A Square & A Cube
Complete, step-by-step solutions to every in-text question, “Figure it Out” exercise, and the closing puzzle — with the chapter’s own diagrams rebuilt for you.
Queen Ratnamanjuri’s 100 Lockers
100 people toggle 100 lockers. Which ones stay open?
Person 1 opens every locker, Person 2 toggles every 2nd, Person 3 every 3rd, and so on up to Person 100. Which lockers remain open at the end — and how did Khoisnam know in advance?
Key idea — count the toggles. Locker number \(N\) is toggled once by every person whose number is a factor of \(N\). So the number of times a locker is toggled equals the number of factors of its locker number.
A locker starts closed. It ends open only if it is toggled an odd number of times. So:
Why squares are special. Factors come in partner pairs whose product is \(N\) (for 6: \(1\times6\) and \(2\times3\)). Each pair contributes 2 factors — an even count. The only way to get an odd count is when one factor pairs with itself, i.e. \(\sqrt{N}\times\sqrt{N}=N\). That happens precisely for perfect squares.
Among 1 to 100 the perfect squares are \(1^2,2^2,\dots,10^2\). Khoisnam simply listed the squares up to 100.
AnswerOpen lockers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
The passcode is the first five locker numbers that were touched exactly twice. Which five lockers are these?
“Touched exactly twice” means the locker number has exactly two factors — namely 1 and the number itself. A number with exactly two factors is a prime number.
The first five primes are 2, 3, 5, 7 and 11.
Code2 – 3 – 5 – 7 – 11
Square Numbers
Every “?” prompt from pages 2–9, solved.
Does every number have an even number of factors?
Not always. Most numbers do, because factors pair up. But when a number is a perfect square, its square root pairs with itself and is counted only once, giving an odd total. For example, 9 has factors 1, 3, 9 — three of them.
AnswerNo.
Can you use this insight to find more numbers with an odd number of factors?
Yes. Any number of the form \(n\times n=n^2\) has the repeated factor \(n\) at its centre, so its factor count is odd. Examples: 1, 4, 9, 16, 25, 36…
AnswerYes — all square numbers have an odd number of factors.
Write the locker numbers that remain open.
These are the perfect squares from 1 to 100.
Answer1, 4, 9, 16, 25, 36, 49, 64, 81, 100.
What patterns do you notice in the table of squares? (Look at the units digit.)
Listing \(1^2\) to \(30^2\), the units digit of a square is always one of 0, 1, 4, 5, 6, 9. A square never ends in 2, 3, 7 or 8.
Also, the squares \(1^2,9^2,11^2,19^2,21^2,29^2\) all end in 1 — numbers ending in 1 or 9 have squares ending in 1. The next two such squares are:
PatternSquares end only in 0, 1, 4, 5, 6 or 9.
If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
No. The units digit can only tell us when a number is not a square. Both 16 and 36 end in 6 and are squares, but 26 also ends in 6 and is not. So the right-most digit is a useful eliminator, not a guarantee.
AnswerNo — e.g. 26 ends in 6 but is not a square.
Write 5 numbers that you can tell are not squares just from their units digit.
Pick any numbers ending in 2, 3, 7 or 8 — squares never do.
Example162, 253, 437, 588, 1027.
Which of these have the digit 6 in the units place?(i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²
The units digit of a square depends only on the units digit of the number. A square ends in 6 exactly when the number ends in 4 or 6 (since \(4^2=16\) and \(6^2=36\)).
| Number | Ends in | Square ends in | 6? |
|---|---|---|---|
| 38 | 8 | 4 | — |
| 34 | 4 | 6 | yes |
| 46 | 6 | 6 | yes |
| 56 | 6 | 6 | yes |
| 74 | 4 | 6 | yes |
| 82 | 2 | 4 | — |
Answer(ii) 34², (iii) 46², (iv) 56² and (v) 74².
If a number ends in 3 zeros, how many zeros will its square end in? And what about parity?
