MathsCBSE Class 8 · Number Play

Divisibility Tests, Digital Roots and Cryptarithms for Class 8

Updated 17 October 2026 · Fact-checked

A number is divisible by 9 (or 3) if its digit sum is; its remainder on division by 9 equals its digital root (with 9 for 0). A number is divisible by 11 if the alternating sum of its digits, from the units digit, is a multiple of 11. For a composite divisor, test coprime factors (3 and 8 for 24). Cryptarithms are solved with unit-digit, size and divisibility reasoning.

For 9, add all digits of 7309; for 11, alternate signs starting with plus from the units digit; each result gives the remainder.

Illustration drawn for QuestionBuddy.

Understand Divisibility Tests, Digital Roots and Cryptarithms

Tests for 2, 5, 10, 4 and 8 (revised from earlier classes) use only the last few digits. Write a number as $\dots + 1000 d + 100 c + 10 b + a$ . Every place value except the units is a multiple of 10, so a number is divisible by 10 if $a = 0$ , by 5 if $a$ is 0 or 5, and by 2 if $a$ is even. Every place value from the hundreds up is a multiple of 4, so divisibility by 4 depends on the last two digits; from the thousands up each place value is a multiple of 8, so divisibility by 8 depends on the last three digits.

Test for 9. Each place value is one more than a multiple of 9: $10 = 9 + 1$ , $100 = 99 + 1$ , $1000 = 999 + 1$ . So the number $7309 = 7 (999 + 1) + 3 (99 + 1) + 0 (9 + 1) + 9$ equals (multiples of 9) $+ (7 + 3 + 0 + 9)$ . Therefore a number leaves the same remainder as its digit sum when divided by 9, and is divisible by 9 if and only if its digit sum is. Repeating the digit sum until one digit remains gives the digital root, which is the remainder on division by 9 (a root of 9 means divisible by 9, and for a positive number a multiple of 9 has digital root 9). For example 427 leaves remainder $4 + 2 + 7 = 13 \to 4$ .

Test for 3. Since $10, 100, \dots$ leave remainder 1 on division by 3 too, a number is divisible by 3 if and only if its digit sum is divisible by 3.

Test for 11. Place values alternate: $1 = 0 + 1$ , $10 = 11 - 1$ , $100 = 99 + 1$ , $1000 = 1001 - 1$ , and so on. So the units, hundreds, ten-thousands, ... positions are "1 more than a multiple of 11" and the tens, thousands, ... positions are "1 less". Compute (sum of digits in the odd positions from the right) $-$ (sum of digits in the even positions). If this difference is a multiple of 11 (including 0), the number is divisible by 11; otherwise it gives the remainder after adding or subtracting multiples of 11. For 3,28,105: $5 - 0 + 1 - 8 + 2 - 3 = - 3$ , so the number is 3 short of a multiple of 11, which is the same as 8 more than a multiple of 11.

Composite divisors. To test divisibility by 6, test 2 and 3 (coprime). To test 24, test 3 and 8, not 4 and 6, because 4 and 6 share the factor 2: 12 is divisible by both 4 and 6 but not by 24. Rule: split the divisor into factors with no common prime factor.

Digital roots in history. The chapter notes that Aryabhata II (c. 950 CE), in the Mahasiddhanta, mentions the method of repeatedly adding digits; the method was used to check arithmetic.

Cryptarithms. In these puzzles each letter is one digit, different letters stand for different digits, and the first digit of a number is never 0. To solve multiplication cryptarithms use: the unit digit of the product depends only on the unit digits of the factors; size (a three-digit answer limits the factors); and divisibility facts.

Parity, remainders and the properties of divisibility are on the previous topic page of this chapter.

Divisibility Tests, Digital Roots and Cryptarithms: formulas to remember

Divisibility by 9 and 3: $9 ∣ N ⟺ 9 ∣ (digit sum); 3 ∣ N ⟺ 3 ∣ (digit sum)$; N and its digit sum leave the same remainder on division by 9.
Digital root: $digital root = N mod 9 (9 instead of 0)$; For a positive number.
Test for 11: $11 ∣ N ⟺ 11 ∣ (a - b + c - d + \dots)$; a is the units digit; alternate signs moving left.
Powers of 10 and 9: $1 0^{n} = (1 0^{n} - 1) + 1, 1 0^{n} - 1 = 99 \dots 9$; Why the remainder is the digit.
Composite divisor test: $mn ∣ N ⟺ m ∣ N and n ∣ N (m, n coprime)$; For 24 use 3 and 8.

