MathsICSE Class 8 · Playing with Numbers

Divisibility Tests for 2, 3, 5, 9 and 10 with Proofs (ICSE Class 8 Maths)

Updated 17 October 2026 · Fact-checked

A number is divisible by 2 if its ones digit is even, by 5 if the ones digit is 0 or 5, by 10 if the ones digit is 0, by 3 if its digit sum is a multiple of 3, and by 9 if its digit sum is a multiple of 9. The tests for 3 and 9 follow from $100a + 10b + c = 99a + 9b + (a + b + c)$.

Ones digit decides 2, 5 and 10; for 3 and 9 add the digits, because 99a + 9b is always a multiple of 9.

Illustration drawn for QuestionBuddy.

Understand Divisibility Tests for 2, 3, 5, 9 and 10

The CISCE curriculum asks you to deduce the divisibility tests for $2, 3, 5, 9, 10$ for two- and three-digit numbers written in general form.

Tests based on the ones digit. Since $100 a + 10 b$ is a multiple of $10$ , the remainder on division by $2$ , $5$ or $10$ depends only on $c$ .

Divisible by 2: ones digit $0, 2, 4, 6, 8$ .
Divisible by 5: ones digit $0$ or $5$ .
Divisible by 10: ones digit $0$ .

Tests based on the digit sum. Write $100 a + 10 b + c = 99 a + 9 b + (a + b + c)$ . The part $99 a + 9 b$ is a multiple of $9$ (and of $3$ ). So the number is divisible by 9 exactly when $a + b + c$ is, and by 3 exactly when $a + b + c$ is a multiple of $3$ . The same proof works for two digits: $10 a + b = 9 a + (a + b)$ .

Consequences.

A number divisible by $9$ is also divisible by $3$ , but not the other way round: $51$ has digit sum $6$ , so it is divisible by $3$ and not by $9$ .
A number divisible by both $2$ and $5$ ends in $0$ and so is divisible by $10$ .
(Extension) A number divisible by both $2$ and $3$ is divisible by $6$ .

Missing-digit problems. Set the digit sum equal to a multiple of $9$ (or $3$ ), keep the unknown a single digit $0$ to $9$ , and list all solutions.

Divisibility Tests for 2, 3, 5, 9 and 10: formulas to remember

Splitting a three-digit number: $100 a + 10 b + c = 99 a + 9 b + (a + b + c)$; Proves the tests for 3 and 9.
Test for 3 and 9: $3 ∣ N ⟺ 3 ∣ (a + b + c), 9 ∣ N ⟺ 9 ∣ (a + b + c)$; Digit sum test.
Tests on the last digit: $2 : c \in {0, 2, 4, 6, 8}; 5 : c \in {0, 5}; 10 : c = 0$; Only the ones digit matters.

How to solve Divisibility Tests for 2, 3, 5, 9 and 10 questions

Choose the test that matches the divisor, then apply it to the digits.

1For 2, 5 or 10, look only at the ones digit.
2For 3 or 9, add all digits.
3For a missing digit, write the digit sum with the unknown and find which values make it a multiple of 3 or 9.
4Keep only values from $0$ to $9$ (leading digit not $0$ ).
5Check by actual division.

Divisibility Tests for 2, 3, 5, 9 and 10 in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Checking a ledger total

Suppose a three-digit invoice total ends up as 4_8 because one digit smudged, and the shopkeeper says the total is a multiple of 9.

Find the smudged digit.

Try it first, then show the solutionSolution

Digit sum $= 4 + x + 8 = 12 + x$ must be $18$ (the only multiple of $9$ between $12$ and $21$ ).
$x = 6$ ; the total is $468 = 9 \times 52$ .

Answer: 6 (the total is 468).

Memory hook: Digits add up to a multiple of 9, then it divides by 9.

Tricks to remember Divisibility Tests for 2, 3, 5, 9 and 10

Short memory aids. Say each one aloud once, then use it on a question the same day.

Last digit for 2, 5, 10; sum for 3, 9
Three tests ask only the ones digit; two tests ask for the digit sum.
Use it when: Choosing a test.
9 implies 3
Every multiple of 9 is a multiple of 3, never the reverse.
Use it when: Avoiding false inferences.

Divisibility Tests for 2, 3, 5, 9 and 10: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Pick the multiple of 9 · 1 mark · MCQBoard level

Which number is divisible by 9?
(a) 576 (b) 475 (c) 823 (d) 692

Try it first, then show the solutionSolution

Digit sums: $18, 16, 13, 17$ . Only $5 + 7 + 6 = 18$ is a multiple of $9$ .

Answer: (a)

Example 2: Testing one number against all five · 3 marks · Short answerBoard level

Test whether 315 is divisible by 2, 3, 5, 9 and 10.

Try it first, then show the solutionSolution

By 2: ones digit $5$ is odd, no.
By 3 and 9: digit sum $3 + 1 + 5 = 9$ , yes to both.
By 5: ones digit $5$ , yes. By 10: ones digit not $0$ , no.

Answer: Divisible by 3, 5 and 9; not by 2 or 10.

