MathsCBSE Class 8 · Number Play

Parity, Remainders and Divisibility Properties for Class 8

Updated 17 October 2026 · Fact-checked

Numbers that leave remainder r when divided by n have the form nk + r. Sums, differences and products can be studied by parity (even or odd) and by remainders, and divisibility properties such as "if a divides M and N then a divides M + N and M - N" are proved with algebra. Class 8 exams ask for always-sometimes-never statements, remainder puzzles solved with the LCM, and proofs with letter-numbers.

Add the remainders 3 + 5 = 8; one 7 joins the multiples of 7, so only 1 is left over.

Illustration drawn for QuestionBuddy.

Understand Parity, Remainders and Divisibility Properties

Parity. A number is even if it has 2 as a factor and odd otherwise. Parity rules, proved with letter-numbers ( $2 m$ is even, $2 m + 1$ is odd):

even $\pm$ even = even, odd $\pm$ odd = even, even $\pm$ odd = odd (so $a + b$ and $a - b$ always have the same parity);
the product is even if either factor is even.

*Four consecutive numbers with signs.* If you place $+$ and $-$ signs between four consecutive numbers (there are eight ways), every result is even. More generally, all expressions $a \pm b \pm c \pm d$ have the same parity, because switching a sign changes the value by an even number ( $2 b$ , $2 c$ or $2 d$ ).

*Algebraic expressions that are always even.* $4 m + 2 q = 2 (2 m + q)$ is even for all integers $m$ and $q$ ; $x^{2} + 2$ is even only when $x$ is even.

Remainders and algebraic forms. Even numbers leave remainder 0 or 2 when divided by 4. A number that leaves remainder $r$ when divided by $n$ has the form $nk + r$ with $0 \leq r < n$ . Numbers that leave remainder 3 when divided by 5 are $5 k + 3$ , which can also be written $5 k - 2$ (for $k \geq 1$ ): "3 more than a multiple of 5" is "2 less than a multiple of 5".

To combine remainders, add the forms: if $M = 7 a + 3$ and $N = 7 b + 5$ then $M + N = 7 (a + b) + 8 = 7 (a + b + 1) + 1$ , so the sum leaves remainder 1.

Properties of divisibility (proved by algebra or by a rows-of-counters picture):
1. If $a$ divides $M$ and $a$ divides $N$ , then $a$ divides $M + N$ and $M - N$ .
2. If $A$ is divisible by $k$ , then every multiple of $A$ is divisible by $k$ .
3. If $A$ is divisible by $k$ , then $A$ is divisible by every factor of $k$ .
4. If $A$ is divisible by both $k$ and $m$ , then $A$ is divisible by $LCM (k, m)$ (not necessarily by $k \times m$ ).
5. Divisibility by a multiple of $k$ does not follow from divisibility by $k$ : 42 is divisible by 7 but not by 28.

Always, sometimes or never true. To settle a statement, give an example (to show "sometimes" or "always" can hold), a non-example (to rule out "always") and an algebraic argument. For instance "an odd number plus an even number is a multiple of 6" is *never* true: the sum is odd and multiples of 6 are even.

Remainder puzzles with the LCM. A number that leaves remainder $r$ on division by each of several numbers is of the form $LCM \times k + r$ . When the remainders are all " $d$ short" (for example 2 short of each divisor), the number is $d$ less than a common multiple.

The divisibility shortcuts (tests) for 3, 9, 11 and digital roots are on the next topic page of this chapter.

Parity, Remainders and Divisibility Properties: formulas to remember

Form of a number with remainder r: $N = nk + r, 0 \leq r < n$; The multiple of n plus the remainder.
Even and odd numbers: $2 m (even), 2 m + 1 (odd)$; Used to prove all parity rules.
Remainder of a sum: $(n A + r_{1}) + (n B + r_{2}) = n (A + B) + (r_{1} + r_{2})$; Add remainders, then reduce modulo n.
Divisibility of sums: $a ∣ M and a ∣ N \Rightarrow a ∣ (M \pm N)$; The symbol a | M means a divides M.
Divisibility by two numbers: $k ∣ A and m ∣ A \Rightarrow LCM (k, m) ∣ A$; Product k x m only if k and m share no factor.
Common remainder: $N = LCM (n_{1}, n_{2}, \dots) \times k + r$; r is the same remainder for every divisor.