How to solve Divisibility Tests, Digital Roots and Cryptarithms questions

Use the tests and reasoning in this order.

1Decide which test fits: units digit (2, 5, 10), last two or three digits (4, 8), digit sum (3, 9), alternating digit sum (11).
2For a composite divisor, split it into coprime factors and test each.
3For a missing digit, write the digit sum with the unknown and list the values that make it a multiple of the divisor.
4For two conditions (for example 18 = 2 x 9), combine them and list every pair that satisfies both.
5For cryptarithms, start with the unit digit, then the size of the product, then test cases.
6Check every answer by dividing or multiplying.

Divisibility Tests, Digital Roots and Cryptarithms 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.

Digital root as a check on arithmetic

Suppose a shopkeeper multiplies 347 by 26 and gets 9022 and wants a quick check, as in the old method of digital roots.

Use digital roots to check the product.

Try it first, then show the solutionSolution

Digital root of 347: $3 + 4 + 7 = 14 \to 1 + 4 = 5$ . Digital root of 26: $2 + 6 = 8$ .
Product of the roots: $5 \times 8 = 40 \to 4 + 0 = 4$ .
Digital root of 9022: $9 + 0 + 2 + 2 = 13 \to 4$ . The two roots agree, so the answer passes the check (and indeed $347 \times 26 = 9022$ ).

Answer: The check agrees: both digital roots are 4.

Memory hook: A matching digital root does not prove the answer, but a mismatch proves an error.

Test for 11

Imagine a ticket number 4,95,627 that must be divisible by 11 for a lucky draw.

Is the ticket number divisible by 11?

Try it first, then show the solutionSolution

From the units digit: $7 - 2 + 6 - 5 + 9 - 4 = 11$ .
11 is a multiple of 11, so the number is divisible by 11: $495627 = 11 \times 45057$ .

Answer: Yes, divisible by 11

Memory hook: Alternate the signs starting from the units digit.

Tricks to remember Divisibility Tests, Digital Roots and Cryptarithms

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

Add the digits for 3 and 9
A digit sum divisible by 3 means the number is; divisible by 9 means the number is.
Use it when: Fast tests and missing digit problems.
Plus, minus, plus, minus for 11
Start from the units digit and alternate signs: +, -, +, -. A result of 0 or a multiple of 11 means divisible.
Use it when: Divisibility by 11 and remainders.
Coprime halves for composite divisors
24 = 3 x 8 (coprime), not 4 x 6. Choose factors with no shared prime.
Use it when: Testing divisibility by 6, 12, 15, 18, 24, 44.
Unit digit first in cryptarithms
Look at what the unit digit of the product must be, then decide the size.
Use it when: Multiplication puzzles.

Divisibility Tests, Digital Roots and Cryptarithms: solved examples

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

Example 1: Divisible by 11 · 1 mark · MCQBoard level

Which number is divisible by 11? (a) 462 (b) 463 (c) 472 (d) 452

Try it first, then show the solutionSolution

Alternating sums from the units digit: 462 gives $2 - 6 + 4 = 0$ ; 463 gives $3 - 6 + 4 = 1$ ; 472 gives $2 - 7 + 4 = - 1$ ; 452 gives $2 - 5 + 4 = 1$ .
Only 462 gives a multiple of 11 (0). Check: $462 = 11 \times 42$ .

Answer: (a) 462

Example 2: Digital root and remainder · 2 marks · Short answerBoard level

Find the digital root of 58426 and the remainder when it is divided by 9.

Try it first, then show the solutionSolution

$5 + 8 + 4 + 2 + 6 = 25$ , and $2 + 5 = 7$ , so the digital root is 7. (1 mark)
The remainder on division by 9 is 7. Check: $58426 = 9 \times 6491 + 7$ . (1 mark)

Answer: Digital root 7; remainder 7

Example 3: A missing digit · 2 marks · Short answerBoard level

If the number 6k27 is divisible by 9, find the digit k. Verify by dividing.

Try it first, then show the solutionSolution

Digit sum $= 6 + k + 2 + 7 = 15 + k$ . For this to be a multiple of 9, $15 + k = 18$ (since $k$ is a digit, $15 + k \leq 24$ ). So $k = 3$ . (1 mark)
Check: $6327 = 9 \times 703$ . (1 mark)

Answer: k = 3

Example 4: Remainder on division by 11 · 3 marks · Short answerHarder board level

Find the remainder when 90,817 is divided by 11, using the alternating digit sum.