Each trailing zero comes from a factor of 10, and squaring doubles every factor. So the number of trailing zeros doubles: \(3\to 6\). For example \(1000^2 = 1\,000\,000\) has six zeros.
This means squares can only ever have an even number of trailing zeros. And for parity: an even number squared is even, an odd number squared is odd.
AnswerSix zeros · zeros always double · even²=even, odd²=odd.
Using the odd-number pattern, find 36² given that 35² = 1225. What is the \(n^{\text{th}}\) odd number?
Adding consecutive odd numbers builds squares: \(1+3+5+\dots\). So 1225 is the sum of the first 35 odd numbers, and to reach \(36^2\) we add just the 36th odd number.
- The \(n^{\text{th}}\) odd number is \(2n-1\).
- The 36th odd number is \(2(36)-1 = \mathbf{71}\).
- Add it on: \(36^2 = 35^2 + 71 = 1225 + 71\).
Answer\(36^2 = 1296\); the \(n^{\text{th}}\) odd number is \(2n-1\).
How many numbers lie between two consecutive perfect squares? How many squares are there in each block of 100 up to 1000, and what is the largest square below 1000?
Between \(n^2\) and \((n+1)^2\) the count of whole numbers strictly in between is
So in general, for consecutive squares \(p\) and \(q\), there are \(q-p-1\) numbers between them.
Counting squares in each hundred:
| 1–100 | 101–200 | 201–300 | 301–400 | 401–500 |
|---|---|---|---|---|
| 10 | 4 | 3 | 3 | 2 |
| 501–600 | 601–700 | 701–800 | 801–900 | 901–1000 |
| 2 | 2 | 2 | 2 | 1 |
That is 31 squares in all (\(1^2\) up to \(31^2\)). The blocks thin out because squares grow farther apart.
Answer\(q-p-1\) (i.e. \(2n\)) numbers between them · largest square below 1000 is \(31^2 = 961\).
Is there a relation between triangular numbers and square numbers? Extend the pattern.
The triangular numbers are 1, 3, 6, 10, 15, …
Adding any two consecutive triangular numbers gives a perfect square:
The next term continues the pattern:
Next\(10+15 = 25 = 5^2\), then \(15+21 = 36 = 6^2\).
What is the square root of 64?
Since \(8\times 8 = 64\) and also \((-8)\times(-8)=64\), there are two integer square roots.
Answer+8 and −8 (in this chapter we take the positive root, 8).
Is 324 a perfect square? Is 156 a perfect square? (Use prime factorisation.)
324: split the prime factors into two equal groups (pairs).
Every prime is paired, so 324 is a perfect square, and one factor from each pair gives the root: \((2\times3\times3)=18\).
156: \(156 = 2\times2\times3\times13\). The 3 and 13 cannot be paired, so 156 is not a perfect square.
Answer324 = 18² (yes) · 156 is not a perfect square.
Find whether 1156 and 2800 are perfect squares using prime factorisation.
- 1156: \(1156 = 2\times2\times17\times17 = (2\times17)^2\). All factors pair up → perfect square with \(\sqrt{1156}=34\).
- 2800: \(2800 = 2\times2\times2\times2\times5\times5\times7 = 2^4\times5^2\times7\). The lone 7 has no partner → not a perfect square.
Answer1156 = 34² (perfect square) · 2800 is not a perfect square.
Figure it Out — Squares
Pages 10–11, fully worked.
Which of the following are not perfect squares?(i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Use the units-digit test first (squares never end in 2, 3, 7, 8):
- 2032 ends in 2 → not a square.
- 2048 ends in 8 → not a square.
- 1027 ends in 7 → not a square.
- 1089 ends in 9 — it could be; indeed \(33^2 = 1089\), so it is a square.
Answer(i), (ii) and (iii) are not perfect squares.
Which one among 64², 108², 292², 36² has last digit 4?
A square ends in 4 when the number ends in 2 or 8 (\(2^2=4\), \(8^2=64\)).
