MathsCBSE Class 8 · Chapter 5

Number Play Class 8: Parity, Remainders, Divisibility Shortcuts and Cryptarithms

Updated 15 October 2026 · Fact-checked

Number Play is Chapter 5 of Ganita Prakash Part I for Class 8. It uses algebra to explain number facts: why parity stays fixed, when sums are multiples, how remainders combine, why the divisibility shortcuts for 2, 3, 4, 5, 8, 9, 10 and 11 work, what a digital root is, and how to solve cryptarithms by reasoning.

What Number Play is about

This chapter is about why number tricks work. Instead of memorising rules, you write numbers in general form (like $2 n$ , $4 p + 2$ or $1000 d + 100 c + 10 b + a$ ) and use algebra to prove what always, sometimes or never happens.

5.1 Is This a Multiple Of?: sums of consecutive numbers; why every expression $a \pm b \pm c \pm d$ has the same parity; expressions that are always even; adding even numbers to get multiples of 4; 'always, sometimes or never true' statements about factors and multiples; numbers with a given remainder, such as $5 k + 3$ .
5.2 Checking Divisibility Quickly: shortcuts for 10, 5, 2, 4 and 8 from place value; the digit-sum test for 9 and 3 and why it works ( $10 = 9 + 1$ , $100 = 99 + 1$ ); the alternating-sum test for 11; combining tests for 6 and 24; digital roots, already known to Aryabhata II (about 950 CE).
5.3 Digits in Disguise: cryptarithms, where letters stand for digits, solved by logical elimination.

School exams test the shortcuts, remainder reasoning with algebra, always/sometimes/never statements with examples, and short cryptarithms. This page explains each idea with an example and gives exam-level and harder questions with full solutions.

Key concepts

Parity of sums and differences

NCERT section 5.1 Is This a Multiple Of?

Parity means whether a number is even or odd. The rules:

odd ± odd = even; even ± even = even; odd ± even = odd.
So $a + b$ and $a - b$ always have the same parity, because they differ by $2 b$ , an even number.

Four numbers with signs: take any four integers $a, b, c, d$ and place '+' or '−' between them in all 8 ways ( $a \pm b \pm c \pm d$ ). Switching one sign changes the value by twice that number (for example $(a + b - c - d) - (a - b - c - d) = 2 b$ ), which is even. So all 8 results have the same parity. With 4 consecutive numbers (two odd, two even) they are all even.

Expressions that are always even: an expression is always even if 2 is a factor of it for all integer values, for example $4 m + 2 q = 2 (2 m + q)$ , $2 u - 4 v$ , $6 m - 4 n$ . An expression like $x^{2} + 2$ is even only when $x$ is even, so it is not always even; give an example ( $x = 6$ gives 38) and a non-example ( $x = 3$ gives 11).

Worked example

Without evaluating, say whether $7 + 12 - 9 + 20$ and $7 - 12 - 9 - 20$ are even or odd. Then decide whether $3 g + 5 h$ is always even.

Try it first, then show the solutionSolution

Both are of the form $a \pm b \pm c \pm d$ with the same numbers, so they have the same parity.
Parity of $7 + 12 + 9 + 20$ : two odd numbers (7, 9) and two even, so the sum is even. Both expressions are even (indeed 30 and $- 34$ ).
$3 g + 5 h$ : with $g = 1, h = 0$ it is 3 (odd), with $g = 1, h = 1$ it is 8 (even). So it is not always even.

Answer: Both even; $3 g + 5 h$ is only sometimes even.

Sums of consecutive numbers

NCERT section 5.1 Is This a Multiple Of?

Consecutive numbers can be written as $n, n + 1, n + 2, \dots$ or, more neatly, centred on a middle value.

Three consecutive numbers: $(n - 1) + n + (n + 1) = 3 n$ , always a multiple of 3, equal to 3 times the middle number.
Five consecutive numbers: $5 n$ , always a multiple of 5.
Four consecutive numbers: $n + (n + 1) + (n + 2) + (n + 3) = 4 n + 6$ , which is even but never a multiple of 4 (it leaves remainder 2).
Two consecutive numbers: $2 n + 1$ , so every odd number above 1 is a sum of two consecutive numbers.

To find consecutive numbers with a given sum, write the general form, set it equal to the sum and solve. If $p$ is the greatest of five consecutive numbers, the others are $p - 1, p - 2, p - 3, p - 4$ .

Worked example

Five consecutive numbers add up to 135. Find them. Is the sum of any five consecutive numbers always a multiple of 5?

Try it first, then show the solutionSolution

Write them as $n - 2, n - 1, n, n + 1, n + 2$ ; their sum is $5 n$ .
$5 n = 135$ gives $n = 27$ .
The numbers are 25, 26, 27, 28, 29.
Since the sum is always $5 n$ , it is always a multiple of 5.

Answer: 25, 26, 27, 28, 29; yes, the sum is always $5 n$ .

Multiples: pairs to make fours, and always/sometimes/never

NCERT section 5.1 Is This a Multiple Of?