Try it first, then show the solutionSolution

From the units digit: $7 - 1 + 8 - 0 + 9 = 23$ . (1 mark)
$23 = 2 \times 11 + 1$ , so the number is 1 more than a multiple of 11. (1 mark)
Check: $11 \times 8256 = 90816$ , so $90817 - 90816 = 1$ . (1 mark)

Answer: Remainder 1

Example 5: Two conditions together · 4 marks · Long answerHarder board level

The five-digit number 5a37b is a multiple of 18. List all possible pairs (a, b) of digits.

Try it first, then show the solutionSolution

$18 = 2 \times 9$ , with 2 and 9 coprime. So $b$ must be even, and $5 + a + 3 + 7 + b = 15 + a + b$ must be a multiple of 9. (1 mark)
$15 + a + b = 18$ or $27$ (36 is too large), so $a + b = 3$ or $12$ . (1 mark)
$a + b = 3$ with $b$ even: $b = 0, a = 3$ and $b = 2, a = 1$ . $a + b = 12$ with $b$ even: $b = 4, a = 8$ ; $b = 6, a = 6$ ; $b = 8, a = 4$ ( $b = 2$ would need $a = 10$ ). (1 mark)
Pairs: (3, 0), (1, 2), (8, 4), (6, 6), (4, 8). Check one: $53370 = 18 \times 2965$ . (1 mark)

Answer: (a, b) = (3, 0), (1, 2), (8, 4), (6, 6), (4, 8)

Where students lose marks in Divisibility Tests, Digital Roots and Cryptarithms

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

Mistake: Using the tests for 4 and 6 to decide divisibility by 24
4 and 6 share the prime 2, so numbers divisible by both need not be divisible by 24.
Wrong working
"36 is divisible by 4 and 6, so by 24."
Right working
Test 3 and 8 (coprime). $36 \div 24$ leaves remainder 12.
Do this: Split the divisor into coprime factors.
Usually costs: All marks of the part.
Mistake: Starting the alternating sum from the left
The signs must start from the units digit for the remainder to be correct.
Wrong working
"3,28,105: 3 - 2 + 8 - 1 + 0 - 5 = 3, so the remainder is 3."
Right working
Start from the units digit: $5 - 0 + 1 - 8 + 2 - 3 = - 3$ , so the number is 3 short of a multiple of 11, and the remainder is 8.
Do this: Number the digits from the right.
Usually costs: The remainder mark.
Mistake: Writing the digital root of a multiple of 9 as 0
Repeated digit addition of a positive multiple of 9 ends at 9, not 0.
Wrong working
"Digital root of 459 is 0."
Right working
$4 + 5 + 9 = 18$ and $1 + 8 = 9$ ; the digital root is 9.
Do this: Say "9 means divisible by 9".
Usually costs: 1 mark.
Mistake: Missing pairs in a two-condition digit problem
Both conditions give several digits; you must list every combination.
Wrong working
"a + b = 3, so (3, 0) only."
Right working
$a + b = 3$ with $b$ even gives $(3, 0)$ and $(1, 2)$ ; and $a + b = 12$ also works.
Do this: List the target digit sums first, then every pair.
Usually costs: 1 to 2 marks.

Exam tips

Write the digit sum with the unknown first (for example 15 + k) and say which multiples of 9 are possible.
For two conditions, say which factor each condition comes from (2 gives an even last digit, 9 gives the digit sum).
Show an alternating sum on one line, signs written above the digits.
In cryptarithms, state what each deduction rules out; examiners reward the elimination.
Verify the final number by actual division or multiplication.