64 ends in 4 (→6), 108 ends in 8 (→4), 292 ends in 2 (→4), 36 ends in 6 (→6).
Answer108² and 292².
Given 125² = 15625, what is the value of 126²?(i) 15625 + 126 (ii) 15625 + 26² (iii) 15625 + 253 (iv) 15625 + 251 (v) 15625 + 51²
Going from one square to the next adds the next odd number, the \(126^{\text{th}}\) odd number:
Answer(iv) 15625 + 251 (which equals 15876).
Find the length of the side of a square whose area is 441 m².
Side \(= \sqrt{\text{area}}\). Factorise: \(441 = 3\times3\times7\times7 = (3\times7)^2 = 21^2\).
Answer21 m.
Find the smallest square number divisible by each of 4, 9 and 10.
- Find the LCM: \(4=2^2,\ 9=3^2,\ 10=2\times5\Rightarrow \text{LCM}=2^2\times3^2\times5 = 180\).
- For a perfect square every prime power must be even. Here 5 appears once (odd), so multiply by 5.
- \(180\times5 = 900 = 2^2\times3^2\times5^2 = 30^2\).
Answer900.
Find the smallest number by which 9408 must be multiplied to make a perfect square. Find the square root of the product.
- Factorise: \(9408 = 2^6 \times 3 \times 7^2\).
- Pairs: \(2^6\) and \(7^2\) are fine, but the single 3 is unpaired. Multiply by 3.
- Product \(= 2^6 \times 3^2 \times 7^2 = 28224\).
- Square root \(= 2^3 \times 3 \times 7 = 8\times3\times7 = 168\).
AnswerMultiply by 3; \(\sqrt{28224}=168\).
How many numbers lie between the squares of (i) 16 and 17, and (ii) 99 and 100?
Between \(n^2\) and \((n+1)^2\) there are exactly \(2n\) numbers.
Answer(i) 32 numbers · (ii) 198 numbers.
Fill in the missing numbers in the pattern.
Each row follows \(a^2 + (a+1)^2 + \big[a(a+1)\big]^2 = \big[a(a+1)+1\big]^2\).
For row 4: \(a=4\Rightarrow a(a+1)=20\), and \(20+1=21\). For row 5: \(a=9\Rightarrow a(a+1)=90\), and \(90+1=91\).
Answer\(4^2+5^2+20^2 = 21^2\) and \(9^2+10^2+90^2 = 91^2\).
How many tiny squares are in the picture? Write the prime factorisation of that number.
The picture is a 9 × 9 arrangement of decorative tiles. Each tile (whether upright or rotated as a diamond) is a 4 × 4 block of tiny squares.
- Tiny squares per tile: \(4\times4 = 16\).
- Number of tiles: \(9\times9 = 81\).
- Total: \(81\times16 = 1296\).
Its prime factorisation:
Answer1296 tiny squares; \(1296 = 2^4 \times 3^4\) (which is also \(6^4\) and the perfect square \(36^2\)).
Cubic Numbers
Cubes, cube roots, taxicab numbers & history (pages 11–15).
How many 1 cm cubes make a cube of side 2 cm? Of side 3 cm? And how many unit cubes fill a cube of edge 4?
A cube of edge \(n\) is filled by \(n\) layers, each an \(n\times n\) square of unit cubes, so it holds \(n\times n\times n = n^3\) cubes.
AnswerSide 2 → 8 cubes · side 3 → 27 cubes · edge 4 → 64 unit cubes.
Is 9 a perfect cube?
The cubes around 9 are \(2^3 = 8\) and \(3^3 = 27\). Since 9 lies strictly between them, no whole number cubed gives 9 (nor any number from 10 to 26).
AnswerNo, 9 is not a perfect cube.
