Chapter 1: A Square and A Cube

Person $$k$$ toggles locker $$n$$ exactly when $$k$$ is a factor of $$n$$. So the number of times locker $$n$$ is toggled equals the number of factors of $$n$$.

A locker starts closed. It ends open only if it is toggled an odd number of times. So a locker is open at the end $$\Leftrightarrow$$ its number has an odd number of factors.

Now, factors normally come in pairs: if $$d$$ divides $$n$$ then so does $$n/d$$, and $$d \ne n/d$$ in general. This gives an even count. The only exception is when $$d = n/d$$, i.e., $$n = d^2$$ — a perfect square. A perfect square has one "self-paired" factor (its square root), giving an odd total.

So Khoisnam knew, without watching the process, that the open lockers would be exactly those whose numbers are perfect squares.

Most numbers do have an even number of factors: each factor $$d$$ pairs with its partner $$n/d$$, and the two together contribute $$2$$ to the count.

But when $$n$$ is a perfect square, say $$n = k^2$$, the factor $$k$$ is paired with itself and counted only once. This gives an odd total.

Example: $$9 = 1 \times 9 = 3 \times 3$$ has factors $$1, 3, 9$$ — three factors, which is odd.

So no — perfect squares have an odd number of factors, so not every number has an even count.

Yes. A factor $$d$$ of $$n$$ is paired with $$n/d$$, contributing $$2$$ to the count — unless $$d = n/d$$, i.e., $$n = d^2$$.

So a number has an odd number of factors precisely when it is a perfect square. Examples: $$1 = 1^2$$, $$4 = 2^2$$, $$9 = 3^2$$, $$16 = 4^2$$, $$25 = 5^2$$, $$36 = 6^2$$, $$49 = 7^2$$, $$64 = 8^2$$, $$81 = 9^2$$, $$100 = 10^2, \ldots$$

You can verify: $$36$$ has factors $$1, 2, 3, 4, 6, 9, 12, 18, 36$$ — that's $$9$$ factors (odd).

Only lockers whose numbers have an odd number of factors remain open — and those are exactly the perfect squares up to $$100$$.

Listing perfect squares from $$1^2$$ to $$10^2$$:

These are the $$10$$ lockers that stay open.

A locker numbered $$n$$ is toggled once for each factor of $$n$$. It is touched exactly twice when $$n$$ has exactly two factors — namely $$1$$ and $$n$$ itself.

A number with exactly two factors is a prime number. The first five primes are

So the passcode is 2-3-5-7-11.

The squares of the first 30 natural numbers are $$1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, \ldots, 900$$.

Several patterns stand out.

• Units digits. Every perfect square ends in one of $$0, 1, 4, 5, 6, 9$$. The digits $$2, 3, 7, 8$$ never appear in the units place of a square.
• Parity is preserved. The square of an even number is even; the square of an odd number is odd.
• Consecutive differences. The differences between successive squares are the odd numbers: $$4-1=3,\ 9-4=5,\ 16-9=7,\ 25-16=9,\ \ldots$$ In general, $$(n+1)^2 - n^2 = 2n+1$$.
• Numbers ending in 5. Their squares always end in $$25$$: $$5^2 = 25$$, $$15^2 = 225$$, $$25^2 = 625$$.
• Trailing zeros. If a number ends in $$k$$ zeros, its square ends in $$2k$$ zeros.

No. Ending in $$0, 1, 4, 5, 6,$$ or $$9$$ is necessary for a number to be a perfect square, but it is not sufficient.

Many numbers with these units digits are not squares. For example, $$26$$ ends in $$6$$, yet $$5^2 = 25$$ and $$6^2 = 36$$, so no natural number squares to $$26$$.

Other examples: $$10, 11, 14, 15, 19, 20, 21, 24, 26, 29, \ldots$$ are all non-squares despite ending in an "allowed" digit.

The units-digit rule only lets us rule out squares (a number ending in $$2, 3, 7,$$ or $$8$$ cannot be a square); it cannot confirm one.

The units digit of a perfect square is always one of $$0, 1, 4, 5, 6, 9$$. So any number ending in $$2, 3, 7,$$ or $$8$$ is definitely not a square.

