Class 8 Maths Ganita Prakash Chapter 1 Solutions (NCERT 2026–27) – A Square and A Cube

These Class 8 Maths Ganita Prakash Chapter 1 solutions cover A Square and A Cube from the new NCF-2023 textbook (Reprint 2026–27). Every Figure it Out question is solved step by step, with worked square roots, cube roots, prime factorisations and number patterns, so you can master the chapter and revise it quickly.

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part I) Chapter: 1 Exercises: Figure it Out (p. 10–11), Figure it Out (p. 16–17) Session: 2026–27
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Chapter 1 Overview

Chapter 1 of Ganita Prakash, A Square and A Cube, opens with Queen Ratnamanjuri’s locker puzzle and uses it to discover why only square numbers have an odd number of factors. From there it builds the ideas of perfect squares (their units digits, the link with consecutive odd numbers, triangular numbers and parity), square roots (by estimation and prime factorisation), perfect cubes and cube roots, the famous taxicab number 1729, and a pinch of history of the Sanskrit words varga, ghana and mula. The Class 8 Maths Ganita Prakash Chapter 1 solutions below work through every Figure it Out, Math Talk and Try This question step by step.

Key Concepts & Definitions

Square number: a number that can be written as the product of a number with itself, e.g. 1, 4, 9, 16, 25, … In general n × n = n2.

Perfect square: the square of a natural number. Every perfect square ends in 0, 1, 4, 5, 6 or 9 (never 2, 3, 7 or 8) and has an even number of zeros at the end.

Square root: the inverse of squaring — if y = x2, then x is a square root of y. The symbol √y means the principal, non-negative square root. For example, √49 = 7, while the equation x2 = 49 has solutions x = 7 and x = −7. Also, √0 = 0 and √(x2) = |x|.

Cube number / perfect cube: a number obtained by multiplying a number by itself three times, e.g. 1, 8, 27, 64, … In general n × n × n = n3.

Cube root: the inverse of cubing — if y = x3, then x is the cube root of y, written 3√y, e.g. 3√8 = 2.

Odd-factor fact: among positive integers, only square numbers have an odd number of factors, because one factor pairs with itself.

Taxicab number: a number expressible as a sum of two positive cubes in two different ways; the smallest is 1729 = 13 + 123 = 93 + 103 (the Hardy–Ramanujan number).

Important Formulas & Patterns (Chapter 1)

Square & cube: n2 = n × n  •  n3 = n × n × n.

Square as a sum of odd numbers: n2 = 1 + 3 + 5 + … + (2n − 1). The k-th odd number is 2k − 1.

Step from one square to the next: (n + 1)2 = n2 + (2n + 1).

Numbers between two squares: between n2 and (n + 1)2 there are exactly 2n whole numbers.

Square root by prime factorisation: split prime factors into two identical groups; the product of one group is √n.

Cube root by prime factorisation: split prime factors into three identical groups; the product of one group is 3√n.

Cubes as a sum of odd numbers: 13 = 1, 23 = 3 + 5, 33 = 7 + 9 + 11, … (consecutive odd numbers).

Figure it Out — Square Numbers & Square Roots (Page 10–11)

1. Which of the following numbers are not perfect squares? (i) 2032   (ii) 2048   (iii) 1027   (iv) 1089

SOLUTION A perfect square can end only in 0, 1, 4, 5, 6 or 9. (i) 2032 ends in 2 → not a perfect square. (ii) 2048 ends in 8 → not a perfect square. (iii) 1027 ends in 7 → not a perfect square. (iv) 1089 ends in 9 (a valid units digit) and 332 = 1089, so it is a perfect square. ∴ (i), (ii) and (iii) are not perfect squares.

2. Which one among 642, 1082, 2922, 362 has last digit 4?

SOLUTION The units digit of a square depends only on the units digit of the number: 42 = 16 (ends 6), 82 = 64 (ends 4), 22 = 4 (ends 4), 62 = 36 (ends 6). 642 ends in 6; 1082 ends in 4; 2922 ends in 4; 362 ends in 6. ∴ 1082 and 2922 have last digit 4.

3. Given 1252 = 15625, what is the value of 1262? (i) 15625 + 126   (ii) 15625 + 262   (iii) 15625 + 253   (iv) 15625 + 251   (v) 15625 + 512

SOLUTION Use (n + 1)2 = n2 + (2n + 1) with n = 125: 1262 = 1252 + (2 × 125 + 1) = 15625 + 251. ∴ the correct option is (iv) 15625 + 251 (= 15876).