Complete the table of cubes and describe the patterns you notice.
| n | n³ | n | n³ |
|---|---|---|---|
| 1 | 1 | 11 | 1331 |
| 2 | 8 | 12 | 1728 |
| 3 | 27 | 13 | 2197 |
| 4 | 64 | 14 | 2744 |
| 5 | 125 | 15 | 3375 |
| 6 | 216 | 16 | 4096 |
| 7 | 343 | 17 | 4913 |
| 8 | 512 | 18 | 5832 |
| 9 | 729 | 19 | 6859 |
| 10 | 1000 | 20 | 8000 |
Patterns: cubes of even numbers are even and cubes of odd numbers are odd; the units digit of a cube is fixed by the units digit of the number (2&8 swap, 3&7 swap, the rest stay); and cubes grow far faster than squares.
AnswerTable completed above (1³ … 20³).
Squares end only in 0, 1, 4, 5, 6, 9. What are the possible last digits of cubes? How many cubes have 1, 2 and 3 digits?
Cubing each of 0–9 gives the units digits 0, 1, 8, 7, 4, 5, 6, 3, 2, 9 — that is all ten digits. So, unlike squares, a cube can end in any digit.
Counting cubes by length: 1-digit cubes are 1, 8 (two); 2-digit cubes are 27, 64 (two); 3-digit cubes are 125, 216, 343, 512, 729 (five). There are noticeably fewer cubes than squares in each range.
AnswerCubes can end in 0–9 (any digit) · 1-digit: 2, 2-digit: 2, 3-digit: 5.
Can a cube end with exactly two zeroes (00)? Explain.
Trailing zeros come from factors of 10, i.e. pairs of 2 and 5. Cubing triples the count of each prime, so the number of trailing zeros in a cube is always a multiple of 3 (0, 3, 6, …). Two is not a multiple of 3.
AnswerNo. Cubes end in 0, 3, 6, … zeros, never exactly two.
The next two taxicab numbers after 1729 are 4104 and 13832. Write each as the sum of two positive cubes in two ways.
A taxicab number can be written as a sum of two cubes in two different ways (as 1729 = 1³ + 12³ = 9³ + 10³).
Answer4104 = 2³+16³ = 9³+15³ · 13832 = 2³+24³ = 18³+20³.
Can you tell the sum 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109 without calculating?
The pattern \(1=1^3\), \(3+5=2^3\), \(7+9+11=3^3\), … uses the next batch of consecutive odd numbers to build each cube. The rows up to row 9 use the first 45 odd numbers, so row 10 begins at the 46th odd number, which is 91, and has 10 terms — exactly the ten odd numbers shown.
Therefore this sum is the 10th cube.
Answer1000 (= 10³).
Is 3375 a perfect cube? Is 500 a perfect cube?
3375: group the prime factors into triplets.
So \(\sqrt[3]{3375} = 15\).
500: \(500 = 2\times2\times5\times5\times5 = 2^2 \times 5^3\). The two 2’s cannot form a triplet, so 500 is not a perfect cube.
Answer3375 = 15³ (yes) · 500 is not a perfect cube.
Find the cube roots: (i) ∛64 (ii) ∛512 (iii) ∛729.
Group each into triplets of primes:
Answer(i) 4 · (ii) 8 · (iii) 9.
Compute successive differences over levels for perfect cubes until all differences at a level are equal. What do you notice?
For squares, the differences become constant after two levels (the constant is 2). For cubes, we keep going one level further.
AnswerAfter three levels every difference equals 6 (i.e. 3! = 6); squares needed only two levels.
Figure it Out — Cubes
Pages 16–17, fully worked.
Find the cube roots of 27000 and 10648.
- \(27000 = 27 \times 1000 = 3^3 \times 10^3 = (3\times10)^3 = 30^3\). So \(\sqrt[3]{27000} = 30\).
- \(10648 = 2^3 \times 11^3 = (2\times11)^3 = 22^3\). So \(\sqrt[3]{10648} = 22\).
Answer∛27000 = 30 · ∛10648 = 22.
What number will you multiply by 1323 to make it a cube number?
