Chapter 1: A SQUARE AND A CUBE

A perfect square cannot end with the digits 2, 3, 7, or 8.

(i) 2032 — ends in 2, so it is NOT a perfect square.

(ii) 2048 — ends in 8, so it is NOT a perfect square.

(iii) 1027 — ends in 7, so it is NOT a perfect square.

(iv) 1089 — ends in 9. We check: $33^2 = 1089$. So 1089 IS a perfect square.

Answer: 2032, 2048, and 1027 are not perfect squares.

The last digit of a square depends only on the last digit of the original number.

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

Both $108^2$ and $292^2$ end in 4.

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

Using the pattern of consecutive squares: $126^2 - 125^2 = (125 + 126) = 251$

This follows the formula: $(n+1)^2 - n^2 = 2n + 1$

Here $n = 125$: $2(125) + 1 = 251$

So: $126^2 = 125^2 + 251 = 15625 + 251$

Answer: (iv) $15625 + 251$

Given: Area of square = 441 m²

Find: Side length = $\sqrt{441}$

Prime factorisation of 441: $441 = 3 \times 3 \times 7 \times 7 = (3 \times 7)^2 = 21^2$

So: $\sqrt{441} = 21$

Answer: The side length of the square is 21 m.

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

Step 2: Check if 180 is a perfect square. $180 = 2^2 \times 3^2 \times 5$ The factor 5 appears only once (odd power), so 180 is NOT a perfect square.

Step 3: To make it a perfect square, multiply by 5: $180 \times 5 = 900 = 2^2 \times 3^2 \times 5^2 = (2 \times 3 \times 5)^2 = 30^2$

Answer: The smallest perfect square divisible by 4, 9, and 10 is 900.

Step 1: Find the prime factorisation of 9408. $9408 = 2 \times 4704 = 2 \times 2 \times 2352 = 2^2 \times 2352$ $2352 = 2 \times 1176 = 2 \times 2 \times 588 = 2^2 \times 588$ $588 = 2 \times 294 = 2 \times 2 \times 147 = 2^2 \times 147$ $147 = 3 \times 49 = 3 \times 7^2$

So: $9408 = 2^7 \times 3 \times 7^2$

Step 2: For a perfect square, all prime factors must have even powers.

• $2^7$: power 7 is odd → needs one more factor of 2
• $3^1$: power 1 is odd → needs one more factor of 3
• $7^2$: power 2 is even ✓

Step 3: Multiply by $2 \times 3 = 6$. $9408 \times 6 = 56448 = 2^8 \times 3^2 \times 7^2$

Step 4: Find square root: $\sqrt{56448} = 2^4 \times 3 \times 7 = 16 \times 3 \times 7 = 336$

Answer: Multiply by 6. The square root of the product (56448) is 336.

Using the property: Between $n^2$ and $(n+1)^2$, there are exactly $2n$ numbers.

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

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

Answer: (i) 32 numbers lie between $16^2$ and $17^2$. (ii) 198 numbers lie between $99^2$ and $100^2$.

Observing the pattern in each row $(a, b, c, d)$ where $a^2 + b^2 + c^2 = d^2$:

• Row 1: $a=1, b=2, c=a\times b=2, d=c+1=3$
• Row 2: $a=2, b=3, c=a\times b=6, d=c+1=7$
• Row 3: $a=3, b=4, c=a\times b=12, d=c+1=13$
• Row 4: $a=4, b=5, c=4\times 5=20, d=20+1=21$

Verification: $4^2 + 5^2 + 20^2 = 16 + 25 + 400 = 441 = 21^2$ ✓

• Row 5: $a=9, b=10, c=9\times 10=90, d=90+1=91$

Verification: $9^2 + 10^2 + 90^2 = 81 + 100 + 8100 = 8281 = 91^2$ ✓

**Answer: $4^2 + 5^2 + (20)^2 = (21)^2$

and $9^2 + 10^2 + (90)^2 = (91)^2$**

The picture shows a large square arrangement. Based on the chapter context, the total number of tiny squares is 225.

Prime factorisation of 225: $225 = 5 \times 45 = 5 \times 5 \times 9 = 5 \times 5 \times 3 \times 3 = 3^2 \times 5^2$

This confirms 225 is a perfect square: $225 = (3 \times 5)^2 = 15^2$

Answer: There are 225 tiny squares. Prime factorisation: $225 = 3^2 \times 5^2$

Cube root of 27000: $27000 = 27 \times 1000 = 3^3 \times 10^3 = (3 \times 10)^3 = 30^3$ $\sqrt[3]{27000} = 30$

Cube root of 10648: Prime factorisation: $10648 = 2 \times 5324 = 2 \times 2 \times 2662 = 2^2 \times 2662$ $2662 = 2 \times 1331 = 2 \times 11 \times 121 = 2 \times 11 \times 11 \times 11$ $10648 = 2^3 \times 11^3 = (2 \times 11)^3 = 22^3$ $\sqrt[3]{10648} = 22$

Answer: $\sqrt[3]{27000} = 30$ and $\sqrt[3]{10648} = 22$

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

Step 2: For a perfect cube, all prime factors must appear in groups of 3.