Five such numbers:

Their units digits are $$2, 3, 7, 8, 7$$ — each forbidden for a square. So none of these can be perfect squares.

Several patterns can be observed by comparing numbers with their squares.

• Units-digit rule. A number's units digit determines the units digit of its square: $$0\to 0,\ 1\to 1,\ 2\to 4,\ 3\to 9,\ 4\to 6,\ 5\to 5,\ 6\to 6,\ 7\to 9,\ 8\to 4,\ 9\to 1$$.
• Numbers ending in 5. Their squares always end in $$25$$: $$5^2 = 25$$, $$15^2 = 225$$, $$25^2 = 625$$.
• Trailing zeros double. If a number ends in $$k$$ zeros, its square ends in $$2k$$ zeros. E.g. $$20^2 = 400$$, $$100^2 = 10\,000$$.
• Parity is preserved. Even → even, odd → odd.
• Consecutive squares differ by odd numbers. $$4-1=3$$, $$9-4=5$$, $$16-9=7$$, $$25-16=9$$, ... so $$(n+1)^2 - n^2 = 2n+1$$.
• Sum of first $$n$$ odd numbers. $$1 + 3 + 5 + \cdots + (2n-1) = n^2$$.

If a number $$n$$ ends in exactly $$3$$ zeros, then $$n = 1000\,m$$ where $$m$$ does not end in $$0$$.

Squaring:

Since $$m$$ does not end in $$0$$, $$m^2$$ does not end in $$0$$ either. So $$n^2$$ ends in exactly $$6$$ zeros.

Example: $$3000^2 = 9\,000\,000$$ — six trailing zeros.

The number of trailing zeros of a square is double the number of trailing zeros of the number itself.

Why. Suppose $$n$$ ends in exactly $$k$$ zeros. Then $$n = 10^k \cdot m$$ where $$m$$ does not end in $$0$$. Squaring:

Since $$m$$ has a non-zero units digit, so does $$m^2$$. Hence $$n^2$$ ends in exactly $$2k$$ zeros.

Yes, this always happens — for every natural number $$n$$.

Because $$2k$$ is always even, a perfect square can only end in an even number of zeros. So a number with $$1, 3, 5, \ldots$$ (odd count) trailing zeros can never be a perfect square.

A number and its square have the same parity.

• Even case. If $$n = 2k$$, then $$n^2 = (2k)^2 = 4k^2 = 2(2k^2)$$, which is even.
• Odd case. If $$n = 2k+1$$, then $$n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$$, which is one more than an even number — hence odd.

So even numbers have even squares, and odd numbers have odd squares.

The sum of the first $$n$$ odd numbers is $$n^2$$. So $$36^2$$ equals $$35^2$$ plus the $$36$$-th odd number.

The $$n$$-th odd number is $$2n - 1$$. Hence the $$36$$-th odd number is

Adding to $$35^2 = 1225$$:

(Check: $$(n+1)^2 = n^2 + (2n+1)$$, so $$36^2 = 35^2 + (2 \cdot 35 + 1) = 1225 + 71 = 1296$$ ✓.)

The odd numbers in order are $$1, 3, 5, 7, 9, \ldots$$ — each obtained by adding $$2$$ to the previous one. The $$1$$st is $$1$$, the $$2$$nd is $$3$$, the $$3$$rd is $$5$$, and so on.

In general, the $$n$$-th odd number is $$2n - 1$$. For $$n = 36$$:

So the $$36$$-th odd number is $$71$$.

Consecutive odd numbers are $$1, 3, 5, 7, \ldots$$ — each is $$2$$ more than the previous.

Starting from $$1 = 2(1) - 1$$ and adding $$2$$ each step gives the $$n$$-th odd number as

Check: $$n=1 \Rightarrow 1$$; $$n=2 \Rightarrow 3$$; $$n=5 \Rightarrow 9$$; $$n=10 \Rightarrow 19$$. ✓

Take two consecutive perfect squares $$n^2$$ and $$(n+1)^2$$. The natural numbers strictly between them are

The count is

So there are $$\mathbf{2n}$$ numbers between $$n^2$$ and $$(n+1)^2$$.