How to solve Parity, Remainders and Divisibility Properties questions

Use this routine for parity and remainder reasoning.

1Name the numbers with letter-numbers: even 2m, odd 2m + 1, remainder form nk + r.
2Expand the expression and collect the multiple of the divisor.
3Reduce the leftover constant by dividing again if it is as large as the divisor.
4For "always, sometimes, never", test two or three numbers first, then prove or disprove with algebra.
5For several remainders, rewrite each as "N plus or minus something is a multiple of n", take the LCM, and list the numbers in the range.
6Check the smallest or largest answer by division.

Parity, Remainders and Divisibility Properties 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.

Remainder puzzle

Suppose a stall owner packs bangles in boxes. When she counts them in 4s she has 3 left over, in 6s she has 5 left, and in 9s she has 8 left.

If there are fewer than 100 bangles, how many could there be?

Try it first, then show the solutionSolution

Each remainder is 1 less than the divisor, so $N + 1$ is a common multiple of 4, 6 and 9.
LCM $= 36$ , so $N + 1 = 36 k$ and $N = 36 k - 1$ .
Below 100: $k = 1$ gives 35 and $k = 2$ gives 71 (and $k = 3$ gives 107, too large).

Answer: 35 or 71 bangles

Memory hook: When every remainder is the same distance short, add that distance back and use the LCM.

Parity check

Imagine a game where a score is made by writing plus or minus signs between the numbers 5, 6, 7 and 8 in any way.

Can the final score ever be odd?

Try it first, then show the solutionSolution

Take $5 + 6 + 7 + 8 = 26$ , which is even.
Switching any sign changes the score by twice a number, which is even, so every score has the same parity as 26.

Answer: No, the score is always even.

Memory hook: A sign flip changes the value by an even amount, so parity never changes.

Tricks to remember Parity, Remainders and Divisibility Properties

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

Same parity for sum and difference
a + b and a - b are always both odd or both even because they differ by 2b.
Use it when: Parity questions with signs.
Remainder as a form
Remainder 3 on division by 5 means 5k + 3. Add and multiply these forms and read off the new remainder.
Use it when: Every remainder-combination question.
LCM, not product
Divisible by 6 and by 4 means divisible by 12, not by 24.
Use it when: Properties of divisibility.
Short by the same amount
If the remainders are 1, 2, 3 short of divisors 2, 3, 4, then the number plus 1 is a multiple of all.
Use it when: Remainder puzzles.

Parity, Remainders and Divisibility Properties: solved examples

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

Example 1: Always-even expression · 1 mark · MCQBoard level

Which expression is even for every integer m? (a) 2m + 1 (b) m squared + m (c) 3m + 2 (d) m + 5

Try it first, then show the solutionSolution

$m^{2} + m = m (m + 1)$ : one of two consecutive numbers is even, so the product is even.
Test the others at $m = 1$ : $2 m + 1 = 3$ , $3 m + 2 = 5$ , $m + 5 = 6$ (even) but at $m = 2$ , $m + 5 = 7$ (odd).

Answer: (b) m squared + m

Example 2: Numbers with a given remainder · 2 marks · Short answerBoard level

Write the general form of numbers that leave remainder 4 when divided by 7, and list all such numbers between 50 and 80.

Try it first, then show the solutionSolution

The form is $7 k + 4$ . (1 mark)
$k = 7, 8, 9, 10$ give $53, 60, 67, 74$ ; $k = 11$ gives $81 > 80$ . (1 mark)

Answer: 7k + 4; the numbers are 53, 60, 67 and 74.

Example 3: Sum of four consecutive numbers · 3 marks · Short answerHarder board level

Show that the sum of four consecutive natural numbers is never a multiple of 4, and find the numbers whose sum is 74.

Try it first, then show the solutionSolution

Let the numbers be $n, n + 1, n + 2, n + 3$ . Sum $= 4 n + 6 = 4 (n + 1) + 2$ . (1 mark)
So the sum always leaves remainder 2 on division by 4 and is never a multiple of 4. (1 mark)
$4 n + 6 = 74$ gives $n = 17$ : the numbers are 17, 18, 19, 20. (1 mark)

Answer: Sum = 4n + 6, remainder 2 on division by 4; the numbers are 17, 18, 19, 20.