Practice questions on Divisibility Tests, Digital Roots and Cryptarithms

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 · Assertion-reasonBoard level
Assertion (A): Every number whose digits add up to 12 is divisible by 3.
Reason (R): A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is 12 divisible by 3?
Answer
(a). 12 is divisible by 3, so by the test of divisibility by 3 the number is too. R is the test that justifies A.
Q2 · 3 marks · Short answerHarder board level
Use the tests for 3 and 8 to decide whether 7368 is divisible by 24. Why would testing 4 and 6 not be enough? Give a counter-example.
Hint
Check the digit sum and the last three digits.
Answer
Digit sum $7 + 3 + 6 + 8 = 24$ , divisible by 3. Last three digits 368: $368 = 8 \times 46$ , so divisible by 8. 3 and 8 are coprime, so $24 ∣ 7368$ (indeed $7368 = 24 \times 307$ ). Testing 4 and 6 is not enough because they share the factor 2: 36 is divisible by both but not by 24 (remainder 12).
Q3 · 3 marks · Short answerHarder board level
The digital root of a number N is 7. Find the digital roots of (i) N + 20 (ii) 4N (iii) N squared. Explain why each is correct.
Hint
Digital roots behave like remainders on division by 9.
Answer
N leaves remainder 7 on division by 9. (i) $N + 20$ leaves $7 + 2 = 9$ , i.e. a multiple of 9, so the digital root is 9. (ii) $4 N$ leaves $28 = 3 \times 9 + 1$ , so the digital root is 1. (iii) $N^{2}$ leaves $49 = 5 \times 9 + 4$ , so the digital root is 4.
Q4 · 4 marks · Long answerHOTS / competency
A three-digit number abc has digits a, b, c, and its reverse is cba. Show that abc - cba is divisible by both 9 and 11. Verify with 853, and explain why a number and its digit-reversal have the same remainder on division by 9.
Hint
Write abc = 100a + 10b + c.
Answer
$ab c - c ba = (100 a + 10 b + c) - (100 c + 10 b + a) = 99 a - 99 c = 99 (a - c)$ . Since $99 = 9 \times 11$ , the difference is divisible by 9 and 11. For 853: $853 - 358 = 495 = 99 \times 5$ , and $495 = 9 \times 55 = 11 \times 45$ . A number and its reversal have the same digits, so the same digit sum, so the same remainder on division by 9.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: in the cryptarithm AB x B = CAB each letter is a different digit and the first digit of a number is not 0.
(i) Show that B must be 5. (2 marks)
(ii) Find all possible values of A and C. (2 marks)
Hint
The unit digit of B x B must be B. Write the equation with A, B and C.
Answer
(i) The unit digit of $B \times B$ must be $B$ , so $B \in {0, 1, 5, 6}$ . $B = 0$ gives a product 0; $B = 1$ gives $A B \times 1 = A B$ , which has only two digits; $B = 6$ : $(10 A + 6) \times 6 = 100 C + 10 A + 6$ gives $50 A + 30 = 100 C$ , i.e. $5 A + 3 = 10 C$ , impossible because the left side ends in 3 or 8. So $B = 5$ . (ii) $(10 A + 5) \times 5 = 100 C + 10 A + 5$ gives $40 A + 20 = 100 C$ , so $2 A + 1 = 5 C$ . $C = 1$ : $A = 2$ (25 x 5 = 125); $C = 3$ : $A = 7$ (75 x 5 = 375); $C = 5$ gives $A = 12$ , not a digit, and $C = 5$ would also repeat the digit 5. Two solutions: 25 x 5 = 125 and 75 x 5 = 375.

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.What is the test for divisibility by 9?Show answer
The digit sum is divisible by 9; the remainder equals the digital root.
2.How do you test for 11?Show answer
Alternately add and subtract the digits from the units digit; a multiple of 11 (including 0) means divisible.
3.Why is 24 tested with 3 and 8?Show answer
They are coprime; 4 and 6 share the factor 2.
4.What is the digital root of a positive multiple of 9?Show answer
9.
5.Who is mentioned in the chapter for using digital roots, and when?Show answer
Aryabhata II (c. 950 CE), in the Mahasiddhanta.
6.What do cryptarithm rules say about the first digit?Show answer
The first digit of a number is never 0, and different letters are different digits.

Divisibility Tests, Digital Roots and Cryptarithms: frequently asked questions

Why does adding the digits work for 9 but not for 8?

Because every power of 10 leaves remainder 1 on division by 9 (10 = 9 + 1, 100 = 99 + 1), but powers of 10 leave different remainders on division by 8, so each digit needs a weight.

Does the test for 3 work like the test for 9?

Yes. Each of 10, 100, 1000, ... is one more than a multiple of 9 (9, 99, 999, ...), which is also a multiple of 3, so each leaves remainder 1 on division by 3 and the digit sum decides divisibility by 3.

Is a matching digital root proof that an addition is correct?

No. It only catches errors that change the remainder on division by 9; the chapter describes it as a check.

Are there tests for divisibility by every number up to 100?

The chapter says there are shortcuts for every number until 100 and for some beyond, and that you will understand them after learning more concepts in higher grades.