Pattern. The counts form the sequence of even numbers:

Each successive gap contains $$2$$ more numbers than the previous one.

The perfect squares up to $$1000$$ are $$1^2, 2^2, \ldots, 31^2$$, since $$31^2 = 961$$ and $$32^2 = 1024 > 1000$$.

Sorting them into blocks of 100:

Total: $$10 + 4 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 1 = 31$$ squares.

The largest perfect square less than $$1000$$ is

The triangular numbers are $$T_1 = 1$$, $$T_2 = 3$$, $$T_3 = 6$$, $$T_4 = 10$$, $$T_5 = 15$$, $$T_6 = 21, \ldots$$ In general,

The pattern shown adds two consecutive triangular numbers:

$$T_1 + T_2 = 1 + 3 = 4 = 2^2,$$
$$T_2 + T_3 = 3 + 6 = 9 = 3^2,$$
$$T_3 + T_4 = 6 + 10 = 16 = 4^2.$$

Next term:

General result. For every $$n \ge 2$$,

So the sum of any two consecutive triangular numbers is a perfect square.

Drawing the next term. Take a triangle of $$10$$ dots ($$T_4$$) and fit an upside-down triangle of $$15$$ dots ($$T_5$$) next to it; together they form a $$5 \times 5$$ square of $$25$$ dots.

The area of a square equals the square of its side length. Let the side be $$s$$ cm. Then

Since $$7 \times 7 = 49$$, we have $$s = 7$$ cm.

We want a number whose square is $$64$$.

Since $$8 \times 8 = 64$$, we have $$8^2 = 64$$. Also, $$(-8)^2 = 64$$. So both $$+8$$ and $$-8$$ are square roots of $$64$$.

By convention, $$\sqrt{64}$$ denotes the positive root, so

Step 1 — Quick elimination. A perfect square cannot end in $$2, 3, 7,$$ or $$8$$, and cannot end in an odd number of zeros. Since $$327$$ ends in $$7$$, it is not a perfect square. So we test only $$576$$.

Step 2 — Prime-factorise.

Every prime appears an even number of times, so the factors can be split into two identical groups:

Hence $$576$$ is a perfect square, and $$\sqrt{576} = 24$$.

Alternative methods.

• List squares $$20^2 = 400, 21^2 = 441, 22^2 = 484, 23^2 = 529, 24^2 = 576$$. Match found — square root is $$24$$.
• Successively subtract odd numbers $$1, 3, 5, \ldots$$ from $$576$$. If it reaches $$0$$, count the subtractions; that count is $$\sqrt{576}$$.

Prime-factorise $$324$$:

Every prime appears an even number of times, so the factors split into two equal groups:

Hence $$324$$ is a perfect square and

Prime-factorise $$156$$:

The prime $$2$$ is paired, but $$3$$ and $$13$$ each appear only once — they have no partners.

Since the factors cannot be split into two identical groups, $$156$$ is not a perfect square.

1156. Prime-factorise:

All primes have even exponents, so

Hence $$1156$$ is a perfect square.

2800. Prime-factorise:

The prime $$7$$ appears only once (odd power). So the factors cannot be split into two identical groups, and $$2800$$ is not a perfect square.

A cube of side $$3$$ cm is built from $$3$$ horizontal layers, each layer a $$3 \times 3$$ square of unit cubes.

Number of unit cubes in one layer: $$3 \times 3 = 9$$. Number of layers: $$3$$.

Total:

A perfect cube is a number of the form $$n^3$$ for some natural number $$n$$.

Check the small cubes: $$1^3 = 1$$, $$2^3 = 8$$, $$3^3 = 27$$. There is no natural number $$n$$ with $$n^3 = 9$$, since $$9$$ lies strictly between $$2^3 = 8$$ and $$3^3 = 27$$.

So $$9$$ is not a perfect cube.

A cube of edge $$4$$ units is made of $$4$$ layers, each layer a $$4 \times 4$$ square of unit cubes.

Number of unit cubes in one layer: $$4 \times 4 = 16$$. Number of layers: $$4$$.

Total:

Compute each missing cube directly.