Even numbers are of two kinds when divided by 4: multiples of 4 ( $4 p$ ) and those leaving remainder 2 ( $4 p + 2$ ).

$4 p + 4 q = 4 (p + q)$ : a multiple of 4.
$(4 p + 2) + (4 q + 2) = 4 (p + q + 1)$ : also a multiple of 4, because the two 2s make a 4.
$4 p + (4 q + 2) = 4 (p + q) + 2$ : not a multiple of 4.

General facts about divisibility:

If $a$ divides $M$ and $N$ , then $a$ divides $M + N$ and $M - N$ .
If $k$ divides $A$ , then $k$ divides every multiple of $A$ , and every factor of $k$ divides $A$ .
If $A$ is divisible by both $k$ and $m$ , then $A$ is divisible by $LCM (k, m)$ . So divisible by 9 and 4 means divisible by 36 (always), but divisible by 6 and 4 only guarantees 12, not 24 (12 itself is a counterexample).

Always, sometimes or never: prove 'always' with algebra; show 'sometimes' with one example and one non-example; prove 'never' by showing it leads to something impossible (an even number equal to an odd one).

Worked example

Decide always, sometimes or never: (i) the sum of three odd numbers is even; (ii) a number divisible by both 6 and 10 is divisible by 60; (iii) the product of two consecutive numbers is even.

Try it first, then show the solutionSolution

(i) odd + odd = even, then even + odd = odd. So the sum is always odd: never true.
(ii) LCM(6, 10) = 30, so the number must be a multiple of 30. 60 works, but 30 is divisible by 6 and 10 and not by 60: sometimes true.
(iii) Of two consecutive numbers one is even, so the product has a factor 2: always true.

Answer: (i) Never (ii) Sometimes (iii) Always.

Remainders in algebra

NCERT section 5.1 Is This a Multiple Of?

A number that leaves remainder $r$ when divided by $d$ can be written $d k + r$ (with $0 \leq r < d$ ). For remainder 3 on division by 5: $5 k + 3$ gives 3, 8, 13, 18, ...; the same numbers can be written $5 k - 2$ (for $k \geq 1$ ), since they are 2 less than multiples of 5.

Combining remainders: if $A = 7 a + 3$ and $B = 7 b + 5$ , then

$A + B = 7 (a + b) + 8 = 7 (a + b + 1) + 1$ : remainder 1;
$A - B = 7 (a - b) - 2$ : 2 less than a multiple of 7, so remainder 5 (when $A > B$ );
$2 A + B = 7 (2 a + b) + 11 = 7 (2 a + b + 1) + 4$ : remainder 4.

So you can find the remainder of a sum or difference from the remainders alone.

Several conditions at once: a number that leaves remainder 2 when divided by 3 and also by 4 is 2 more than a multiple of both, so it is $12 k + 2$ . A number leaving remainders 3, 4, 5 on division by 4, 5, 6 is 1 less than a multiple of 4, 5 and 6, so it is $60 k - 1$ .

Worked example

A leaves remainder 3 and B leaves remainder 5 when divided by 7. Find the remainders of $A + B$ , $A - B$ (with $A > B$ ) and $2 A + B$ . Check with $A = 24$ , $B = 12$ .

Try it first, then show the solutionSolution

$A + B = 7 (a + b) + 8$ , and $8 = 7 + 1$ : remainder 1. Check: $36 = 35 + 1$ .
$A - B = 7 (a - b) - 2$ , which is 5 more than a multiple of 7: remainder 5. Check: $12 = 7 + 5$ .
$2 A + B = 7 (2 a + b) + 11$ , and $11 = 7 + 4$ : remainder 4. Check: $60 = 56 + 4$ .

Answer: Remainders 1, 5 and 4.

Shortcuts for 10, 5, 2, 4 and 8 from place value

NCERT section 5.2 Checking Divisibility Quickly

Any number $\dots d c ba$ (digits) equals $\dots + 1000 d + 100 c + 10 b + a$ .

10, 5 and 2: every place value except the units is a multiple of 10 (hence of 5 and 2). So divisibility by 10, 5 or 2 depends only on the units digit $a$ : 0 for 10; 0 or 5 for 5; even for 2.
4: 100, 1000, ... are multiples of 4, so only $10 b + a$ , the number formed by the last two digits, matters.
8: 1000, 10000, ... are multiples of 8, so only the number formed by the last three digits matters.

Example: 7,29,624: the last three digits 624 $= 8 \times 78$ , so the number is divisible by 8; 24 is divisible by 4, so it is divisible by 4.

Worked example

Using algebra, explain why a number is divisible by 4 when its last two digits form a multiple of 4. Then test 5,37,316.

Try it first, then show the solutionSolution

Write the number as $100 \times (the rest) + (last two digits)$ .
$100 = 4 \times 25$ , so the first part is always a multiple of 4.
So the whole number is a multiple of 4 exactly when the last two digits form one.
5,37,316: last two digits 16 $= 4 \times 4$ , so it is divisible by 4.

Answer: Because 100 is a multiple of 4; 5,37,316 is divisible by 4.

