A Square and A Cube — NCERT Solutions

Note: The figure is not visible in the OCR. Based on the context of the chapter (squares and prime factorisation), a common version of this problem uses a $12 \times 12$ grid.

Assumed grid: $12 \times 12$

Total tiny squares $= 12 \times 12 = 144$

Prime factorisation of 144:
$$144 = 2 \times 72 = 2 \times 2 \times 36 = 2 \times 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 2 \times 9 = 2^4 \times 3^2$$

Answer: There are $\mathbf{144}$ tiny squares, and its prime factorisation is $2^4 \times 3^2$.

Given: A cube of side 3 cm is to be filled with unit cubes (side 1 cm).

Volume of big cube $= 3^3 = 27\text{ cm}^3$

Volume of each small cube $= 1^3 = 1\text{ cm}^3$

Number of small cubes $= \dfrac{27}{1} = 27$

Answer: $\mathbf{27}$ cubes of side 1 cm are needed to make a cube of side 3 cm.

Cube root of 27000:

Prime factorisation:
$$27000 = 27 \times 1000 = 3^3 \times 10^3 = 3^3 \times 2^3 \times 5^3$$

$$\sqrt[3]{27000} = 3 \times 2 \times 5 = \mathbf{30}$$

Cube root of 10648:

Prime factorisation:
$$10648 \div 2 = 5324,\quad 5324 \div 2 = 2662,\quad 2662 \div 2 = 1331$$
$$1331 \div 11 = 121,\quad 121 \div 11 = 11,\quad 11 \div 11 = 1$$
$$10648 = 2^3 \times 11^3$$

$$\sqrt[3]{10648} = 2 \times 11 = \mathbf{22}$$

Step 1: Prime factorisation of 1323.
$$1323 \div 3 = 441,\quad 441 \div 3 = 147,\quad 147 \div 3 = 49,\quad 49 \div 7 = 7,\quad 7 \div 7 = 1$$
$$1323 = 3^3 \times 7^2$$

Step 2: For a perfect cube, every prime factor must appear a multiple of 3 times.

• $3^3$ → already a perfect cube ✓
• $7^2$ → needs one more 7 to become $7^3$ ✗

Step 3: Multiply by $7$:
$$1323 \times 7 = 9261 = 3^3 \times 7^3 = (3 \times 7)^3 = 21^3$$

Answer: Multiply 1323 by $\mathbf{7}$ to get the perfect cube $9261 = 21^3$.

(i) The cube of any odd number is even.

False. Odd $\times$ Odd $\times$ Odd = Odd. For example, $3^3 = 27$, which is odd.

(ii) There is no perfect cube that ends with 8.

False. $2^3 = 8$ ends in 8. Also $12^3 = 1728$ ends in 8. The cube of any number ending in 2 ends in 8.

(iii) The cube of a 2-digit number may be a 3-digit number.

False. The smallest 2-digit number is 10, and $10^3 = 1000$, which is a 4-digit number. So the cube of any 2-digit number is at least 4 digits.

(iv) The cube of a 2-digit number may have seven or more digits.

False. The largest 2-digit number is 99, and $99^3 = 970299$, which has 6 digits. So the cube of a 2-digit number has at most 6 digits.

(v) Cube numbers have an odd number of factors.

False. Perfect squares have an odd number of factors (not cubes in general). For example, $8 = 2^3$ has factors 1, 2, 4, 8 — that is 4 factors (even). However, if a cube is also a perfect square (a perfect sixth power), it may have an odd number of factors. In general, cube numbers do not necessarily have an odd number of factors.

Method: Use the fact that the units digit of a cube determines the units digit of the cube root, and estimate the tens digit from the thousands group.

Cube root of 1331:

• Units digit of 1331 is 1 → units digit of cube root is 1 (since $1^3=1$).
• Remaining after removing last 3 digits: $1$ → $1^3 = 1$, so tens digit is 1.
• $\sqrt[3]{1331} = \mathbf{11}$ ✓ (Check: $11^3 = 1331$)

Cube root of 4913:

• Units digit is 3 → units digit of cube root is 7 (since $7^3 = 343$, ends in 3).
• Remaining group: $4$ → $1^3=1 \leq 4 < 8=2^3$, so tens digit is 1.
• $\sqrt[3]{4913} = \mathbf{17}$ ✓ (Check: $17^3 = 4913$)

Cube root of 12167:

• Units digit is 7 → units digit of cube root is 3 (since $3^3=27$, ends in 7).
• Remaining group: $12$ → $2^3=8 \leq 12 < 27=3^3$, so tens digit is 2.
• $\sqrt[3]{12167} = \mathbf{23}$ ✓ (Check: $23^3 = 12167$)

Cube root of 32768:

• Units digit is 8 → units digit of cube root is 2 (since $2^3=8$).
• Remaining group: $32$ → $3^3=27 \leq 32 < 64=4^3$, so tens digit is 3.
• $\sqrt[3]{32768} = \mathbf{32}$ ✓ (Check: $32^3 = 32768$)

Concept/Identities used:
$$a^2 - b^2 = (a+b)(a-b)$$
$$a^3 - b^3 = (a-b)(a^2+ab+b^2)$$

For consecutive integers $n$ and $n-1$ (so $a-b=1$):
$$n^2-(n-1)^2 = 2n-1$$
$$n^3-(n-1)^3 = n^2 + n(n-1)+(n-1)^2 = 3n^2-3n+1$$

(iii) $67^2 - 66^2$:
$$= 2(67)-1 = 134-1 = 133$$

(iv) $43^2 - 42^2$:
$$= 2(43)-1 = 86-1 = 85$$

(ii) $43^3 - 42^3$:
$$= 3(43)^2 - 3(43)+1 = 3(1849)-129+1 = 5547-129+1 = 5419$$

(i) $67^3 - 66^3$:
$$= 3(67)^2 - 3(67)+1 = 3(4489)-201+1 = 13467-201+1 = 13267$$

Comparison:
$$13267 > 5419 > 133 > 85$$

Answer: $\mathbf{67^3 - 66^3}$ (option i) is the greatest, with a value of $13267$.