• $$6^3 = 6 \times 6 \times 6 = 36 \times 6 = 216$$
• $$7^3 = 7 \times 7 \times 7 = 49 \times 7 = 343$$
• $$8^3 = 8 \times 8 \times 8 = 64 \times 8 = 512$$
• $$9^3 = 9 \times 9 \times 9 = 81 \times 9 = 729$$
• $$10^3 = 10 \times 10 \times 10 = 1000$$
• $$12^3 = 144 \times 12 = 1728$$
• $$15^3 = 225 \times 15 = 3375$$
• $$16^3 = 256 \times 16 = 4096$$
• $$20^3 = 400 \times 20 = 8000$$

Completed table:

Several patterns emerge from the table of cubes.

• Parity. Cubes of odd numbers are odd; cubes of even numbers are even (same parity as the base).
• Last-digit variety. The last digits of $$1^3, 2^3, \ldots, 10^3$$ are $$1, 8, 7, 4, 5, 6, 3, 2, 9, 0$$ — every digit $$0$$ through $$9$$ appears exactly once. This is unlike squares, where only $$0, 1, 4, 5, 6, 9$$ occur.
• Cubes of multiples of 10. $$10^3 = 1000$$, $$20^3 = 8000$$ — a number ending in $$k$$ zeros has a cube ending in $$3k$$ zeros.
• Cubes of numbers ending in 5. Always end in $$5$$ (in fact in $$25$$ or $$75$$): $$5^3 = 125$$, $$15^3 = 3375$$.
• Rapid growth. Cubes grow much faster than squares — $$10^3 = 1000$$ already has $$4$$ digits, whereas $$10^2 = 100$$ has $$3$$; $$20^3 = 8000$$ is $$20$$ times larger than $$20^2 = 400$$.
• Reverse-digit curiosity. $$11^3 = 1331$$, $$101^3 = 1\,030\,301$$ show a nice palindromic pattern for cubes of $$1\underbrace{0\ldots0}_{k}1$$.

The units digit of $$n^3$$ depends only on the units digit of $$n$$. Check each case:

Every digit from $$0$$ to $$9$$ appears exactly once. So a cube can end in any digit.

List cubes and count digits in each range.

1-digit cubes (between $$1$$ and $$9$$): $$1^3 = 1$$, $$2^3 = 8$$. That's 2 cubes.

2-digit cubes (between $$10$$ and $$99$$): $$3^3 = 27$$, $$4^3 = 64$$. ($$5^3 = 125$$ is already 3-digit.) That's 2 cubes.

3-digit cubes (between $$100$$ and $$999$$): $$5^3 = 125$$, $$6^3 = 216$$, $$7^3 = 343$$, $$8^3 = 512$$, $$9^3 = 729$$. ($$10^3 = 1000$$ is 4-digit.) That's 5 cubes.

Observation. Cubes are far sparser than squares. Between $$1$$ and $$100$$ there are $$10$$ squares but only $$4$$ cubes ($$1, 8, 27, 64$$). Cubes grow rapidly, so relatively few of them fall in each range of a fixed width.

Suppose a cube $$n^3$$ ends in at least one zero. Then $$n^3$$ is divisible by $$10 = 2 \times 5$$. Since $$2$$ and $$5$$ are primes, they must each appear in the prime factorisation of $$n$$ itself.

So $$n$$ is divisible by $$10$$, i.e., $$n = 10\,k$$ for some natural number $$k$$. Cubing:

Hence $$n^3$$ is divisible by $$1000$$, meaning it ends in at least three zeros. It can never end in exactly two zeros.

(By the same reasoning, cubes always have a number of trailing zeros that is a multiple of $$3$$: $$0, 3, 6, 9, \ldots$$)

4104. Try subtracting small cubes and check whether the remainder is a cube.

• $$4104 - 2^3 = 4104 - 8 = 4096 = 16^3$$. So $$4104 = 2^3 + 16^3$$.
• $$4104 - 9^3 = 4104 - 729 = 3375 = 15^3$$. So $$4104 = 9^3 + 15^3$$.

Verification: $$8 + 4096 = 4104$$ ✓ and $$729 + 3375 = 4104$$ ✓.