Divisibility by 9 and 3: digit sums

NCERT section 5.2 Checking Divisibility Quickly

Every place value is 1 more than a multiple of 9: $10 = 9 + 1$ , $100 = 99 + 1$ , $1000 = 999 + 1$ . So a digit $c$ in the hundreds place contributes $99 c + c$ : a multiple of 9 plus $c$ . Adding over all places:

a number and the sum of its digits leave the same remainder when divided by 9.

So a number is divisible by 9 if and only if its digit sum is divisible by 9, and you can keep adding digits until one digit remains; that digit is the remainder (with 9 meaning remainder 0).

Example: $7309$ : $7 + 3 + 0 + 9 = 19$ , $1 + 9 = 10$ , $1 + 0 = 1$ : remainder 1.

Since $9 = 3 \times 3$ , the same argument works for 3 ( $10 = 9 + 1$ is also 1 more than a multiple of 3): a number is divisible by 3 if and only if its digit sum is.

Divisibility by 6: a number is divisible by 6 exactly when it is divisible by both 2 and 3 (because 2 and 3 have no common factor). For 24, test 3 and 8, not 4 and 6: 12 passes 4 and 6 but is not a multiple of 24.

Worked example

Find the remainder when 58,246 is divided by 9 and by 3, without dividing.

Try it first, then show the solutionSolution

Digit sum: $5 + 8 + 2 + 4 + 6 = 25$ ; then $2 + 5 = 7$ .
So 58,246 leaves remainder 7 when divided by 9.
For 3: 25 leaves remainder 1 when divided by 3, so 58,246 leaves remainder 1.

Answer: Remainder 7 (by 9) and 1 (by 3).

Divisibility by 11: alternating sums

NCERT section 5.2 Checking Divisibility Quickly

Place values alternate between 1 more and 1 less than multiples of 11:

$1 = 0 + 1$ , $10 = 11 - 1$ , $100 = 99 + 1$ , $1000 = 1001 - 1$ , $10000 = 9999 + 1$ , ...

So digits in the units, hundreds, ten-thousands places count as excess, and digits in the tens, thousands, lakhs places count as shortfall.

Shortcut: starting from the units digit, add and subtract the digits alternately (+units − tens + hundreds − ...). The result is the remainder on division by 11 (add 11 if it is negative); the number is divisible by 11 exactly when this alternating sum is 0 or a multiple of 11.

Example: 3,28,105: $5 - 0 + 1 - 8 + 2 - 3 = - 3$ , so it is 3 short of a multiple of 11, that is, remainder 8.

Nice consequence: every 4-digit palindrome $abba$ equals $1001 a + 110 b = 11 (91 a + 10 b)$ , so it is always divisible by 11.

Worked example

Is 9,07,192 divisible by 11? Use the alternating sum.

Try it first, then show the solutionSolution

Digits from the units: 2, 9, 1, 7, 0, 9.
$2 - 9 + 1 - 7 + 0 - 9 = - 22$ .
$- 22$ is a multiple of 11, so the number is divisible by 11.

Answer: Yes, 9,07,192 is divisible by 11.

Digital roots and cryptarithms

NCERT section 5.3 Digits in Disguise

Digital root: add the digits of a number again and again until one digit is left. For 4,89,710: $29 \to 11 \to 2$ . The digital root equals the remainder on division by 9, except that multiples of 9 have digital root 9. Aryabhata II's *Mahasiddhanta* (about 950 CE) describes this, and it was used to check calculations: the digital root of a product equals the digital root of the product of the digital roots.

Useful facts: adding 9 does not change the digital root; the digital roots of consecutive numbers cycle 1, 2, ..., 9, 1, 2, ...

Cryptarithms are puzzles where each letter stands for a different digit and no number starts with 0. Solve them by reasoning about size, units digits and carries, not by guessing blindly.

Example idea: in $P Q \times 8 = R S$ (two-digit answer), PQ must be at most 12, because $13 \times 8 = 104$ has three digits; 10 and 11 fail because letters must be different digits, leaving $12 \times 8 = 96$ .

Worked example

Solve $P Q \times 6 = QQQ$ , where P and Q are different digits.

Try it first, then show the solutionSolution

$QQQ = 111 \times Q = 3 \times 37 \times Q$ , and it must be divisible by 6, so Q is even.
$P Q = \frac{111 Q}{6} = 18.5 Q$ . Try even Q: $Q = 2$ gives 37 (units digit 7, not 2); $Q = 4$ gives 74 (units digit 4 ✓); $Q = 6$ gives 111 (three digits); $Q = 8$ gives 148.
Only $Q = 4$ , $P = 7$ works: $74 \times 6 = 444$ .

Answer: $74 \times 6 = 444$ (P = 7, Q = 4).

Maths in real life: Number Play

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.

Divisibility by 9

Suppose 5,346 chairs are to be arranged in rows of 9 for a school function.

Will every row be full?

Try it first, then show the solutionSolution

Digit sum $5 + 3 + 4 + 6 = 18$ , a multiple of 9.
So 5,346 is divisible by 9: 594 full rows.

