 9), by expressing the number in place-value form and separating out the part built from powers of 10 that are always one more than a multiple of 9.

Step 1. Take any 3-digit number $\overline{abc}=100a+10b+c$ (the argument extends the same way to any number of digits, since $10^k = \underbrace{99\ldots9}_{k}+1$ for every power $k$ ).

Step 2. Write each power of 10 as a multiple of 9 plus 1: $100=99+1$ and $10=9+1$ . Substituting,

100a+10b+c = (99+1)a+(9+1)b+c = 99a+9b+(a+b+c)

Step 3. Group the terms: $99a+9b$ is manifestly a multiple of 9, since $99=9\times11$ and $9b=9\times b$ . It is therefore also a multiple of 3. So the original number equals (a multiple of 9) $+\ (a+b+c)$ .

Step 4. Since adding a multiple of 9 never changes whether a sum is divisible by 3 or by 9, the number $100a+10b+c$ is divisible by 3 exactly when $a+b+c$ is divisible by 3, and divisible by 9 exactly when $a+b+c$ is divisible by 9. This is the digit-sum test, and it holds for numbers of any length by the same grouping.

Worked example

Problem. Check whether 4,572 is divisible by 3, 4, 6, 9 and 11, and then find the digit $x$ so that $\overline{47x2}$ is divisible by 9.

Step 1. Find the digit sum of 4,572: $4+5+7+2=18$ .

Step 2. Test 3 and 9: since 18 is a multiple of both 3 and 9, 4,572 is divisible by 3 and by 9.

Step 3. Test 4: look at the last two digits, 72. Since $72\div4=18$ exactly, 4,572 is divisible by 4.

Step 4. Test 6: the units digit 2 is even (passes the test for 2) and the digit sum 18 is a multiple of 3 (passes the test for 3), so 4,572 is divisible by 6.

Step 5. Test 11: the alternating digit sum, working from the units digit, is $(2+5)-(7+4)=7-11=-4$ . Since $-4$ is not 0 or a multiple of 11, 4,572 is not divisible by 11.

Step 6. For the digit puzzle $\overline{47x2}$ , apply the digit-sum test for 9: the digit sum is $4+7+x+2=13+x$ . The smallest multiple of 9 that is at least 13 is 18, so $13+x=18$ , giving $x=5$ . Checking, $13+14=27$ would need $x=14$ , not a valid digit, so $x=5$ is the only answer.

Final answer: 4,572 is divisible by 2, 3, 4, 6 and 9, but not by 11; and $x=5$ , giving $\overline{4752}$ , whose digit sum $4+7+5+2=18$ confirms divisibility by 9 (and $52\div4=13$ confirms it is also divisible by 4).

Competitive stretch

Digit puzzles sometimes ask for divisibility by 11 instead of 9. For $\overline{47x2}$ to be divisible by 11, the alternating sum $(2+7)-(x+4)=5-x$ must be 0 or a multiple of 11. Since $x$ is a single digit, $5-x=0$ gives $x=5$ as the only solution in range — a useful check that different tests can sometimes agree on the same missing digit, though this need not happen in general.

Common traps

Using the digit-sum test (meant for 3 or 9) to check divisibility by 4, 5, 10 or 11 — each of those tests looks at a different, specific part of the number, not the full digit sum.
Checking only the last digit for the test of 4 — e.g., 1,714 ends in an even digit, but its last two digits, 14, are not divisible by 4, so 1,714 fails the test for 4.
Getting the alternating sum for 11 backwards by mixing up which digits are added and which are subtracted — always work outward from the units digit consistently, as (units + hundreds + …) − (tens + thousands + …).
Assuming divisibility by 3 implies divisibility by 9 — 9 is a stricter condition on the same digit sum; 12 has digit sum 3 (divisible by 3) but not by 9.

Ask the tutor: Want to practise deriving a fresh divisibility rule (say, for 8 or 25) from the place-value split, or work through more digit-puzzle problems step by step? Ask the AI tutor.

Why does the divisibility test for 3 use the digit sum instead of just the last digit, unlike the tests for 2, 5 and 10

The tests for 2, 5 and 10 work off the last digit because every power of 10 from $10^1$ upward is already a multiple of those numbers — so only the units place matters. But no power of 10 is a multiple of 3 or 9; instead, every power of 10 is exactly 1 more than a multiple of 9 (and hence of 3), as in $100=99+1$ . That "+1" from each place value is what forces the digit sum, rather than a single digit, to carry the divisibility information.

Does the last-two-digits rule for 4 also work for other divisors like 8 or 16

Yes, the same reasoning extends: since $1000=8\times125$ , any multiple of 1000 is automatically divisible by 8, so a number is divisible by 8 exactly when its last three digits are. In general, for a divisor that is a power of 2 or 5, you only need to check as many final digits as match the power (one digit for 2 or 5, two for 4 or 25, three for 8 or 125), because that is where $10^k$ first becomes a clean multiple of the divisor.

If a number fails the divisibility test for 11, is there a quick way to know how far off it is

The alternating digit sum itself tells you how far off you are, though not directly as a remainder. If the alternating sum is, say, $-4$ as in 4,572, it means the number is 11 short of or 11 over a multiple of 11 relative to that adjustment — you would still need to divide to get the exact remainder. The test is designed to answer yes-or-no quickly, not to compute the remainder; for that, ordinary division is still the reliable method.

Divisibility test rules

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads