SSC CGL Maths · Number System

Divisibility Tests

A divisibility test reads a few digits, or a sum of digits, and tells you whether a number divides exactly without doing the division.

All Number System notes

Why this matters

Twenty-nine questions in the bank, twenty-six EASY and three MODERATE, all from the 2023 and 2024 shifts. Most hand you four numbers and ask which one passes, so a test you can run in your head in a few seconds is worth more than any long division.

Concept 1 of 4: Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25

10, 100 and 1000 are already multiples of 2, 4, 5, 8 and 25, so only the last few digits can spoil divisibility by those. And 10 leaves remainder 1 on division by 3 and by 9, so a number behaves like the sum of its digits there.

Definition

Tests on the last digits: 2, 5 and 10 look at the last digit, 4 and 25 at the last two, 8 at the last three.
Tests on the digit sum: 3 and 9. Every multiple of 9 is a multiple of 3, but not the other way round: a digit sum of 21 means divisible by 3 and not by 9.
11 uses the difference between the sums of alternate digits.
7 has no quick test that beats dividing. One shortcut: take twice the last digit away from the number formed by the other digits, and repeat; the number is divisible by 7 exactly when the result is.
For a 'NOT divisible' question, run the test on every option and pick the one that fails. Run the cheapest test first: a last digit before a digit sum.
When the digits are given only by a rule, write the digit sum in letters. Six consecutive digits starting at $a$ add to $6a + 15 = 3(2a + 5)$ , so such a number is always divisible by 3, in any order.

Divisor	Test	Example
2	last digit is 0, 2, 4, 6 or 8	3574
3	digit sum is divisible by 3	5124: digit sum 12
4	last two digits form a multiple of 4	7316: 16
5	last digit is 0 or 5	6845
7	twice the last digit taken from the rest gives a multiple of 7	2149: $214 - 18 = 196 = 7 \times 28$
8	last three digits form a multiple of 8	41136: $136 = 8 \times 17$ Three digits, not two: 8 is where most slips happen.
9	digit sum is divisible by 9	7236: digit sum 18
10	last digit is 0	4590
11	the two sums of alternate digits differ by 0 or a multiple of 11	2816: $(6 + 8) - (1 + 2) = 11$
25	last two digits are 00, 25, 50 or 75	3475

Each test says whether the number divides exactly. The same digits also give the remainder, which a later concept on this page uses.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 27 Jul 2023, Shift 3 · Q52Easy

Example 1 · Number System · Divisibility Tests

How many of the following numbers are divisible by 3 but NOT by 9? 5826, 5964, 6039, 6336, 6489, 6564, 6867 and 6960

Testing 8 on the last two digits

4 needs the last two digits; 8 needs the last three. 316 ends in 16, a multiple of 8, yet

316 = 8 \times 39 + 4

Treating divisible by 3 as divisible by 9

A digit sum of 12, 21 or 24 gives divisibility by 3 only. For 9 the digit sum itself must be 9, 18, 27 and so on.

Drill 7 more on divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25

Concept 2 of 4: The alternating-sum test for 11

10 is one less than 11, so each place value swings between one more and one less than a multiple of 11. That is why the digits count with alternating signs: plus, minus, plus, minus.

Definition

You met the rule in the table above; here is how to run it without slips.
Number the places from the right: units is place 1, tens place 2, and so on. Add the digits in the odd places and the digits in the even places separately.
Let $D$ = (odd-place sum) $-$ (even-place sum). The number is divisible by 11 exactly when $D$ is 0, 11, $-11$ , 22 and so on.
Fast route: write the signs straight onto the digits from the right, $+\,-\,+\,-$ , and add as you go.
A zero is a digit: it holds its place, so never skip it.
A six-digit number made of a three-digit block written twice, like 458458, equals the block $\times 1001 =$ the block $\times 7 \times 11 \times 13$ , so it is divisible by 7, 11 and 13 at once.

Test for 11

D = (\text{sum of digits in odd places}) - (\text{sum of digits in even places}), \quad 11 \text{ divides } N \text{ exactly when } D = 0, \pm 11, \pm 22, \ldots \qquad \overline{abcabc} = \overline{abc} \times 7 \times 11 \times 13

Dthe alternating difference, places counted from the right
$\overline{abc}$ the three-digit number with digits a, b, c

Worked example

Is 849376 divisible by 11?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 19 Sep 2024, Shift 2 · Q69Easy

Example 2 · Number System · Divisibility Tests

From the given numbers A, B, C and D, which number is NOT divisible by 11? A = 712712 B = 177210 C = 64614 D = 756148

Rejecting a difference of 11

The difference does not have to be 0. A difference of 11,

-11

or 22 also means the number is divisible by 11.

Skipping a zero digit

In 30712 the 0 sits in place 4. Leave it out and every digit after it changes side, and the test gives the wrong verdict.

Drill 4 more on the alternating-sum test for 11

Concept 3 of 4: Testing a composite divisor by its co-prime parts

There is no separate test for 36 or 44. Split the divisor into parts that share no factor, run the test for each part, and the number passes only if it passes every part.