Answer: Yes, all rows are full (594 rows).

Memory hook: Digit sum a multiple of 9 means no chair is left over.

Remainders in cycles

Suppose today is a Monday and a school event is exactly 100 days away.

On which day of the week will it fall?

Try it first, then show the solutionSolution

Days repeat every 7: $100 = 7 \times 14 + 2$ .
Two days after Monday is Wednesday.

Answer: Wednesday.

Memory hook: Only the remainder by 7 moves the weekday.

Digital root check

Suppose a shopkeeper works out $123 \times 45$ by hand and gets 5,535.

Use digital roots to check that the answer is believable.

Try it first, then show the solutionSolution

Digital roots: 123 gives 6, 45 gives 9; $6 \times 9 = 54$ , which gives 9.
5,535 gives $5 + 5 + 3 + 5 = 18$ , which gives 9: they match.

Answer: The check passes (and indeed $123 \times 45 = 5535$ ).

Memory hook: Roots of the factors multiply to the root of the product.

Pairs to make fours

Suppose two shops each pack pencils in boxes of 4 and each has exactly 2 pencils left over.

If they combine their leftover pencils, can they fill whole boxes of 4?

Try it first, then show the solutionSolution

Each leftover count is 2, so together they have 4.
In algebra, $(4 p + 2) + (4 q + 2) = 4 (p + q + 1)$ .

Answer: Yes, the leftovers make exactly one more box.

Memory hook: Two remainders of 2 make a four.

Topics in this chapter

Number Play formula sheet

Even and odd: $even = 2 n, odd = 2 n + 1$
Remainder form: $N = d k + r, 0 \leq r < d$
Three consecutive numbers: $(n - 1) + n + (n + 1) = 3 n$
Divisibility of sums: $a ∣ M, a ∣ N \Rightarrow a ∣ (M \pm N)$
Two divisors: $k ∣ A, m ∣ A \Rightarrow LCM (k, m) ∣ A$
Place value: $\overline{d c ba} = 1000 d + 100 c + 10 b + a$
Test for 9 (and 3): $1 0^{n} = (1 0^{n} - 1) + 1 \Rightarrow N \equiv digit sum$; Same remainder on division by 9 or 3.
Test for 11: $a - b + c - d + \dots (from the units digit)$
4-digit palindrome: $\overline{abba} = 1001 a + 110 b = 11 (91 a + 10 b)$

How the school exam asks Number Play

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Divisibility shortcuts	Test numbers for divisibility by 2, 3, 4, 5, 6, 8, 9, 10, 11 (often in a table); find a missing digit.	Apply the right test; for composite divisors use co-prime factors (6 = 2 × 3, 24 = 3 × 8); for a missing digit, write the test as an equation in the digit.
Remainders without dividing	Find the remainder by 9 or 11; find remainders of sums and differences.	Use digit sums or alternating sums; for combinations, write each number as $d k + r$ and collect.
Always, sometimes, never	Classify statements about parity, multiples and factors, with justification.	Use algebra for 'always', an example and non-example for 'sometimes', and an impossible equality for 'never'.
Consecutive numbers	Find consecutive numbers with a given sum; describe them in terms of the largest.	Write a general form centred on the middle, set up an equation and solve.
Explain why a shortcut works	Use algebra to explain the test for 4, 8, 9 or 11.	Write the number in expanded form and split each place value into a multiple of the divisor plus a small remainder.
Cryptarithms and digital roots	Solve a short letter-digit puzzle; find the digital root of an expression.	Use size limits, units digits and carries; for digital roots, use that adding 9 changes nothing.

Tricks to remember Number Play

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

Nines hide in every place
10, 100, 1000 are 'nine-ish plus one' (9 + 1, 99 + 1, 999 + 1), so only the digits survive: add them.
Use it when: Tests for 9 and 3.
Eleven flips plus and minus
For 11, place values go 'plus one, minus one, plus one ...' so the digits go + − + − from the units.
Use it when: Test for 11.
Last 1, 2, 3 digits for 2, 4, 8
2 looks at 1 digit, 4 at 2 digits, 8 at 3 digits (because $2^{1}$ , $2^{2}$ , $2^{3}$ ).
Use it when: Tests for powers of 2.
Co-prime pairs only
To test 24 use 3 and 8 (no common factor), not 4 and 6 (share 2).
Use it when: Combining tests.
Remainders add like coins
Add remainders, then take away full groups of the divisor.
Use it when: Remainder of a sum or difference.

Number Play: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Divisible by 9 · 1 mark · MCQExam level

Which of these is divisible by 9?
(a) 4,356
(b) 5,374
(c) 6,123
(d) 2,431

Try it first, then show the solutionSolution

Digit sums: 18, 19, 12, 10.
Only 18 is a multiple of 9.

Answer: (a) 4,356

Example 2: Sum of multiples · 1 mark · Assertion-reasonExam level

Assertion (A): The sum of a multiple of 12 and a multiple of 15 is always a multiple of 3.
Reason (R): If 3 divides M and 3 divides N, then 3 divides $M + N$ .
(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.