4. Find the length of the side of a square whose area is 441 m2.

SOLUTION Side = √(area) = √441. 441 = 3 × 3 × 7 × 7 = (3 × 7)2 = 212. ∴ side = 21 m.
A square of area 441 square metres has side length 21 metres.
Area is measured in square metres; the side length is measured in metres.

5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.

SOLUTION First find the LCM of 4, 9 and 10. LCM = 22 × 32 × 5 = 180. For a perfect square every prime must appear an even number of times. In 180 = 22 × 32 × 5, the prime 5 appears once, so multiply by 5. Required number = 180 × 5 = 900, and 900 = 302. It is the smallest square divisible by 4, 9 and 10.

6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.

SOLUTION Prime factorise: 9408 = 26 × 3 × 72. Pairing the primes: 26 (three pairs) and 72 (one pair) are paired, but 3 is unpaired. So multiply by 3: product = 9408 × 3 = 28224 = 26 × 32 × 72. √28224 = 23 × 3 × 7 = 8 × 3 × 7 = 168.
Prime factors of 9408 form three pairs of 2 and one pair of 7; add a 3 to complete its pair.
The smallest positive multiplier is 3. Taking one factor from each pair gives the square root 168.

7. How many numbers lie between the squares of the following numbers? (i) 16 and 17   (ii) 99 and 100

SOLUTION Between n2 and (n + 1)2 there are exactly 2n whole numbers. (i) Between 162 = 256 and 172 = 289: 2 × 16 = 32 numbers. (ii) Between 992 = 9801 and 1002 = 10000: 2 × 99 = 198 numbers.

8. In the following pattern, fill in the missing numbers: 12 + 22 + 22 = 32 22 + 32 + 62 = 72 32 + 42 + 122 = 132 42 + 52 + 202 = (___)2 92 + 102 + (___)2 = (___)2

SOLUTION Look at each row: the third base is the product of the first two bases, and the right-hand base is (that product + 1). For bases n and n + 1, n × (n + 1) is the third number and n(n + 1) + 1 is the answer. Row 4 (n = 4): third number = 4 × 5 = 20, answer = 20 + 1 = 21. So 42 + 52 + 202 = 212 (16 + 25 + 400 = 441 = 212). ✓ Row 5 (n = 9): third number = 9 × 10 = 90, answer = 90 + 1 = 91. So 92 + 102 + 902 = 912 (81 + 100 + 8100 = 8281 = 912). ✓

9. How many tiny squares are there in the picture? Write the prime factorisation of the number of tiny squares.

SOLUTIONThere are 9 rows of 9 groups, so 9 × 9 = 81 groups.Each group, whether upright or tilted, contains a 5 × 5 grid of 25 tiny squares.Total = 81 × 25 = 2025 tiny squares.Prime factorisation: 2025 = 3 × 3 × 3 × 3 × 5 × 5 = 34 × 52. This counts the smallest cells shown, not every possible larger square.
Nine rows of nine groups, each with a 5 by 5 grid, contain 2025 tiny squares in total.
Textbook page 11, Question 9: 9 × 9 × 5 × 5 = 2025 = 3⁴ × 5². Rotating a group does not change its count.

Math Talk & Try This — Answered

These are the in-text reflective and short tasks in the chapter; the determinate ones are answered, the open ones are guided.