Definition

Two numbers are co-prime when no number other than 1 divides both: 4 and 9, 3 and 8, 9 and 11.
If $m$ and $n$ are co-prime, a number is divisible by $m \times n$ exactly when it is divisible by $m$ and by $n$ .
The parts must be co-prime. 24 is $3 \times 8$ , never $4 \times 6$ : 12 passes both 4 and 6, yet 24 does not divide 12.
A divisor with three parts works the same way: $132 = 3 \times 4 \times 11$ .
When one divisor in a list is a multiple of the others, the others add nothing: 'divisible by 2, 4 and 8' is just divisible by 8.
If a number divides two numbers, it divides their sum and their difference. A multiple of 21 is a multiple of 3, and so is 45, so their sum is a multiple of 3 (but not always of 7).

Common co-prime splits

6 = 2 \times 3 \quad 12 = 3 \times 4 \quad 18 = 2 \times 9 \quad 24 = 3 \times 8 \quad 33 = 3 \times 11 \quad 36 = 4 \times 9 \quad 44 = 4 \times 11 \quad 45 = 5 \times 9 \quad 72 = 8 \times 9 \quad 99 = 9 \times 11

Worked example

Is 61272 divisible by 72?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 12 Sep 2024, Shift 3 · Q53Easy

Example 3 · Number System · Divisibility Tests

Which of the following numbers is divisible by 2, 5, 10 and 11?

Splitting a divisor into parts that share a factor

Testing 4 and 6 does not test 24, because 4 and 6 share the factor 2. Use co-prime parts:

24 = 3 \times 8

Running only one of the two tests

For 36 a number must pass both 4 and 9. 1242 has digit sum 9 but ends in 42, which is not a multiple of 4, so 36 does not divide it.

Drill 11 more on testing a composite divisor by its co-prime parts

Concept 4 of 4: Remainders from the digit tests

A test does more than say yes or no. The digits it reads leave the same remainder as the whole number, so the test also tells you how far the number is from a multiple.

Definition

9 and 3: a number leaves the same remainder as its digit sum.
8, 4, 2, 5 and 10: a number leaves the same remainder as its last three, last two or last digit.
11: the remainder is $D$ = (odd-place sum) $-$ (even-place sum), with places counted from the right. If $D$ is negative, add 11; if it is 11 or more, take 11 away.
Then, as on the division page: the least number to subtract is the remainder, and the least number to add is the divisor minus the remainder.

Remainders from digits

N \div 9 \text{ leaves the remainder of (digit sum)} \div 9 \qquad N \div 11 \text{ leaves the remainder of } D \div 11

Worked example

What is the least number that must be added to 352817 to make it divisible by 11?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 23 Sep 2024, Shift 2 · Q73Easy

Example 4 · Number System · Divisibility Tests

The least number which must be subtracted from 7278745 so as to obtain a sum that is divisible by 11 is:

Reading the 11 difference from the left

For the remainder, count places from the right. From the left, a number with an even count of digits gives the difference with the wrong sign: 4072 read from the left gives 9, but its remainder is 2.

Giving a negative difference as the remainder

A remainder is never negative. A difference of

-9

on division by 11 means a remainder of

-9 + 11 = 2

Drill 3 more on remainders from the digit tests

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The alternating-sum test for 11
Test for 11
$D = (\text{sum of digits in odd places}) - (\text{sum of digits in even places}), \quad 11 \text{ divides } N \text{ exactly when } D = 0, \pm 11, \pm 22, \ldots \qquad \overline{abcabc} = \overline{abc} \times 7 \times 11 \times 13$
Testing a composite divisor by its co-prime parts
Common co-prime splits
$6 = 2 \times 3 \quad 12 = 3 \times 4 \quad 18 = 2 \times 9 \quad 24 = 3 \times 8 \quad 33 = 3 \times 11 \quad 36 = 4 \times 9 \quad 44 = 4 \times 11 \quad 45 = 5 \times 9 \quad 72 = 8 \times 9 \quad 99 = 9 \times 11$
Remainders from the digit tests
Remainders from digits
$N \div 9 \text{ leaves the remainder of (digit sum)} \div 9 \qquad N \div 11 \text{ leaves the remainder of } D \div 11$

Reference tables (1)

Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 2510 rows

Divisor	Test	Example
2	last digit is 0, 2, 4, 6 or 8	3574
3	digit sum is divisible by 3	5124: digit sum 12
4	last two digits form a multiple of 4	7316: 16
5	last digit is 0 or 5	6845
7	twice the last digit taken from the rest gives a multiple of 7	2149: $214 - 18 = 196 = 7 \times 28$
8	last three digits form a multiple of 8	41136: $136 = 8 \times 17$ Three digits, not two: 8 is where most slips happen.
9	digit sum is divisible by 9	7236: digit sum 18
10	last digit is 0	4590
11	the two sums of alternate digits differ by 0 or a multiple of 11	2816: $(6 + 8) - (1 + 2) = 11$
25	last two digits are 00, 25, 50 or 75	3475

Each test says whether the number divides exactly. The same digits also give the remainder, which a later concept on this page uses.

Watch out for (8)

Testing 8 on the last two digits
→ Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25
Treating divisible by 3 as divisible by 9
→ Divisibility tests for 2, 3, 4, 5, 7, 8, 9, 10, 11 and 25
Rejecting a difference of 11
→ The alternating-sum test for 11
Skipping a zero digit
→ The alternating-sum test for 11
Splitting a divisor into parts that share a factor
→ Testing a composite divisor by its co-prime parts
Running only one of the two tests
→ Testing a composite divisor by its co-prime parts
Reading the 11 difference from the left
→ Remainders from the digit tests
Giving a negative difference as the remainder
→ Remainders from the digit tests

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