Try it first, then show the solutionSolution

Multiples of 12 and of 15 are both multiples of 3 (since $12 = 3 \times 4$ and $15 = 3 \times 5$ ).
By R, their sum is a multiple of 3, so R explains A.

Answer: (a)

Example 3: Remainder by 9 · 2 marks · Very short answerExam level

Find the remainder when 73,415 is divided by 9, without dividing.

Try it first, then show the solutionSolution

Digit sum $= 7 + 3 + 4 + 1 + 5 = 20$ , and $2 + 0 = 2$ .
A number and its digit sum leave the same remainder by 9.

Answer: 2

Example 4: Test for 11 · 2 marks · Very short answerHarder exam level

Use the alternating-sum test to decide whether 8,17,069 is divisible by 11.

Try it first, then show the solutionSolution

From the units: $9 - 6 + 0 - 7 + 1 - 8 = - 11$ .
$- 11$ is a multiple of 11, so yes ( $817069 = 11 \times 74279$ ).

Answer: Yes, it is divisible by 11.

Example 5: Always, sometimes, never · 3 marks · Short answerHarder exam level

Classify each as always, sometimes or never true, with a reason: (i) the sum of two consecutive odd numbers is a multiple of 4; (ii) a number divisible by 4 and 6 is divisible by 24; (iii) the square of an odd number leaves remainder 3 when divided by 4.

Try it first, then show the solutionSolution

(i) $(2 n + 1) + (2 n + 3) = 4 n + 4 = 4 (n + 1)$ : always.
(ii) Only LCM(4, 6) = 12 is guaranteed; 24 works, 12 does not: sometimes.
(iii) $(2 m + 1)^{2} = 4 m^{2} + 4 m + 1 = 4 (m^{2} + m) + 1$ , so the remainder is always 1, never 3: never.

Answer: (i) Always (ii) Sometimes (iii) Never

Example 6: Combining remainders · 5 marks · Long answerHarder exam level

A leaves remainder 3 and B leaves remainder 5 when divided by 7, with $A > B$ . Find the remainders when $A + B$ , $A - B$ and $2 A + B$ are divided by 7. Show your algebra.

Try it first, then show the solutionSolution

Write $A = 7 a + 3$ and $B = 7 b + 5$ .
$A + B = 7 (a + b) + 8 = 7 (a + b + 1) + 1$ : remainder 1.
$A - B = 7 (a - b) - 2 = 7 (a - b - 1) + 5$ : remainder 5.
$2 A + B = 7 (2 a + b) + 11 = 7 (2 a + b + 1) + 4$ : remainder 4.
Check with $A = 24$ , $B = 12$ : 36, 12 and 60 leave 1, 5 and 4.

Answer: 1, 5 and 4.

Example 7: Missing digits · 3 marks · Short answerHarder exam level

Find the digit $x$ so that $5 x 38$ is divisible by 9, and the digit $y$ so that $5 y 38$ is divisible by 11.

Try it first, then show the solutionSolution

For 9: $5 + x + 3 + 8 = 16 + x$ must be a multiple of 9, so $x = 2$ (sum 18).
For 11: from the units, $8 - 3 + y - 5 = y$ must be 0 or 11, so $y = 0$ .
Check: $5238 = 9 \times 582$ and $5038 = 11 \times 458$ .

Answer: $x = 2$ ; $y = 0$

Example 8: A remainder puzzle · 4 marks · Case-basedHOTS / competency

Find the smallest number that leaves remainder 1 when divided by 2, 3, 4, 5 or 6, and is exactly divisible by 7.

Try it first, then show the solutionSolution

The number is 1 more than a multiple of 2, 3, 4, 5 and 6, so it is 1 more than a multiple of LCM = 60: $60 k + 1$ .
Try $k = 1, 2, \dots$ : 61, 121, 181, 241, 301.
Check divisibility by 7: 61, 121, 181, 241 are not; $301 = 7 \times 43$ .
So the smallest is 301.

Answer: 301

Example 9: A multiplication cryptarithm · 3 marks · Short answerHOTS / competency

Solve $P Q \times 3 = R QQ$ , where P, Q and R are different digits.

Try it first, then show the solutionSolution

Units: $3 \times Q$ ends in Q, so $2 Q$ ends in 0: $Q = 0$ or $Q = 5$ .
$Q = 0$ : $30 P = 100 R$ , so $3 P = 10 R$ , impossible for digits. $Q = 5$ : $30 P + 15 = 100 R + 55$ , so $3 P = 10 R + 4$ .
Only $P = 8$ , $R = 2$ fits: $85 \times 3 = 255$ .

Answer: P = 8, Q = 5, R = 2 ( $85 \times 3 = 255$ )

Example 10: Palindromes and 11 · 3 marks · Short answerHOTS / competency

Prove that every 4-digit palindrome (a number like 7887) is divisible by 11. Is every 5-digit palindrome divisible by 11?