Math Talk — Units digit of squares Which of 382, 342, 462, 562, 742 and 822 have 6 in the units place? Answer. A square ends in 6 when its base ends in 4 or 6. Therefore 342, 462, 562 and 742 end in 6. The squares of 38 and 82 end in 4. A possible last digit alone does not prove that an arbitrary number is a square: 26 is a counterexample.
Math Talk — Zeros at the end of a square If a number ends in 3 zeros, how many zeros does its square end in? What is the link between the zeros of a number and of its square? Answer. A number with 3 trailing zeros is (a number) × 103; squaring gives 106, i.e. 6 zeros. In general a number with k trailing zeros has a square with 2k trailing zeros. So a perfect square can end only in an even number of zeros — this always happens.
Math Talk — Parity What can you say about the parity of a number and of its square? Answer. The square of an even number is even and the square of an odd number is odd — squaring preserves parity, because even × even is even and odd × odd is odd.
Try This — Numbers not squares by units digit Write 5 numbers that you can tell are not squares just by looking at the units digit. Answer. Any numbers ending in 2, 3, 7 or 8 work, e.g. 32, 53, 87, 128 and 2043 — no perfect square ends in 2, 3, 7 or 8.
Math Talk — Numbers between consecutive squares How many numbers lie between two consecutive perfect squares? Do you notice a pattern? Answer. Yes — between n2 and (n + 1)2 there are exactly 2n numbers (e.g. between 32 = 9 and 42 = 16 there are 6 numbers: 10, 11, 12, 13, 14, 15).
Try This — Cube ending in two zeros Can a cube end with exactly two zeros (00)? Explain. Answer. No. Trailing zeros come from factors of 10 = 2 × 5. In a cube each prime appears a multiple of 3 times, so the number of trailing zeros is always a multiple of 3 (0, 3, 6, …). Exactly two zeros is impossible.
Math Talk — Cubes as a sum of consecutive odd numbers Later in the odd-number pattern for cubes we reach 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109. Can you tell what this sum is without doing the calculation? Answer. This is the tenth group in the pattern: it starts at 102 − 10 + 1 = 91 and ends at 102 + 10 − 1 = 109. There are 10 terms, with average (91 + 109)/2 = 100. The sum is therefore 10 × 100 = 1000 = 103. Ten arbitrary consecutive odd numbers need not have this sum.
Successive groups of 1, 2, 3, 4 and 10 odd numbers sum to their corresponding cubes; the tenth group runs from 91 to 109.
The tenth row has 10 terms with average 100, so its sum is 1000. Any arbitrary ten consecutive odd numbers need not sum to 1000.
Try This — Cube roots by factorisation Find the cube roots of these numbers: 64, 512, 729. Answer. 64 = 26 = (2 × 2)3, so ∛64 = 4. Also, 512 = 29 = (2 × 2 × 2)3, so ∛512 = 8; 729 = 36 = (3 × 3)3, so ∛729 = 9. Each group of three identical prime factors supplies one factor of the cube root.
Try This — Taxicab numbers Express 4104 and 13832 as sums of two positive cubes in two different ways. Answer. 4104 = 23 + 163 = 93 + 153, because 8 + 4096 = 729 + 3375 = 4104.
13832 = 23 + 243 = 183 + 203, because 8 + 13824 = 5832 + 8000 = 13832. The chapter’s first example is 1729 = 13 + 123 = 93 + 103.

Figure it Out — Cube Numbers & Cube Roots (Page 16–17)

1. Find the cube roots of 27000 and 10648.

SOLUTION 27000 = 27 × 1000 = 33 × 103 = (3 × 10)3 = 303, so 3√27000 = 30. 10648 = 23 × 113 = (2 × 11)3 = 223, so 3√10648 = 22.

2. What number will you multiply by 1323 to make it a cube number?

SOLUTION Prime factorise: 1323 = 33 × 72. For a perfect cube every prime must appear a multiple of 3 times. Here 33 is complete, but 7 appears only twice, so we need one more 7. Multiply by 7: 1323 × 7 = 9261 = 33 × 73 = 213, a perfect cube.
1323 contains three factors of 3 and two factors of 7; adding a 7 makes two complete triplets.
The smallest positive multiplier is 7. Other multipliers, such as 56 = 7 × 2³, also make a cube.

3. State true or false. Explain your reasoning. (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may be a 3-digit number. (iv) The cube of a 2-digit number may have seven or more digits. (v) Cube numbers have an odd number of factors.

SOLUTION (i) False. odd × odd × odd is odd (e.g. 33 = 27), so the cube of an odd number is odd. (ii) False. 23 = 8 and 123 = 1728 both end in 8; cubes can end in any digit 0–9. (iii) False. The smallest 2-digit number is 10 and 103 = 1000 (4 digits), so the cube of a 2-digit number always has at least 4 digits — never just 3. (iv) False. The largest 2-digit number is 99 and 993 = 970299 (6 digits), so a 2-digit cube has at most 6 digits, never seven or more. (v) False. A positive integer has an odd number of factors exactly when it is a square. Some cubes, such as 64 = 82 = 43, are also squares, but not all cubes are. For example 27 has the factors 1, 3, 9, 27 — an even number (4) of factors.