13832. Similarly:

• $$13832 - 2^3 = 13832 - 8 = 13824 = 24^3$$. So $$13832 = 2^3 + 24^3$$.
• $$13832 - 18^3 = 13832 - 5832 = 8000 = 20^3$$. So $$13832 = 18^3 + 20^3$$.

Verification: $$8 + 13824 = 13832$$ ✓ and $$5832 + 8000 = 13832$$ ✓.

Recall the pattern from the chapter — consecutive odd numbers group into cubes:

The $$k$$-th group contains $$k$$ consecutive odd numbers and sums to $$k^3$$.

The given sum has $$10$$ consecutive odd numbers ($$91, 93, \ldots, 109$$) — exactly the $$10$$-th group. Therefore

(Quick check: the average of the terms is $$\dfrac{91 + 109}{2} = 100$$, and there are $$10$$ terms, so the sum is $$10 \times 100 = 1000$$.)

Prime-factorise $$3375$$:

The primes can be split into three identical groups:

So $$3375$$ is a perfect cube, and

Prime-factorise $$500$$:

For a number to be a perfect cube, every prime in its factorisation must appear an exact multiple of $$3$$ times. Here $$5$$ appears $$3$$ times (good), but $$2$$ appears only $$2$$ times.

The $$2$$s cannot be split into three identical groups, so $$500$$ is not a perfect cube.

Start with the cubes and take differences of consecutive terms, then differences of those, and so on.

Level 0 (cubes): $$1,\ 8,\ 27,\ 64,\ 125,\ 216,\ 343,\ \ldots$$

Level 1 (first differences):

$$8-1=7,\ 27-8=19,\ 64-27=37,\ 125-64=61,\ 216-125=91,\ 343-216=127,\ \ldots$$

giving $$7,\ 19,\ 37,\ 61,\ 91,\ 127,\ \ldots$$

Level 2 (second differences):

$$19-7=12,\ 37-19=18,\ 61-37=24,\ 91-61=30,\ 127-91=36,\ \ldots$$

giving $$12,\ 18,\ 24,\ 30,\ 36,\ \ldots$$

Level 3 (third differences):

$$18-12=6,\ 24-18=6,\ 30-24=6,\ 36-30=6,\ \ldots$$

All values are $$6$$.

Observation. For perfect cubes, the differences become constant at Level $$3$$, and the constant value is $$6$$. (Compare: for perfect squares the second-level differences are constant at $$2$$. In general, for the sequence of $$k$$-th powers, the $$k$$-th level differences are constant at $$k!$$ — here $$3! = 6$$.)

For each number $$1$$ through $$17$$, list the other numbers with which it sums to a perfect square (the only possible sums are $$4, 9, 16, 25$$):

Notice that $$16$$ has just one partner (namely $$9$$), and $$17$$ has just one partner (namely $$8$$). Since each end of the row has only one neighbour and the other numbers each have at least two, $$16$$ and $$17$$ must occupy the two ends.

Starting from $$16$$, every next number is forced (each number has only one remaining unused partner at each step):

Check every adjacent sum: $$16+9=25,\ 9+7=16,\ 7+2=9,\ 2+14=16,\ 14+11=25,\ 11+5=16,\ 5+4=9,\ 4+12=16,\ 12+13=25,\ 13+3=16,\ 3+6=9,\ 6+10=16,\ 10+15=25,\ 15+1=16,\ 1+8=9,\ 8+17=25$$. All perfect squares. ✓

Because every choice along the way was forced, the arrangement is essentially unique — the only other one is its reverse. So it cannot be arranged in more than one essentially different way.

We want to arrange $$1, 2, \ldots, 32$$ around a circle so that every pair of adjacent numbers (including the pair closing the circle) sums to a perfect square. The possible squares are $$4, 9, 16, 25, 36, 49, 64$$.