Try it first, then show the solutionSolution

$\overline{abba} = 1000 a + 100 b + 10 b + a = 1001 a + 110 b = 11 (91 a + 10 b)$ .
So it is always a multiple of 11.
For 5 digits, 12,321 has alternating sum $1 - 2 + 3 - 2 + 1 = 1$ , so it is not divisible by 11: not always.

Answer: 4-digit: always; 5-digit: not always (12,321 fails).

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Packing laddoos

Suppose a sweet shop has fewer than 100 laddoos. Packed in boxes of 4, 3 are left over; in boxes of 5, 4 are left over; in boxes of 6, 5 are left over.

(i) · 1 mark
Explain why one more laddoo would make the total a multiple of 4, 5 and 6.
Answer
Answer: Each leftover is one short of a full box.
Remainders 3, 4, 5 are each 1 less than 4, 5, 6, so adding 1 fills every box exactly.
(ii) · 1 mark
How many laddoos are there?
Answer
Answer: 59
Total + 1 is a multiple of LCM(4, 5, 6) = 60, so total $= 60 k - 1$ ; under 100 gives 59.
(iii) · 2 marks
The shop later has more than 100 but fewer than 200 laddoos with the same three leftovers, and they fit exactly into boxes of 7. How many are there?
Answer
Answer: 119
Totals with these leftovers are $60 k - 1$ : 59, 119, 179 (1 mark). Between 100 and 200: 119 and 179; $119 = 7 \times 17$ while 179 is not a multiple of 7, so 119 (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Without dividing, find the remainder when 6,47,315 is divided by 11. (2 marks)

Step	Marks
Alternating sum from the units: $5 - 1 + 3 - 7 + 4 - 6 = - 2$	1
2 short of a multiple of 11, so remainder $11 - 2 = 9$	1

Show that the sum of any three consecutive numbers is a multiple of 3, and find three consecutive numbers whose sum is 141. (3 marks)

Step	Marks
Write them as $n - 1, n, n + 1$	1
Sum $= 3 n$ , which always has 3 as a factor	1
$3 n = 141$ gives $n = 47$ : the numbers are 46, 47, 48	1

Prove that a 3-digit number is a multiple of 9 exactly when its digit sum is a multiple of 9. Use your proof to test 738. (5 marks)

Step	Marks
Write the number as $100 c + 10 b + a$	1
Split: $100 c + 10 b + a = (99 c + 9 b) + (c + b + a)$	1
$99 c + 9 b = 9 (11 c + b)$ is always a multiple of 9	1
So the number is a multiple of 9 exactly when $a + b + c$ is	1
738: $7 + 3 + 8 = 18$ , a multiple of 9, so 738 is divisible by 9 ( $738 = 9 \times 82$ )	1

Where students lose marks in Number Play

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

Mistake: Testing 24 by checking 4 and 6
Any two factors whose product is 24 seem enough.
Wrong working
36 is divisible by 4 and 6, so it is divisible by 24.
Right working
4 and 6 share the factor 2, so they only guarantee LCM(4, 6) = 12. Test 3 and 8: 36 is not divisible by 8, so not by 24.
Do this: Use factors with no common factor.
Usually costs: The answer and reason.
Mistake: Starting the 11-test from the wrong end
The alternating sum is begun at the leftmost digit and the sign is then misread.
Wrong working
For 6,47,315: $6 - 4 + 7 - 3 + 1 - 5 = 2$ , so remainder 2.
Right working
From the units: $5 - 1 + 3 - 7 + 4 - 6 = - 2$ , so remainder $11 - 2 = 9$ .
Do this: Start with + at the units digit; turn a negative result into a remainder by adding 11.
Usually costs: The remainder mark.
Mistake: Proving 'always' with examples only
A few true cases feel like enough.
Wrong working
$3 + 4 + 5 = 12$ and $6 + 7 + 8 = 21$ , so the sum of three consecutive numbers is always a multiple of 3.
Right working
$(n - 1) + n + (n + 1) = 3 n$ , a multiple of 3 for every $n$ . Examples only illustrate.
Do this: Use algebra for every 'always' claim.
Usually costs: 1-2 marks for the justification.
Mistake: Reading a remainder as negative
The algebra gives $7 k - 2$ .
Wrong working
$A - B = 7 (a - b) - 2$ , so the remainder is $- 2$ .
Right working
2 less than a multiple of 7 is 5 more than the previous multiple: remainder 5.
Do this: Remainders are always between 0 and the divisor minus 1.
Usually costs: 1 mark.
Mistake: Treating a digital root of 9 as remainder 9
The digital root and the remainder are mixed up for multiples of 9.
Wrong working
Digital root of 4,356 is 9, so 4,356 ÷ 9 leaves remainder 9.
Right working
Digital root 9 means the number is a multiple of 9: remainder 0.
Do this: Remainder = digital root, except root 9 gives remainder 0.
Usually costs: 1 mark.

NCERT exercise guide: Number Play

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Figure it Out (multiples and remainders)

What it tests: Consecutive numbers with a given sum, describing consecutive numbers from the greatest, always/sometimes/never statements on multiples, numbers with given remainders, a pebble riddle, sums of numbers with a fixed remainder, and remainders of sums and differences.