4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

SOLUTION Method (for a known perfect cube with a two-digit cube root): the units digit of a cube fixes the units digit of its root (1→1, 8→2, 7→3, 4→4, 5→5, 6→6, 3→7, 2→8, 9→9, 0→0). The tens digit of the root comes from the part of the number to the left of the last three digits. 1331: ends in 1 → root ends in 1; left part “1” lies in 13, so tens digit 1 → 3√1331 = 11. 4913: ends in 3 → root ends in 7; left part “4” lies between 13=1 and 23=8, so tens digit 1 → 3√4913 = 17. 12167: ends in 7 → root ends in 3; left part “12” lies between 23=8 and 33=27, so tens digit 2 → 3√12167 = 23. 32768: ends in 8 → root ends in 2; left part “32” lies between 33=27 and 43=64, so tens digit 3 → 3√32768 = 32.
Separate 32768 into 32 and 768; 27 is the largest cube not exceeding 32, while the last digit 8 gives root units digit 2.
The two clues give 32; confirm by cubing: 32³ = 32768.

5. Which of the following is the greatest? Explain your reasoning. (i) 673 − 663   (ii) 433 − 423   (iii) 672 − 662   (iv) 432 − 422

SOLUTION Use the identities a3 − b3 = (a − b)(a2 + ab + b2) and a2 − b2 = (a − b)(a + b). Here a − b = 1 in every part. (i) 673 − 663 = 672 + 67 × 66 + 662 = 4489 + 4422 + 4356 = 13267. (ii) 433 − 423 = 432 + 43 × 42 + 422 = 1849 + 1806 + 1764 = 5419. (iii) 672 − 662 = 67 + 66 = 133. (iv) 432 − 422 = 43 + 42 = 85. ∴ (i) 673 − 663 is the greatest — the cube differences dominate, and the larger bases (67, 66) make (i) bigger than (ii).

Textbook activities and visual explanations

Locker puzzle and passcode (pages 1–3)

Locker n is toggled once for each positive factor of n. Most factors occur in distinct pairs, but a square has one unpaired middle factor. Thus the open lockers are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. The first five lockers toggled exactly twice are the first five primes, giving the passcode 2–3–5–7–11.

A 10 by 10 locker chart highlights 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100.
An odd number of toggles leaves a locker open. Exactly the square-numbered lockers stay open.

For 36, the pairs are 1 × 36, 2 × 18, 3 × 12, 4 × 9 and 6 × 6. Its nine distinct positive factors are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Five factor pairs of 36 show the repeated factor 6, giving nine distinct factors.
For a positive integer, factors normally form pairs. In a square, the square root pairs with itself.

Square table and last-digit patterns (pages 3–5)

nn²nn²nn²
111112121441
241214422484
391316923529
4161419624576
5251522525625
6361625626676
7491728927729
8641832428784
9811936129841
101002040030900

The next squares ending in 1 after 29² are 31² = 961 and 39² = 1521. Bases ending in 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 produce square last digits 0, 1, 4, 9, 6, 5, 6, 9, 4, 1 respectively.

Squaring also works for non-integers: a square of side 3/5 has area 9/25 square units; side 2.5 gives area 6.25 square units. These are squares of numbers, but not perfect squares of integers.

Odd-number layers and subtraction (pages 5–6)

A 5 by 5 square is divided into L-shaped layers of 1, 3, 5, 7 and 9 cells.
Adding the next odd-number layer changes n² into (n + 1)².

The nth positive odd number is 2n − 1. Thus 36² = 35² + 71 = 1225 + 71 = 1296. Subtracting 1, 3, 5, 7, 9 from 25 reaches zero in five steps, so √25 = 5. Starting from 38 instead gives 37, 34, 29, 22, 13, 2, −11; crossing zero shows that 38 is not a perfect square.

Square counts in blocks of 100 (page 7)

Inclusive intervalSquare basesNumber of squares
1–1001–1010
101–20011–144
201–30015–173
301–40018–203
401–50021–222
501–60023–242
601–70025–262
701–80027–282
801–90029–302
901–1000311

The largest square less than 1000 is 31² = 961; the next is 32² = 1024.

Triangular numbers (page 7)

The pattern continues with 10 + 15 = 25 = 5². In general, n(n − 1)/2 + n(n + 1)/2 = n².

Dot squares show 1 plus 3 equals 4, 3 plus 6 equals 9, 6 plus 10 equals 16, and 10 plus 15 equals 25.
The next diagram has 10 blue or gold dots in one triangle and 15 in the other: 10 + 15 = 25.