Forcing. Several numbers have very few partners. In particular, each of these has exactly two partners, so both partners must lie next to it in the circle:

• $$16$$: $$\{9, 20\}$$
• $$18$$: $$\{7, 31\}$$
• $$25$$: $$\{11, 24\}$$
• $$26$$: $$\{10, 23\}$$
• $$27$$: $$\{9, 22\}$$
• $$28$$: $$\{8, 21\}$$
• $$29$$: $$\{7, 20\}$$
• $$30$$: $$\{6, 19\}$$
• $$31$$: $$\{5, 18\}$$
• $$32$$: $$\{4, 17\}$$

Piecing the forced neighbourhoods together and joining the segments through the remaining numbers gives the circular arrangement:

(then back to $$22$$, closing the circle).

Verification (every adjacent sum): $$22+14=36,\ 14+2=16,\ 2+23=25,\ 23+26=49,\ 26+10=36,\ 10+15=25,\ 15+1=16,\ 1+8=9,\ 8+28=36,\ 28+21=49,\ 21+4=25,\ 4+32=36,\ 32+17=49,\ 17+19=36,\ 19+30=49,\ 30+6=36,\ 6+3=9,\ 3+13=16,\ 13+12=25,\ 12+24=36,\ 24+25=49,\ 25+11=36,\ 11+5=16,\ 5+31=36,\ 31+18=49,\ 18+7=25,\ 7+29=36,\ 29+20=49,\ 20+16=36,\ 16+9=25,\ 9+27=36,\ 27+22=49$$. All are perfect squares ✓.

The units digit of $$a^2$$ depends only on the units digit of $$a$$. Squares of the units digits:

So $$a^2$$ ends in $$4$$ exactly when $$a$$ ends in $$2$$ or $$8$$.

• $$64$$ ends in $$4 \Rightarrow 64^2$$ ends in $$6$$.
• $$108$$ ends in $$8 \Rightarrow 108^2$$ ends in $$4$$. ✓
• $$292$$ ends in $$2 \Rightarrow 292^2$$ ends in $$4$$. ✓
• $$36$$ ends in $$6 \Rightarrow 36^2$$ ends in $$6$$.

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

Use the identity $$(n+1)^2 = n^2 + (2n+1)$$ with $$n = 125$$:

This matches option (iv) — $$15625 + 251$$.

(Equivalently, going from $$125^2$$ to $$126^2$$ means adding the $$126$$-th odd number, which is $$2(126) - 1 = 251$$.)

Let the side of the square be $$s$$ metres. Then

Prime-factorise $$441$$:

So $$s = \sqrt{441} = 21$$ m.

Any number divisible by $$4$$, $$9$$, and $$10$$ must be a multiple of their LCM.

Prime factorisations: $$4 = 2^{2}$$, $$9 = 3^{2}$$, $$10 = 2 \times 5$$. Take the highest power of each prime:

Every multiple of $$180$$ is divisible by $$4, 9, 10$$. Now, for the multiple to also be a perfect square, every prime in its factorisation must appear an even number of times.

In $$180 = 2^{2} \times 3^{2} \times 5$$, the primes $$2$$ and $$3$$ are paired, but $$5$$ appears just once. To pair the $$5$$, multiply by another $$5$$:

Check: $$900 \div 4 = 225$$ ✓, $$900 \div 9 = 100$$ ✓, $$900 \div 10 = 90$$ ✓; and $$900 = 30^2$$ is a perfect square.

No smaller square works: the only smaller multiple of $$180$$ is $$180$$ itself, which is not a square.

Hence the smallest such square is $$\boxed{900}$$.

Prime-factorise $$9408$$:

So $$9408 = 2^{6} \times 3 \times 7^{2}$$.

For a perfect square, every prime needs an even exponent. Here $$2$$ has exponent $$6$$ (even) and $$7$$ has exponent $$2$$ (even), but $$3$$ has exponent $$1$$ (odd). The smallest fix is to multiply by $$3$$:

Hence the smallest multiplier is $$\boxed{3}$$, the product is $$28224$$, and $$\sqrt{28224} = 168$$.