How to approach it: Write numbers in general form ( $n, n + 1$ ; $d k + r$ ), simplify, and read off the factor or remainder; test riddles with the combined form (such as $12 k + 2$ ).

Like the consecutive-number questions in Section 5.1

The sum of four consecutive numbers is 90. Find them, and show that such a sum can never be a multiple of 4.

Show the solutionSolution

Let them be $n, n + 1, n + 2, n + 3$ ; sum $= 4 n + 6$ .
$4 n + 6 = 90$ gives $n = 21$ : 21, 22, 23, 24.
$4 n + 6 = 4 (n + 1) + 2$ always leaves remainder 2 on division by 4, so it is never a multiple of 4.

Answer: 21, 22, 23, 24; the sum always leaves remainder 2.

Figure it Out (divisibility by 9)

What it tests: Testing numbers for divisibility by 9 without dividing, the smallest multiple of 9 with no odd digits, the multiple of 9 nearest a given number, and counting multiples of 9 in a range.

How to approach it: Use digit sums; for 'nearest' use the remainder to step down or up; for counting, find the first and last multiples in the range.

Like the nearest-multiple question in Section 5.2

Find the multiple of 9 closest to 8,500.

Show the solutionSolution

Digit sum of 8500 is 13, and $1 + 3 = 4$ , so 8500 leaves remainder 4: $8500 = 9 k + 4$ .
Going down by 4 gives 8,496, a multiple of 9; going up by $9 - 4 = 5$ gives 8,505, also a multiple of 9.
8,496 is 4 away and 8,505 is 5 away, so 8,496 is closer.

Answer: 8,496

Figure it Out (digital roots)

What it tests: Digital root after adding 10, sequences formed by adding 11, the digital root of an algebraic expression, and patterns linking parity, digital roots and remainders by 3 and 9.

How to approach it: Use that the digital root behaves like the remainder by 9: adding 9 leaves it unchanged, and roots of sums and products follow the sums and products of the roots.

Like the digital-root questions in Section 5.2

A number has digital root 5. What is the digital root of (a) 4 times the number, (b) the number plus 13?

Show the solutionSolution

(a) The number is $9 k + 5$ ; four times it is $36 k + 20$ , and 20 has digital root 2.
(b) $9 k + 5 + 13 = 9 k + 18$ , a multiple of 9: digital root 9.

Answer: (a) 2 (b) 9

Practice questions on Number Play

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 · MCQExam level
Which number is divisible by 11?
(a) 1,234
(b) 2,453
(c) 3,456
(d) 5,432
Hint
Alternating sum from the units.
Answer
(b) 2,453: $3 - 5 + 4 - 2 = 0$
Q2 · 1 mark · MCQExam level
Which number is of the form $4 k + 3$ ?
(a) 14
(b) 15
(c) 16
(d) 17
Hint
Remainder 3 on division by 4.
Answer
(b) 15 $= 4 \times 3 + 3$
Q3 · 1 mark · Very short answerExam level
Find the remainder when 7,777 is divided by 9.
Hint
Digit sum.
Answer
$28 \to 10 \to 1$ : remainder 1
Q4 · 2 marks · Very short answerExam level
Is 4,170 divisible by 6? Give the reason using shortcuts.
Hint
Test 2 and 3.
Answer
Yes: it ends in 0 (even) and its digit sum 12 is a multiple of 3.
Q5 · 2 marks · Very short answerHarder exam level
The sum of four consecutive numbers is 130. Find them.
Hint
$4 n + 6$ .
Answer
$4 n + 6 = 130$ , so $n = 31$ : 31, 32, 33, 34
Q6 · 2 marks · Very short answerHarder exam level
p is the largest of five consecutive odd numbers. Write the other four in terms of p, and their total sum.
Hint
Odd numbers are 2 apart.
Answer
$p - 2, p - 4, p - 6, p - 8$ ; sum $= 5 p - 20$ .
Q7 · 3 marks · Short answerHarder exam level
Is 7,29,624 divisible by 24? Use shortcuts, then verify.
Hint
Test 3 and 8.
Answer
Digit sum 30 (divisible by 3); last three digits 624 $= 8 \times 78$ (divisible by 8). So yes: $729624 = 24 \times 30401$ .
Q8 · 3 marks · Short answerHarder exam level
Explain, using place value, why a number is divisible by 8 when the number formed by its last three digits is.
Hint
$1000 = 8 \times 125$ .
Answer
Write the number as $1000 \times (rest) + (last three digits)$ . Since $1000 = 8 \times 125$ , the first part is a multiple of 8, so the number is a multiple of 8 exactly when the last three digits form one.
Q9 · 2 marks · Very short answerHarder exam level
Find the remainder when 5,82,913 is divided by 11, and the smallest number to add to make it divisible by 11.
Hint
Alternating sum from the units.
Answer
$3 - 1 + 9 - 2 + 8 - 5 = 12$ , and $12 = 11 + 1$ : remainder 1; add 10.
Q10 · 5 marks · Long answerHarder exam level
Numbers A and B each leave remainder 1 when divided by 4. (a) What remainder does $A + B$ leave? (b) Show that $A \times B$ also leaves remainder 1. (c) Check with $A = 5$ , $B = 9$ .
Hint
Write $A = 4 a + 1$ , $B = 4 b + 1$ .
Answer
(a) $4 (a + b) + 2$ : remainder 2. (b) $(4 a + 1) (4 b + 1) = 16 ab + 4 a + 4 b + 1 = 4 (4 ab + a + b) + 1$ : remainder 1. (c) $5 + 9 = 14$ leaves 2; $5 \times 9 = 45$ leaves 1.
Q11 · 3 marks · Short answerHOTS / competency
Solve the cryptarithm $A B + A B + A B = C A B$ , where A, B, C are different digits.
Hint
$3 \times A B = 100 C + A B$ .
Answer
$2 \times A B = 100 C$ , so $A B = 50 C$ ; AB has two digits, so $C = 1$ , $A B = 50$ : $50 + 50 + 50 = 150$ .
Q12 · 3 marks · Short answerHOTS / competency
A number has digital root 7. Without knowing the number, find the digital root of (a) the number plus 20, (b) 3 times the number. Explain.
Hint
Write it as $9 k + 7$ .
Answer
(a) $9 k + 27 = 9 (k + 3)$ : digital root 9. (b) $27 k + 21$ , and 21 has digital root 3: digital root 3.