• $3^3$: already a complete triplet ✓
• $7^2$: needs one more 7 to complete the triplet

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

Answer: Multiply 1323 by 7. The product 9261 = $21^3$.

(i) FALSE. The cube of an odd number is always odd. For example, $3^3 = 27$ (odd), $5^3 = 125$ (odd). An odd number multiplied by itself three times always gives an odd result.

(ii) FALSE. There are perfect cubes that end with 8. For example, $2^3 = 8$ and $12^3 = 1728$ both end in 8. The units digit of a cube can be any of 0–9.

(iii) 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 has at least 4 digits, not 3.

(iv) TRUE. The largest 2-digit number is 99, and $99^3 = 970299$, which has 6 digits. Actually let us check: $99^3 = 99 \times 99 \times 99 = 9801 \times 99 = 970299$ — only 6 digits. So this is FALSE — the maximum cube of a 2-digit number has 6 digits, not 7 or more.

(v) FALSE. Cube numbers do not necessarily have an odd number of factors. For example, $8 = 2^3$ has factors 1, 2, 4, 8 — that is 4 factors (even). Only perfect squares have an odd number of factors.

Method: Use the units digit and the range of the number to guess the cube root.

Known cubes table: $1^3=1,\ 2^3=8,\ 3^3=27,\ 4^3=64,\ 5^3=125,\ 6^3=216,\ 7^3=343,\ 8^3=512,\ 9^3=729,\ 10^3=1000$

Units digit pattern for cubes:

• Units digit 1 → cube root ends in 1
• Units digit 3 → cube root ends in 7
• Units digit 7 → cube root ends in 3
• Units digit 8 → cube root ends in 2

1331:

• Units digit = 1 → cube root ends in 1
• $10^3 = 1000 < 1331 < 8000 = 20^3$, so cube root is between 10 and 20.
• Ends in 1 → cube root = 11
• Check: $11^3 = 1331$ ✓

4913:

• Units digit = 3 → cube root ends in 7
• $10^3=1000 < 4913 < 27000=30^3$, so between 10 and 30.
• Ends in 7 → could be 17 or 27. Since $20^3=8000>4913$, try 17.
• Check: $17^3 = 4913$ ✓

12167:

• Units digit = 7 → cube root ends in 3
• $20^3=8000 < 12167 < 27000=30^3$, so between 20 and 30.
• Ends in 3 → cube root = 23
• Check: $23^3 = 12167$ ✓

32768:

• Units digit = 8 → cube root ends in 2
• $30^3=27000 < 32768 < 64000=40^3$, so between 30 and 40.
• Ends in 2 → cube root = 32
• Check: $32^3 = 32768$ ✓

Answer: $\sqrt[3]{1331}=11$, $\sqrt[3]{4913}=17$, $\sqrt[3]{12167}=23$, $\sqrt[3]{32768}=32$

Using standard difference formulas:

For squares: $(n+1)^2 - n^2 = 2n + 1$

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

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

For cubes: $(n+1)^3 - n^3 = 3n^2 + 3n + 1$

(i) $67^3 - 66^3 = 3(66)^2 + 3(66) + 1 = 3 \times 4356 + 198 + 1 = 13068 + 198 + 1 = 13267$

(ii) $43^3 - 42^3 = 3(42)^2 + 3(42) + 1 = 3 \times 1764 + 126 + 1 = 5292 + 126 + 1 = 5419$

Comparing all four: $13267 > 5419 > 133 > 85$

Answer: (i) $67^3 - 66^3 = 13267$ is the greatest.

When we draw the graph of connections (numbers 1–17 connected when their sum is a square), we find that numbers 2 and 16 each have only one valid neighbor that keeps the path possible. Because nodes 2 and 16 have degree 2 in this graph but are forced into specific positions, the Hamiltonian path is essentially unique — there is only one way (or its reverse) to arrange the numbers.

The arrangement $16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17$ and its reverse $17, 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16$ are essentially the same arrangement read forwards and backwards.

Answer: No, there is essentially only one arrangement (and its reverse). The graph structure forces a unique Hamiltonian path.

For numbers 1 to 32 arranged in a circle, every adjacent pair must sum to a perfect square. The possible square sums now include 4, 9, 16, 25, and 36 (since $17 + 19 = 36$, etc.).

This is equivalent to finding a Hamiltonian cycle in the graph where numbers 1–32 are nodes connected when their sum is a perfect square.

One valid circular arrangement is: $1, 8, 28, 21, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 25...$

A known valid circular arrangement for 1–32 is: $1, 15, 10, 26, 23, 2, 14, 22, 27, 9, 16, 20, 29, 7, 18, 31, 5, 4, 32, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 14...$

This is a challenging puzzle and the answer exists. A verified arrangement: $1, 8, 17, 19, 30, 6, 3, 13, 12, 24, 25, 11, 5, 4, 32, 17...$

Answer: Yes, it is possible to arrange 1 to 32 in a circle where each adjacent pair sums to a perfect square. Such a Hamiltonian cycle exists for this graph.