Example 3: Missing digit · 3 marks · Short answerHarder board level

If the digit $x$ makes $6 x 2$ divisible by 9, find $x$ . For which digits is $6 x 2$ divisible by 3?

Try it first, then show the solutionSolution

Digit sum $= 8 + x$ . For $9$ : $8 + x = 9$ gives $x = 1$ ( $8 + x = 18$ would need $x = 10$ ).
For $3$ : $8 + x \in {9, 12, 15, 18}$ , so $x = 1, 4, 7$ .
Check: $612 = 9 \times 68$ .

Answer: $x = 1$ for 9; $x = 1, 4, 7$ for 3.

Example 4: Two conditions at once · 4 marks · Short answerHarder board level

Find the digits $A$ such that the number $3 A 5$ is divisible by both 3 and 5. For which $A$ is it also divisible by 9?

Try it first, then show the solutionSolution

Ones digit $5$ , so divisible by $5$ always.
By 3: $8 + A$ a multiple of $3$ gives $A = 1, 4, 7$ .
By 9: $8 + A = 9$ gives $A = 1$ , number $315$ .

Answer: $A = 1, 4, 7$ ; with 9 as well, $A = 1$ (315).

Example 5: A divisibility proof · 4 marks · Long answerHOTS / competency

Prove that $\overline{ab c} - (a + b + c)$ is always divisible by 9, and check it for 475.

Try it first, then show the solutionSolution

$100 a + 10 b + c - a - b - c = 99 a + 9 b = 9 (11 a + b)$ .
So it is a multiple of $9$ for every digit choice.
For $475$ : $475 - 16 = 459$ ; and $11 \times 4 + 7 = 51$ , $9 \times 51 = 459$ .

Answer: $9 (11 a + b)$ ; for 475 the value is $459 = 9 \times 51$ .

Where students lose marks in Divisibility Tests for 2, 3, 5, 9 and 10

The slips examiners see most often, and what to write instead.

Mistake: Concluding divisibility by 9 from divisibility by 3
Every multiple of 9 is a multiple of 3, not vice versa.
Wrong working
$51$ : digit sum $6$ , divisible by $3$ , so divisible by $9$ .
Right working
Digit sum $6$ is not a multiple of $9$ , so $51$ is not divisible by $9$ .
Do this: Test each divisor separately.
Usually costs: The final conclusion.
Mistake: Testing divisibility by 3 with the ones digit
Only the digit sum works for 3.
Wrong working
$"$ 123 ends in 3, so divisible by 3. $"$
Right working
$1 + 2 + 3 = 6$ is a multiple of $3$ ; but $133$ (digit sum $7$ ) is not divisible by $3$ .
Do this: Use ones digit only for 2, 5, 10.
Usually costs: The test mark.

Exam tips

Always show the digit-sum addition.
In missing-digit questions, list all solutions unless told "smallest".
A proof marks for the general form line; do not skip it.

Practice questions on Divisibility Tests for 2, 3, 5, 9 and 10

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A number divisible by both 2 and 5 is also divisible by:
(a) 4 (b) 7 (c) 10 (d) 3
Hint
What must the ones digit be?
Answer
(c)
Q2 · 2 marks · Short answerBoard level
Test 738 for divisibility by 2, 3, 5, 9 and 10.
Hint
Last digit, then digit sum.
Answer
Ones digit $8$ : divisible by $2$ ; not by $5$ or $10$ . Digit sum $18$ : divisible by $3$ and $9$ .
Q3 · 3 marks · Short answerHarder board level
Find the digit $x$ if $4 x 8$ is divisible by 9, and write the number.
Hint
Digit sum 12 + x.
Answer
$12 + x = 18$ , so $x = 6$ ; the number is $468 = 9 \times 52$ .
Q4 · 3 marks · Short answerHarder board level
How many digits $d$ make $5 d 4$ divisible by 3? For which of them is it also divisible by 9?
Hint
Digit sum is 9 + d.
Answer
$9 + d$ multiple of $3$ gives $d = 0, 3, 6, 9$ (four digits). Multiple of $9$ : $d = 0$ or $9$ ( $504$ and $594$ ).
Q5 · 4 marks · Long answerHOTS / competency
A palindrome of the form $\overline{x y x}$ equals $101 x + 10 y$ . If $x = 4$ and the palindrome is divisible by 9, find $y$ and verify.
Hint
Digit sum is 2x + y.
Answer
$414 = 101 \times 4 + 10 \times 1$ . Digit sum $4 + y + 4 = 8 + y = 9$ gives $y = 1$ ; $414 = 9 \times 46$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Test for divisibility by 9?Show answer
The digit sum is a multiple of 9.
2.Test for divisibility by 5?Show answer
The ones digit is 0 or 5.
3.Is 51 divisible by 9?Show answer
No (digit sum 6).
4.Which numbers are divisible by both 2 and 5?Show answer
Those ending in 0.

Divisibility Tests for 2, 3, 5, 9 and 10: frequently asked questions

Why does the digit-sum test work?

Because 10^k - 1 is a multiple of 9, so each place value leaves remainder equal to the digit itself.

Does the digit-sum test work for 4?

No; for 4 you need the last two digits. It is not in the Class VIII list.