A Square & A Cube

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.

“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

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.

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.

These are the perfect squares from 1 to 100.

Answer1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

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.

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.

Pick any numbers ending in 2, 3, 7 or 8 — squares never do.

Example162, 253, 437, 588, 1027.

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 \(6^2=36\)).

Answer(ii) 34², (iii) 46², (iv) 56² and (v) 74².

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.

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.

1. The \(n^{\text{th}}\) odd number is \(2n-1\).
2. The 36th odd number is \(2(36)-1 = \mathbf{71}\).
3. Add it on: \(36^2 = 35^2 + 71 = 1225 + 71\).

Answer\(36^2 = 1296\); the \(n^{\text{th}}\) odd number is \(2n-1\).

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:

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 \(31^2 = 961\).

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 \(15+21 = 36 = 6^2\).

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).

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.

1. 1156: \(1156 = 2\times2\times17\times17 = (2\times17)^2\). All factors pair up → perfect square with \(\sqrt{1156}=34\).
2. 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.

Use the units-digit test first (squares never end in 2, 3, 7, 8):

1. 2032 ends in 2 → not a square.
2. 2048 ends in 8 → not a square.
3. 1027 ends in 7 → not a square.
4. 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.

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.

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).

Side \(= \sqrt{\text{area}}\). Factorise: \(441 = 3\times3\times7\times7 = (3\times7)^2 = 21^2\).

Answer21 m.

1. Find the LCM: \(4=2^2,\ 9=3^2,\ 10=2\times5\Rightarrow \text{LCM}=2^2\times3^2\times5 = 180\).
2. For a perfect square every prime power must be even. Here 5 appears once (odd), so multiply by 5.
3. \(180\times5 = 900 = 2^2\times3^2\times5^2 = 30^2\).

Answer900.

1. Factorise: \(9408 = 2^6 \times 3 \times 7^2\).
2. Pairs: \(2^6\) and \(7^2\) are fine, but the single 3 is unpaired. Multiply by 3.
3. Product \(= 2^6 \times 3^2 \times 7^2 = 28224\).
4. Square root \(= 2^3 \times 3 \times 7 = 8\times3\times7 = 168\).

AnswerMultiply by 3; \(\sqrt{28224}=168\).

Between \(n^2\) and \((n+1)^2\) there are exactly \(2n\) numbers.

Answer(i) 32 numbers · (ii) 198 numbers.

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 \(9^2+10^2+90^2 = 91^2\).

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.

1. Tiny squares per tile: \(4\times4 = 16\).
2. Number of tiles: \(9\times9 = 81\).
3. 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 \(36^2\)).