Square roots and estimation (pages 7–10)

A square of area 49 cm² has side 7 cm. Also, √64 = 8; the two solutions of x² = 64 are ±8. Since 729 = 27², subtracting the first 27 odd numbers (1 through 53) from 729 reaches zero, so √729 = 27.

1156 = 2² × 17² = 34², so it is a perfect square. In 2800 = 2⁴ × 5² × 7, the factor 7 is unpaired, so 2800 is not a perfect square. For √1936, first locate it between 40 and 50; its last digit is 4 or 6. Since 45² = 2025 exceeds 1936, test 44: 44² = 1936.

On a number line from 15 to 16, the square root of 250 is about 15.81, closer to 16.
Comparing areas first places the square root between two consecutive whole numbers.

Akhil’s cloth has area 125 cm². A 15 cm square needs 225 cm², so it cannot be cut from this cloth. Since 11² = 121 ≤ 125 < 144 = 12², the largest permitted integer side is 11 cm.

Unit cubes and the cube table (pages 11–13)

Cubes with side lengths 2, 3 and 4 units contain 8, 27 and 64 unit cubes.
For side n, each layer has n² unit cubes and there are n layers, giving n³ unit cubes.
nn³nn³
11111331
28121728
327132197
464142744
5125153375
6216164096
7343174913
8512185832
9729196859
101000208000

Positive one-digit cubes: 1, 8 (2 numbers). Two-digit cubes: 27, 64 (2). Three-digit cubes: 125, 216, 343, 512, 729 (5). The cube of an odd integer is odd, and the cube of an even integer is even. Every units digit from 0 to 9 can occur in a cube.

A cube of side 4 units contains 64 unit cubes. Negative and fractional numbers can also be cubed: (−6)³ = −216; (4/6)³ = (2/3)³ = 8/27.

Successive differences (page 15)

A difference table for cubes 1, 8, 27, 64, 125, 216 gives third differences 6, 6, 6.
Subtract each entry from the entry to its right. Repeating three times gives the constant 6.

Consecutive squares have constant second difference 2. Consecutive cubes have constant third difference 6, after first differences 7, 19, 37, 61, 91 and second differences 12, 18, 24, 30.

Square Pairs! (page 18)

Arrange 1–17 in a row

A valid row is 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17.
Read across the first line and then the second. The join is 12 + 13 = 25. Reversing the row also works.

A valid order is 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17. Reverse it for the same row in the opposite direction.

There is only one row up to reversal. The numbers 16 and 17 must be endpoints: their only partners are 9 and 8. The degree-two partner constraints force 17–8–1–15–10–6–3–13–12–4–5–11–14–2–7–9–16. Every number other than 1 and 3 has at most two possible partners, so both partners of each non-endpoint are forced. This gives the displayed full chain; 1 and 3 already have two neighbours in it, so the optional link 1–3 cannot be used.

Arrange 1–32 in a circle

A circular arrangement of 1 through 32 has a perfect-square sum at every pair of neighbours, including 15 plus 1 at the closing edge.
Start at 1 at the top and follow clockwise. Every link, including the final link back to 1, gives a perfect square.

One clockwise order is 1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 31, 18, 7, 29, 20, 16, 9, 27, 22, 14, 2, 23, 26, 10, 15, returning to 1. Check neighbouring sums: 9, 36, 49, 25, 36, 49, 36, 49, 36, 9, 16, 25, 36, 49, 36, 16, 36, 49, 25, 36, 49, 36, 25, 36, 49, 36, 16, 25, 49, 36, 25, 16. Each is a perfect square, including the closing sum 15 + 1 = 16.

Common Mistakes to Avoid

Watch out for these

  • Thinking a units digit of 0, 1, 4, 5, 6 or 9 proves a number is a square — it only tells you when a number is not a square (e.g. 26 ends in 6 but is not a square).
  • For the smallest-multiplier questions, forgetting to prime factorise first — supply only the primes needed to complete pairs (squares) or triplets (cubes).
  • Mixing up the tests: two identical groups for a square root, three identical groups for a cube root.
  • Saying every cube has an odd number of factors. A positive cube has an odd factor count only if it is also a square, as 64 is.
  • Assuming a cube can end in any even count of zeros; a perfect cube’s trailing zeros are always a multiple of 3.
  • In “numbers between squares”, counting the endpoints — the count strictly between n2 and (n + 1)2 is 2n.

