Number Play – Class 8 Maths Chapter 5 Notes & NCERT PDF

Introduction

In Number Play, the book invites us to play with numbers — and then to prove what we notice. We explore sums of consecutive numbers, parity (odd/even), remainders, quick divisibility tests, digital roots and cryptarithms. The big idea: algebra and visual (dot-grid) reasoning let us prove a pattern for all numbers, instead of checking cases one by one.

Key Ideas

5.1 Is This a Multiple Of?

Sums of consecutive numbers. Anshu writes numbers as sums of consecutive numbers: 7 = 3 + 4, 10 = 1 + 2 + 3 + 4, and 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5. Every odd number can be written as a sum of two consecutive numbers. Can every number? Can 0 (using negative numbers)?

Plus and minus signs. Put '+' and '–' between 4 consecutive numbers, e.g. 3, 4, 5, 6. There are 8 possible expressions (3 + 4 – 5 + 6 = 8, 3 – 4 – 5 – 6 = –12, …) and every result is even.

For any 4 numbers a, b, c, d, all the expressions a ± b ± c ± d have the same parity.

Why? Switching one sign changes the value by twice that number (e.g. switching +b to –b changes it by 2b), which is even — and two numbers that differ by an even number have the same parity.

Breaking Even. Use parity rules (odd ± odd = even, even ± even = even, odd ± even = odd) to spot even results without calculating. 4m + 2q = 2(2m + q) is always even; x² + 2 is not (x = 6 gives 38, but x = 3 gives 11).

Pairs to Make Fours. Even numbers are of two kinds: multiples of 4 (form 4p) and those leaving remainder 2 (form 4p + 2).

Always, Sometimes, or Never?

If A is divisible by k and by m, then A is divisible by the LCM of k and m.

What Remains? Numbers leaving remainder 3 when divided by 5 are 3, 8, 13, 18, 23, … = 5k + 3 (also 5k – 2 for k ≥ 1).

5.2 Checking Divisibility Quickly

We write a number as a sum of place values: …dcba = … + 1000d + 100c + 10b + a. Every place value except units is a multiple of 10 — so a number is divisible by 10 exactly when its units digit is 0.

Divisibility by 9. 10 = 9 + 1, 100 = 99 + 1, 1000 = 999 + 1, … so each digit is the remainder its place value leaves on division by 9.

A number is divisible by 9 if and only if the sum of its digits is divisible by 9. Repeatedly adding digits gives the remainder.

Example: 427 → 4 + 2 + 7 = 13 → 1 + 3 = 4, so 427 ÷ 9 leaves remainder 4.

Divisibility by 3 works the same way (sum of digits divisible by 3).

Divisibility by 11. Place values are alternately 1 more and 1 less than a multiple of 11: 1 = 0 + 1, 10 = 11 – 1, 100 = 99 + 1, 1000 = 1001 – 1. So put alternating signs on the digits starting with + at the units digit and add.

Other numbers. Divisibility by 6 = by 2 and by 3. For 24, checking 4 and 6 is not enough (12 is divisible by both but not by 24); check 3 and 8 instead.

Digital roots. Add digits repeatedly until one digit is left: 489710 → 29 → 11 → 2. Multiples of 9 always have digital root 9. Āryabhaṭa II (c. 950 CE), in Mahāsiddhānta, described this method, which was used to check arithmetic.

5.3 Digits in Disguise

Cryptarithms: each letter stands for a digit, each digit is represented by at most one letter, and the first digit of a number is never 0. Reason with sizes and units digits: in PQ × 8 = RS, PQ must be 12 (12 × 8 = 96), because 13 × 8 = 104 has three digits.

Worked Examples

Figure it Out — Hints

Maths Words

Historical and Indian Connections

Common Mistakes

Maths in Daily Life

Divisibility tests help check bills and share things equally. Digital roots were used as quick error-checks in arithmetic. Parity arguments appear in puzzles, games and computer error-checking codes.