- Factorise: \(1323 = 3 \times 441 = 3 \times 3^2 \times 7^2 = 3^3 \times 7^2\).
- For a cube, every prime power must be a multiple of 3. Here \(7^2\) is short by one 7.
- Multiply by 7: \(1323 \times 7 = 3^3 \times 7^3 = (3\times7)^3 = 21^3 = 9261\).
Answer7 (giving 9261 = 21³).
State true or false, with reasoning.
- False(i) The cube of any odd number is even. An odd × odd × odd is always odd (e.g. \(3^3=27\)).
- False(ii) There is no perfect cube that ends with 8. \(2^3=8\) and \(12^3=1728\) both end in 8.
- False(iii) The cube of a 2-digit number may be a 3-digit number. The smallest 2-digit number is 10, and \(10^3 = 1000\) already has 4 digits. So a 2-digit cube can never be only 3 digits.
- False(iv) The cube of a 2-digit number may have seven or more digits. The largest is \(99^3 = 970299\), which has just 6 digits.
- False(v) Cube numbers have an odd number of factors. Only perfect squares do. For instance 8 has factors 1, 2, 4, 8 — four of them (even).
AnswerAll five statements are False.
Given that 1331 is a perfect cube, guess its cube root without factorising. Then guess ∛4913, ∛12167 and ∛32768.
The shortcut. Use two clues: the units digit of the cube fixes the units digit of the root, and grouping the number into thousands fixes the tens digit.
| cube ends in | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| root ends in | 0 | 1 | 8 | 7 | 4 | 5 | 6 | 3 | 2 | 9 |
- 1331 ends in 1 → root ends in 1; it lies in \([10^3,20^3]\) → tens digit 1 → 11.
- 4913 ends in 3 → root ends in 7; lies in \([10^3,20^3]\) → 17.
- 12167 ends in 7 → root ends in 3; lies in \([20^3,30^3]\) → 23.
- 32768 ends in 8 → root ends in 2; lies in \([30^3,40^3]\) → 32.
Answer∛1331 = 11, ∛4913 = 17, ∛12167 = 23, ∛32768 = 32.
Which is greatest?(i) 67³ − 66³ (ii) 43³ − 42³ (iii) 67² − 66² (iv) 43² − 42²
Two handy identities for consecutive numbers:
- \(67^3-66^3 = 3(67)^2 – 3(67) + 1 = 13467 – 201 + 1 = \mathbf{13267}\).
- \(43^3-42^3 = 3(43)^2 – 3(43) + 1 = 5547 – 129 + 1 = 5419\).
- \(67^2-66^2 = 67+66 = 133\).
- \(43^2-42^2 = 43+42 = 85\).
Cube differences grow with \(n^2\), so they dwarf the square differences, and the larger base (67) wins.
Answer(i) 67³ − 66³ = 13267 is the greatest.
Square Pairs
Arrange numbers so that every neighbouring pair adds to a square.
Arrange the numbers 1 to 17 in a row so that every adjacent pair adds to a perfect square. Is more than one arrangement possible?
One valid arrangement (read left to right):
Why it is essentially the only one. Look at which partners each number can have. Among 1–17, the number 16 can only pair with 9 (16 + 9 = 25), and 17 can only pair with 8 (17 + 8 = 25). A number with a single possible partner must sit at an end of the row, so 16 and 17 are forced to be the two ends. From there each next number is forced, giving this chain (and its mirror image).
Answer16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17 — unique up to reversing it.
Can you arrange the numbers 1 to 32 in a circle so that every adjacent pair adds to a perfect square?
Yes — 32 is in fact the smallest number for which a full circle is possible. Here is one such circular arrangement (going around, and the last number links back to the first):
Reading clockwise from the top: 14, 2, 23, 26, 10, 15, 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 — and 22 + 14 = 36 closes the loop.
AnswerYes — see the circle above (one of several valid arrangements).
End of Chapter 1 · A Square and A Cube · solutions by EduGrown