NCERT Solutions for Class 8th Maths Chapter 1 A Square and A Cube
Updated on 2026-09-19
About this chapter
A locker is toggled once for each factor of its number, so it ends up open only when the factor count is odd. Factors come in partner pairs, and a pair repeats itself only in a square — so exactly the square-numbered lockers stay open. A perfect square is n × n = n 2 for a natural number n. Perfect squares end only in 0, 1, 4, 5, 6 or 9, and can carry only an even number of terminal zeros. The sum of the first n odd numbers is n 2 . Subtracting 1, 3, 5, … in turn is therefore a test: if you land exactly on 0 at the k-th step, the number is k 2 . Prime factorisation settles both questions. A number is a perfect square when its prime factors split into two identical groups, and a perfect cube when they split into three. Square root is the inverse of squaring; every perfect square has two int
- Queen Ratnamanjuri's puzzle
- Factors and partner factors
- Square Numbers
- Patterns and Properties of Perfect Squares
- Perfect Squares and Odd Numbers
- Perfect Squares and Triangular Numbers; Square Roots
- Square Roots
- Square Roots by prime factorisation
- Cubic Numbers
- Taxicab Numbers
Quick revision
| Idea | Meaning | Test / rule | Example |
|---|---|---|---|
| Square number | n × n, written n2 | Factors split into two identical groups | 182 = 324 = (2×3×3)2 |
| Units digit | Last digit of a square | Only 0, 1, 4, 5, 6, 9 possible | 1027 ends in 7 → not a square |
| Terminal zeros | Zeros at the end of a square | Always an even count | 4002 = 160000 (4 zeros) |
| Odd-number sum | 1 + 3 + 5 + … + (2n–1) | Equals n2 | 1+3+5+7+9 = 25 = 52 |
| n-th odd number | 2n – 1 | 36th odd number = 71 | 352 + 71 = 362 = 1296 |
| Gap between squares | Count of numbers between n2 and (n+1)2 | 2n | Between 162 and 172: 32 numbers |
| Triangular link | Tn–1 + Tn | Equals n2 | 10 + 15 = 25 = 52 |
| Square root | x with x2 = y | √y; two integer roots, ± | √441 = 21 |
| Cube number | n × n × n, written n3 | Factors split into three identical groups | 3375 = (3×5)3 = 153 |
| Cube units digit | Last digit of a cube | All ten digits 0–9 occur | 23 = 8, 83 = 512 |
| Cube root | x with x3 = y | ∛y | ∛10648 = 22 |
| Successive differences | Repeated differencing | Constant at level 2 for squares, level 3 for cubes | Cubes: 6, 6, 6, … |
Exercises
- Queen Ratnamanjuri's puzzle — In-text Questions Page 1
- Factors and partner factors — In-text Questions Page 2
- In-text Questions — Square Numbers Page 3
- Patterns and Properties of Perfect Squares — In-text Questions Page 4
- Patterns and Properties of Perfect Squares — In-text Questions Page 5
- Perfect Squares and Odd Numbers — In-text Questions Page 6
- Perfect Squares and Triangular Numbers; Square Roots — In-text Questions Page 7
- Square Roots — In-text Questions Page 8
- Square Roots by prime factorisation — In-text Questions Page 9
- Figure it Out — Square Numbers Page 10
- In-text Questions — Cubic Numbers Page 11–12
- In-text Questions — Cubic Numbers Page 13
- Taxicab Numbers — Try This Page 13
- Perfect Cubes and Consecutive Odd Numbers; Cube Roots — In-text Questions Page 14
- Cube Roots; Successive Differences — In-text Questions Page 15
- In-text Questions — A Pinch of History Page 16
- Figure it Out — A Pinch of History Page 16
- Square Pairs! — Puzzle Time — Square Pairs! Page 18