Look at the general pattern. Writing row $$n$$ ($$n = 1, 2, 3, \ldots$$):

Check on the given rows:

• $$n = 1$$: $$1 + 4 + 4 = 9 = 3^2$$, and $$1 + 1 + 1 = 3$$. ✓
• $$n = 2$$: $$4 + 9 + 36 = 49 = 7^2$$, and $$4 + 2 + 1 = 7$$. ✓
• $$n = 3$$: $$9 + 16 + 144 = 169 = 13^2$$, and $$9 + 3 + 1 = 13$$. ✓

Row 4 ($$n = 4$$): the right side is $$(4^{2} + 4 + 1)^{2} = 21^{2}$$. So

Row 5 ($$n = 9$$): the middle blank is $$n(n+1) = 9 \times 10 = 90$$, and the right blank is $$n^{2} + n + 1 = 81 + 9 + 1 = 91$$. So

Completed pattern:

The picture is a $$9 \times 9$$ arrangement (nine rows and nine columns) of pattern-squares, in which two types of small patterns alternate. On carefully counting the tiny squares inside every pattern-square and adding them up, the total comes out to $$900$$ tiny squares.

Prime-factorise $$900$$:

Because every prime appears twice, the factors can be split into two identical groups:

So the number of tiny squares is $$900$$, and its prime factorisation is $$2^{2} \times 3^{2} \times 5^{2}$$. In particular, $$900$$ is itself a perfect square — it is $$30^{2}$$.

Cube root of 27000. Factorise:

Hence $$\sqrt[3]{27000} = 30$$.

Cube root of 10648. Factorise step by step:

Recognise $$1331 = 11 \times 11 \times 11 = 11^{3}$$. Therefore

Hence $$\sqrt[3]{10648} = 22$$.

Prime-factorise $$1323$$:

For a perfect cube, every prime must have an exponent that is a multiple of $$3$$. Here $$3$$ has exponent $$3$$ (good) but $$7$$ has exponent $$2$$. To reach $$7^{3}$$, multiply by one more $$7$$.

Hence the required multiplier is $$\boxed{7}$$, and the resulting cube is $$1323 \times 7 = 9261 = 21^{3}$$.

The units digit of a cube uniquely determines the units digit of its cube root. Combining this with the size of the cube pins down the tens digit.

1331. Ends in $$1 \Rightarrow$$ cube root ends in $$1$$. Also $$10^{3} = 1000$$ and $$20^{3} = 8000$$, so the root lies between $$10$$ and $$20$$. Only $$11$$ fits. Check: $$11^{3} = 1331$$ ✓.

4913. Ends in $$3 \Rightarrow$$ cube root ends in $$7$$. Also $$10^{3} = 1000$$ and $$20^{3} = 8000$$, so the root is between $$10$$ and $$20$$. Only $$17$$ fits. Check: $$17^{3} = 4913$$ ✓.

12167. Ends in $$7 \Rightarrow$$ cube root ends in $$3$$. Also $$20^{3} = 8000$$ and $$30^{3} = 27000$$, so the root is between $$20$$ and $$30$$. Only $$23$$ fits. Check: $$23^{3} = 12167$$ ✓.

32768. Ends in $$8 \Rightarrow$$ cube root ends in $$2$$. Also $$30^{3} = 27000$$ and $$40^{3} = 64000$$, so the root is between $$30$$ and $$40$$. Only $$32$$ fits. Check: $$32^{3} = 32768$$ ✓.

Use the identities

In every option the difference is between two consecutive integers, so $$a - b = 1$$. This simplifies each expression:

Compute each option:

• (i) $$67^{3} - 66^{3} = 67^{2} + 67 \cdot 66 + 66^{2} = 4489 + 4422 + 4356 = 13\,267$$.
• (ii) $$43^{3} - 42^{3} = 43^{2} + 43 \cdot 42 + 42^{2} = 1849 + 1806 + 1764 = 5419$$.
• (iii) $$67^{2} - 66^{2} = 67 + 66 = 133$$.
• (iv) $$43^{2} - 42^{2} = 43 + 42 = 85$$.

Compare: $$13267 > 5419 > 133 > 85$$.

Reasoning. The gap between consecutive squares is $$2n + 1$$, whereas the gap between consecutive cubes is $$3n^{2} + 3n + 1$$ — much larger for the same $$n$$. Between two options with the same base numbers, the cube-difference always beats the square-difference. And a larger base makes the gap even bigger. So the winner is the cube-difference with the larger base — that is option (i).