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 parity?Show answer
Whether a number is even or odd.
2.Why do $a + b$ and $a - b$ have the same parity?Show answer
They differ by $2 b$ , which is even.
3.Write all numbers leaving remainder 3 when divided by 5.Show answer
$5 k + 3$ for whole numbers $k$ (or $5 k - 2$ for $k \geq 1$ ).
4.If 8 divides M and N, what else must 8 divide?Show answer
$M + N$ and $M - N$ .
5.Which digits decide divisibility by 8?Show answer
The last three digits.
6.State the test for 9.Show answer
A number is a multiple of 9 exactly when its digit sum is a multiple of 9.
7.State the test for 11.Show answer
The alternating sum of digits from the units place is 0 or a multiple of 11.
8.Which two tests together check divisibility by 24?Show answer
Divisibility by 3 and by 8.
9.What is a digital root?Show answer
The single digit you get by repeatedly adding the digits of a number.
10.Who described digital roots around 950 CE?Show answer
Aryabhata II, in the Mahasiddhanta.

Best order to study it

Explore sums of consecutive numbers and the 8 sign patterns; learn the parity argument.
Practise deciding which algebraic expressions are always even, with examples and non-examples.
Learn the 'pairs to make fours' algebra and the always/sometimes/never statements.
Practise remainder forms like $5 k + 3$ and combining remainders (first Figure it Out set).
Learn why the tests for 10, 5, 2, 4, 8 work, then the digit-sum tests for 9 and 3 (second set).
Learn the alternating-sum test for 11 and combined tests for 6 and 24.
Finish with digital roots (third set) and cryptarithms in 5.3.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain why all $a \pm b \pm c \pm d$ have the same parity?
Can you show the sum of four consecutive numbers is never a multiple of 4?
Can you prove that two even non-multiples of 4 add to a multiple of 4?
Can you decide always/sometimes/never with algebra and counterexamples?
Can you find the remainder of a sum from the remainders of its parts?
Can you explain the tests for 4 and 8 using place value?
Can you find a remainder on division by 9 from the digit sum?
Can you use the alternating sum to test for 11?
Can you say why 3 and 8 test for 24 but 4 and 6 do not?
Can you solve a two-letter cryptarithm by reasoning?

Number Play: frequently asked questions

Where is Number Play in the CBSE Class 8 textbook?

Number Play is Chapter 5 of Ganita Prakash Part I, the NCERT Grade 8 Mathematics book (Reprint 2026-27). Its three sections are Is This a Multiple Of?, Checking Divisibility Quickly, and Digits in Disguise.

Do I have to prove the divisibility rules or just use them?

Both. The chapter's main aim is to explain why the shortcuts work using place value and algebra, so expect questions like 'explain why the test for 9 works'. Practise writing the expanded form and splitting each place value.

What answer is expected in an always, sometimes or never question?

State your choice and justify it. For 'always', give an algebraic argument; for 'sometimes', give one example where it holds and one where it fails; for 'never', show the statement would force something impossible, such as an even number equalling an odd one.

Is the test for 7 in the syllabus?

No. Ganita Prakash covers shortcuts for 2, 3, 4, 5, 6, 8, 9, 10 and 11 and combinations such as 24, and mentions that shortcuts exist for other numbers in higher grades. For 7, divide directly or reason with remainders.

How do I start a cryptarithm?

Look for the strongest limits first: how many digits the answer has, which letter must be 1 because of a carry, and what the units digits allow. Then test the few remaining cases and check that all letters stand for different digits.

Based on: NCERT Ganita Prakash, Mathematics Textbook for Grade 8, Part I (Reprint 2026-27); NCERT chapter PDF: Number Play