Example 4: Largest three-digit number with equal remainders · 3 marks · Short answerHarder board level

Find the largest three-digit number that leaves remainder 1 when divided by 4, 6 and 9.

Try it first, then show the solutionSolution

The number minus 1 is a common multiple of 4, 6, 9: LCM $= 36$ . (1 mark)
So $N = 36 k + 1$ . (1 mark)
$36 \times 27 + 1 = 973$ and $36 \times 28 + 1 = 1009$ has four digits, so the answer is 973. Check: $972 = 4 \times 243 = 6 \times 162 = 9 \times 108$ . (1 mark)

Answer: 973

Example 5: Always, sometimes or never · 4 marks · Long answerHarder board level

State whether each is always, sometimes or never true, and justify with algebra: (a) the sum of three consecutive numbers is a multiple of 3; (b) if a number is divisible by 6, it is divisible by 12; (c) the sum of an odd and an even number is a multiple of 4.

Try it first, then show the solutionSolution

(a) Always: $n + (n + 1) + (n + 2) = 3 n + 3 = 3 (n + 1)$ . (1 mark)
(b) Sometimes: 12 is divisible by 6 and by 12, but 18 is divisible by 6 and not by 12. (1 mark)
(c) Never: the sum $2 a + (2 b + 1) = 2 (a + b) + 1$ is odd, and every multiple of 4 is even. (1 mark)
Each verdict has an example or non-example plus an algebraic reason. (1 mark)

Answer: (a) Always, (b) Sometimes, (c) Never.

Where students lose marks in Parity, Remainders and Divisibility Properties

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

Mistake: Assuming divisibility by 6 and 4 means divisibility by 24
The LCM of 4 and 6 is 12, not 24.
Wrong working
"Divisible by 6 and 4, so divisible by 6 x 4 = 24."
Right working
The number is divisible by $LCM (4, 6) = 12$ ; for example 12 is divisible by both but not by 24.
Do this: Compute the LCM, not the product, unless the numbers share no factor.
Usually costs: The reasoning marks.
Mistake: Not reducing the leftover remainder
A remainder must be smaller than the divisor.
Wrong working
"Three numbers with remainder 5 on division by 8: the sum has remainder 15."
Right working
$15 = 8 + 7$ , so the remainder is 7.
Do this: After adding remainders, divide again by the divisor.
Usually costs: The answer.
Mistake: Proving "always true" from a few examples
Examples do not prove a general claim; algebra does.
Wrong working
"Checked 3 + 4 + 5, 6 + 7 + 8; both multiples of 3, so always true."
Right working
$n + (n + 1) + (n + 2) = 3 (n + 1)$ proves it.
Do this: Give an algebraic argument for "always" and a counter-example for "not always".
Usually costs: 1 to 2 marks.
Mistake: Using 5k + 3 with k starting from 1 only
Allow k = 0, so that 3 itself is included, unless the question restricts k.
Wrong working
"Numbers: 8, 13, 18, ..."
Right working
$k = 0$ gives 3, $k = 1$ gives 8, and so on.
Do this: Check the smallest case separately.
Usually costs: 1 mark.

Exam tips

Name numbers with letter-numbers (2m, 2m + 1, nk + r) as the first line of a proof.
For puzzles with several remainders, note whether each remainder is the same or each is the same amount short of its divisor.
Write the LCM computation by prime factorisation and show the form N = LCM x k + r.
Always include one example and one counter-example in always-sometimes-never answers.
State the range and list all numbers when the question says "between".