Practice MCQs & Assertion–Reason

1. Which of these is a perfect square?

(a) 1024    (b) 1027    (c) 2048    (d) 2032

2. A perfect square can never end in:

(a) 4    (b) 6    (c) 7    (d) 9

3. How many numbers lie between 202 and 212?

(a) 20    (b) 40    (c) 41    (d) 21

4. The square root of 1296 is:

(a) 34    (b) 36    (c) 38    (d) 46

5. Using 1252 = 15625, the value of 1242 is:

(a) 15625 − 249    (b) 15625 − 124    (c) 15625 + 249    (d) 15625 − 250

6. Which number has an odd number of factors?

(a) 12    (b) 18    (c) 36    (d) 40

7. The cube root of 1728 is:

(a) 8    (b) 12    (c) 14    (d) 18

8. The smallest number that 1323 must be multiplied by to become a perfect cube is:

(a) 3    (b) 7    (c) 9    (d) 21

9. The smallest taxicab number (Hardy–Ramanujan number) is:

(a) 1024    (b) 1331    (c) 1728    (d) 1729

10. The number of trailing zeros in a perfect cube is always:

(a) odd    (b) a multiple of 2    (c) a multiple of 3    (d) exactly 2

Answer key: 1-(a), 2-(c), 3-(b), 4-(b), 5-(a), 6-(c), 7-(b), 8-(b), 9-(d), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 2048 is not a perfect square.

Reason: A perfect square cannot end in the digit 8.

A-R 2. Assertion: Only square numbers have an odd number of factors.

Reason: In a square, one factor pairs with itself, leaving an odd factor count.

A-R 3. Assertion: The cube of every odd number is even.

Reason: The product of three odd numbers is odd.

A-R 4. Assertion: 1729 = 13 + 123 = 93 + 103.

Reason: 1729 is the smallest number expressible as a sum of two positive cubes in two different ways.

A-R 5. Assertion: The smallest positive integer multiplier that makes 1323 a perfect cube is 7.

Reason: 1323 = 33 × 72, and the factor 7 needs one more to form a triplet.

Answer key: 1-(A), 2-(A), 3-(D), 4-(B), 5-(A).

Quick Revision Summary

  • A square number is n × n = n2; squares of natural numbers are perfect squares.
  • Perfect squares end only in 0, 1, 4, 5, 6 or 9, and have an even number of trailing zeros.
  • Only square numbers have an odd number of factors (one factor pairs with itself).
  • Every square is the sum of consecutive odd numbers from 1; (n + 1)2 = n2 + (2n + 1).
  • Between n2 and (n + 1)2 there are exactly 2n numbers.
  • Square root: split prime factors into two identical groups; cube root: split into three identical groups.
  • A cube number is n3; its trailing-zero count is a multiple of 3; 1729 is the smallest taxicab number.

How to score full marks in this chapter

Always prime factorise before answering smallest-multiplier and root questions, and write the factors in pairs (for squares) or triplets (for cubes). For “is it a square/cube” checks, state the units-digit or trailing-zero rule you are using. Use (n + 1)2 = n2 + 2n + 1 and the identity a3 − b3 = (a − b)(a2 + ab + b2) to avoid long multiplication, and keep your working tidy so each step earns its mark.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 1 about?

Chapter 1, A Square and A Cube, covers square numbers and their properties, square roots, perfect cubes and cube roots, finding roots by estimation and prime factorisation, number patterns (odd numbers, triangular numbers) and the famous taxicab number 1729.

How many Figure it Out exercises are there in Chapter 1?

There are two main “Figure it Out” sets — one on square numbers and square roots (pages 10–11) and one on cube numbers and cube roots (pages 16–17) — plus several Math Talk and Try This tasks, with worked explanations on this page.

Why do only square numbers have an odd number of factors?

Factors usually come in partner pairs whose product is the number. For a perfect square, one factor is the square root, which pairs with itself, so it is counted once — making the total number of factors odd.

Are these Class 8 Maths Ganita Prakash Chapter 1 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash (Part I) textbook for the 2026–27 session, with worked answers checked against the textbook and mathematical calculations.

Textbook reference: NCERT Ganita Prakash, Chapter 1 (official PDF). ClearStudy diagrams and worked explanations are original teaching aids.

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