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NCERT Solutions Ganita Prakash (Part 1) Chapter 1 Figure it Out — A Pinch of History

Book page 16 Updated on2026-09-05

Q1.
Find the cube roots of 27000 and 10648.
Answer

Factorise into triplets.

27000 = 27 × 1000 = 33 × 103 = (3 × 10)3 = 303
∛27000 = 30
10648 = 2 × 5324 = 2 × 2 × 2662 = 2 × 2 × 2 × 1331
1331 = 11 × 11 × 11
10648 = 23 × 113 = (2 × 11)3 = 223
∛10648 = 22
Why you could also guess these: 10648 ends in 8, so its cube root ends in 2. It lies between 8000 = 203 and 27000 = 303, so the root is between 20 and 30 — and 22 is the only number in that range ending in 2.
Q2.
What number will you multiply by 1323 to make it a cube number?
Answer

Factorise and see which prime falls short of a multiple of 3.

1323 = 3 × 441 = 3 × 3 × 147 = 3 × 3 × 3 × 49
= 33 × 72

The 3s already form a triplet; the 7s are one short. So multiply by 7.

1323 × 7 = 9261 = 33 × 73 = (3 × 7)3 = 213
∛9261 = 21
Why 7 is the smallest such multiplier: To make a cube, every exponent must become a multiple of 3. Here 7 has exponent 2, so it needs exactly one more 7 to reach 3. Nothing smaller can do it, and any larger multiplier would introduce fresh primes to fix.
Q3.
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.
Answer

All five are false. Here is why in each case.

StatementVerdictReason
(i)Cube of any odd number is evenFalseOdd × odd × odd is odd: 33 = 27, 73 = 343
(ii)No perfect cube ends with 8False23 = 8, 123 = 1728, 223 = 10648
(iii)Cube of a 2-digit number may have 3 digitsFalseThe smallest is 103 = 1000, already 4 digits
(iv)Cube of a 2-digit number may have 7 or more digitsFalseThe largest is 993 = 970299, only 6 digits
(v)Cube numbers have an odd number of factorsFalse8 has factors 1, 2, 4, 8 — four of them
Why (i) and (ii) fail: An odd number is 2k + 1, and any product of odd numbers is odd, so no cube of an odd number can be even. For (ii), the last digit of a cube is decided by the last digit of the base, and 2 cubed ends in 8 — in fact every number ending in 2 has a cube ending in 8.
Why (iii) and (iv) fail: Two-digit numbers run from 10 to 99, so their cubes run from 103 = 1000 to 993 = 970299. Every cube in that range has 4, 5 or 6 digits — never 3, never 7.
Why (v) fails: An odd number of factors is the mark of a square, not a cube. A cube n3 has an odd number of factors only when it is also a square, as with 64 = 43 = 82 (factors 1, 2, 4, 8, 16, 32, 64 — seven of them). For 8 or 27 the count is even.
Q4.
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.
Answer

Yes — use the last digit to fix the units digit, and the size to fix the tens digit.

CubeLast digit → units digit of rootRangeCube root
13311 → 11000 = 103, 8000 = 203 → root in 10–2011
49133 → 7root in 10–2017
121677 → 38000 = 203, 27000 = 303 → root in 20–3023
327688 → 227000 = 303, 64000 = 403 → root in 30–4032
Why the guess is certain, not lucky: Cubing sends each digit to a different last digit, so the last digit of a cube pins down the units digit of its root uniquely (1→1, 3→7, 7→3, 8→2). The size of the number then pins down the tens digit. Together these leave exactly one candidate.
Check it yourself: 113 = 1331, 173 = 4913, 233 = 12167, 323 = 32768. ✓
Q5.
Which of the following is the greatest? Explain your reasoning. (i) 67³ – 66³ (ii) 43³ – 42³ (iii) 67² – 66² (iv) 43² – 42²
Answer

(i) 673 – 663 is the greatest. Use the two difference formulas rather than the full powers.

a3 – (a – 1)3 = 3a2 – 3a + 1
a2 – (a – 1)2 = 2a – 1
ExpressionWorkingValue
(i)673 – 6633(672) – 3(67) + 1 = 13467 – 201 + 113267
(ii)433 – 4233(432) – 3(43) + 1 = 5547 – 129 + 15419
(iii)672 – 6622(67) – 1 = 134 – 1133
(iv)432 – 4222(43) – 1 = 86 – 185
Why you can see the answer before computing: A difference of consecutive cubes grows like 3a2, while a difference of consecutive squares grows only like 2a. So (i) and (ii) are far bigger than (iii) and (iv), and between the two cube differences the one with the larger a wins. Since 67 > 43, option (i) is greatest.
Check it yourself: 673 = 300763 and 663 = 287496, and 300763 – 287496 = 13267. ✓
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