A Square and A Cube

Concept: A perfect square can only end in the digits 0, 1, 4, 5, 6, or 9. A number ending in 2, 3, 7, or 8 is never a perfect square.

(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 (possible perfect square). Check: $33^2 = 1089$. ✓ → Is a perfect square.

Answer: (i) 2032, (ii) 2048, and (iii) 1027 are not perfect squares.

Concept: The last digit of a square depends only on the last digit of the base number.

- $64^2$: last digit of base = 4 → last digit of square = $4 \times 4 = 16$ → last digit 6
- $108^2$: last digit of base = 8 → last digit of square = $8 \times 8 = 64$ → last digit 4 ✓
- $292^2$: last digit of base = 2 → last digit of square = $2 \times 2 = 4$ → last digit 4 ✓
- $36^2$: last digit of base = 6 → last digit of square = $6 \times 6 = 36$ → last digit 6

Both $108^2$ and $292^2$ end in 4. Among the options, $108^2$ and $292^2$ have last digit 4.

Answer: $108^2$ and $292^2$ have last digit 4.

Concept: We use the identity
$$n^2 - (n-1)^2 = 2n - 1$$
or equivalently
$$(n+1)^2 = n^2 + 2n + 1$$

Here $n = 125$, so:
$$126^2 = 125^2 + 2(125) + 1 = 15625 + 250 + 1 = 15625 + 251$$

Verification: $15625 + 251 = 15876$ and $126^2 = 15876$ ✓

Answer: Option (iv) $15625 + 251$

Given: Area of square $= 441\text{ m}^2$

Formula: Side of square $= \sqrt{\text{Area}}$

Prime factorisation of 441:
$$441 = 3 \times 147 = 3 \times 3 \times 49 = 3^2 \times 7^2$$

Square root:
$$\sqrt{441} = \sqrt{3^2 \times 7^2} = 3 \times 7 = 21$$

Answer: The length of the side of the square is $\mathbf{21\text{ m}}$.

Step 1: Find the LCM of 4, 9, and 10.
$$4 = 2^2,\quad 9 = 3^2,\quad 10 = 2 \times 5$$
$$\text{LCM} = 2^2 \times 3^2 \times 5 = 180$$

Step 2: For a number to be a perfect square, every prime factor must appear an even number of times.

Prime factorisation of $180 = 2^2 \times 3^2 \times 5^1$.

The factor $5$ appears only once (odd power). Multiply by $5$ to make it even:
$$180 \times 5 = 900 = 2^2 \times 3^2 \times 5^2$$

Verification: $\sqrt{900} = 30$ ✓, and $900$ is divisible by 4, 9, and 10 ✓.

Answer: The smallest such perfect square is $\mathbf{900}$.

Step 1: Prime factorisation of 9408.
$$9408 \div 2 = 4704$$
$$4704 \div 2 = 2352$$
$$2352 \div 2 = 1176$$
$$1176 \div 2 = 588$$
$$588 \div 2 = 294$$
$$294 \div 2 = 147$$
$$147 \div 3 = 49$$
$$49 \div 7 = 7$$
$$7 \div 7 = 1$$

$$9408 = 2^6 \times 3^1 \times 7^2$$

Step 2: For a perfect square, all prime factors must have even exponents.
- $2^6$ → even ✓
- $3^1$ → odd ✗ (need one more 3)
- $7^2$ → even ✓

Step 3: Multiply by $3$:
$$9408 \times 3 = 28224 = 2^6 \times 3^2 \times 7^2$$

Step 4: Square root of the product:
$$\sqrt{28224} = 2^3 \times 3 \times 7 = 8 \times 3 \times 7 = 168$$

Answer: The smallest multiplier is $\mathbf{3}$, and the square root of the product is $\mathbf{168}$.

Concept: Between the squares of two consecutive integers $n$ and $(n+1)$, the number of integers lying strictly between them is:
$$n^2 + 1,\ n^2 + 2,\ \ldots,\ (n+1)^2 - 1$$
Count $= (n+1)^2 - n^2 - 1 = 2n + 1 - 1 = 2n$

(i) Between $16^2$ and $17^2$:
$$16^2 = 256,\quad 17^2 = 289$$
Numbers between them $= 289 - 256 - 1 = 32$

Using formula: $2 \times 16 = 32$ ✓

Answer: 32 numbers lie between $16^2$ and $17^2$.

(ii) Between $99^2$ and $100^2$:
$$99^2 = 9801,\quad 100^2 = 10000$$
Numbers between them $= 10000 - 9801 - 1 = 198$

Using formula: $2 \times 99 = 198$ ✓

Answer: 198 numbers lie between $99^2$ and $100^2$.

Observing the pattern:

Row 1: $1^2 + 2^2 + (1\times2)^2 = (1\times2+1)^2$, i.e., $n=1$: $1^2+2^2+2^2=3^2$
Row 2: $n=2$: $2^2+3^2+6^2=7^2$ (third term $= 2\times3=6$, RHS $= 2\times3+1=7$)
Row 3: $n=3$: $3^2+4^2+12^2=13^2$ (third term $= 3\times4=12$, RHS $= 3\times4+1=13$)

General pattern: $n^2 + (n+1)^2 + [n(n+1)]^2 = [n(n+1)+1]^2$

Row 4 ($n=4$):
$$4^2 + 5^2 + (4\times5)^2 = (4\times5+1)^2$$
$$4^2 + 5^2 + 20^2 = 21^2$$
Verification: $16 + 25 + 400 = 441 = 21^2$ ✓

Missing number: $\mathbf{21}$

Row for $n=9$:
$$9^2 + 10^2 + (9\times10)^2 = (9\times10+1)^2$$
$$9^2 + 10^2 + 90^2 = 91^2$$
Verification: $81 + 100 + 8100 = 8281 = 91^2$ ✓

Missing numbers: $\mathbf{90}$ and $\mathbf{91}$

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.

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(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.

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(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.

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(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.

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(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: $11^3 = 1331$)

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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: $17^3 = 4913$)

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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: $23^3 = 12167$)

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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 $13267$.