Practice questions on Parity, Remainders and Divisibility Properties

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): 4m + 2q is even for all integers m and q.
Reason (R): 4m + 2q = 2(2m + q), which has 2 as a factor.
(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
Factorise the expression.
Answer
(a). $2 (2 m + q)$ is 2 times an integer, so it is even for all integers $m$ and $q$ , and R gives the reason.
Q2 · 2 marks · Very short answerBoard level
The number 100 leaves remainder 2 when divided by 7, and 53 leaves remainder 4. Without dividing, find the remainders when 153 (= 100 + 53) and 47 (= 100 - 53) are divided by 7.
Hint
Write 100 = 7a + 2 and 53 = 7b + 4.
Answer
$100 + 53 = 7 (a + b) + 6$ , so the remainder is 6. $100 - 53 = 7 (a - b) + (2 - 4) = 7 (a - b) - 2 = 7 (a - b - 1) + 5$ , so the remainder is 5. Check: $153 = 7 \times 21 + 6$ and $47 = 7 \times 6 + 5$ .
Q3 · 3 marks · Short answerHarder board level
Three numbers each leave remainder 5 when divided by 8. What remainder does their sum leave when divided by 8? Prove it and give an example.
Hint
Write each as 8k + 5 and add.
Answer
$(8 a + 5) + (8 b + 5) + (8 c + 5) = 8 (a + b + c) + 15 = 8 (a + b + c + 1) + 7$ . The remainder is 7. Example: $5 + 13 + 21 = 39 = 8 \times 4 + 7$ .
Q4 · 4 marks · Long answerHOTS / competency
Prove that the product of three consecutive integers is always divisible by 6. Why does this not hold for divisibility by 4?
Hint
Show a factor of 2 and a factor of 3 separately. Try 1 x 2 x 3.
Answer
Let the numbers be $n, n + 1, n + 2$ . Among any two consecutive integers one is even, so the product is divisible by 2. Write $n = 3 k$ , $3 k + 1$ or $3 k + 2$ : in the first case $n$ is a multiple of 3; in the second $n + 2 = 3 k + 3 = 3 (k + 1)$ ; in the third $n + 1 = 3 k + 3 = 3 (k + 1)$ . So one factor is a multiple of 3 and the product is divisible by 3. As 2 and 3 share no factor, 6 divides the product. For 4 it fails: $1 \times 2 \times 3 = 6$ is not divisible by 4.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a stall owner packs bangles. Counting in 4s leaves 3, in 6s leaves 5, in 9s leaves 8.
(i) Show that N + 1 is a multiple of 4, 6 and 9. (1 mark)
(ii) Write all possible N in the form 36k - 1. (1 mark)
(iii) List all possible N below 100. (1 mark)
(iv) If N is also a multiple of 7, find N. (1 mark)
Hint
Every remainder is one less than its divisor.
Answer
(i) $N = 4 a + 3 = 6 b + 5 = 9 c + 8$ , so $N + 1 = 4 (a + 1) = 6 (b + 1) = 9 (c + 1)$ . (ii) LCM of 4, 6, 9 is 36, so $N + 1 = 36 k$ , $N = 36 k - 1$ . (iii) $k = 1$ : 35; $k = 2$ : 71; $k = 3$ : 107 (too big). (iv) Of 35 and 71, only 35 is a multiple of 7: $N = 35$ .

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 general form of numbers leaving remainder r when divided by n?Show answer
nk + r.
2.What is the parity of a + b and a - b?Show answer
The same: both even or both odd.
3.If a divides M and N, what else does a divide?Show answer
M + N and M - N.
4.If A is divisible by k and by m, what divides A?Show answer
The LCM of k and m.
5.Is 4m + 2q always even?Show answer
Yes, it equals 2(2m + q).
6.Can an odd plus an even number be a multiple of 4?Show answer
Never: the sum is odd.

Parity, Remainders and Divisibility Properties: frequently asked questions

Why are 5k + 3 and 5k - 2 the same set of numbers?

5k - 2 = 5(k - 1) + 3, so both describe numbers 3 more than a multiple of 5 (the second form with k starting from 1).

How do I decide between sometimes and always true?

If you can find a case where the statement fails, it is not always true. If it also holds for some case, it is sometimes true.

When is the sum of two multiples of 3 a multiple of 6?

Write them as 3a and 3b. The sum 3(a + b) is a multiple of 6 exactly when a + b is even, that is when a and b are both even or both odd. Otherwise the sum is an odd multiple of 3.

Is 0 even?

Yes. Zero equals 2 times 0, so it has 2 as a factor.