Chapter 1: A Square and a Cube (Squares and Cubes)

A locker is toggled once for each factor of its number. It ends open if it is toggled an odd number of times.

Factors come in partner pairs (like 1 × 6 and 2 × 3 for 6), so most numbers have an even number of factors. The exception is when a pair has the same number twice: $36=6\times 6$. The factors of 36 are 1 & 36, 2 & 18, 3 & 12, 4 & 9, and 6 on its own: 9 factors (odd).

So only the square numbers have an odd number of factors, and the open lockers are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100.

The lockers touched exactly twice are those with exactly two factors, the primes; the first five give the code 2-3-5-7-11.

Open lockers: the squares 1, 4, 9, …, 100. Not every number has an even number of factors: squares have an odd number (36 has 9).

Yes: areas ${\left(\frac{3}{5}\right)}^2=\frac{9}{25}$ and $2.5^2=6.25$ square units.

Patterns: squares end only in 0, 1, 4, 5, 6 or 9; numbers ending in 5 have squares ending in 25; the units digits repeat in the pattern 1, 4, 9, 6, 5, 6, 9, 4, 1, 0; consecutive squares differ by consecutive odd numbers.

Yes (9/25 and 6.25); squares end only in 0, 1, 4, 5, 6 or 9.

Not squares (they end in 2, 3, 7 or 8): 42, 153, 297, 1008, 2222.

Next squares ending in 1: $31^2=961$ and $39^2=1521$.

A square ends in 6 when the number ends in 4 or 6: 34², 46², 56², 74² (1156, 2116, 3136, 5476). 38² and 82² end in 4.

E.g. 42, 153, 297, 1008, 2222; 31² = 961, 39² = 1521; 34², 46², 56², 74² end in 6.

Each zero at the end is a factor of 10, and squaring doubles the factors: 3 zeros → 6 zeros. A square always ends with an even number of zeros (double the number of zeros of its root), so yes, squares cannot end in an odd number of zeros.

Parity: an even number has an even square and an odd number an odd square (same parity).

6 zeros; squares end in an even number of zeros; a number and its square have the same parity.

$36^2=1225+(\text{36th odd number})=1225+71=$ 1296.

Between $n^2$ and $(n+1{)}^2$ there are $(n+1{)}^2-n^2-1=$ 2n numbers (e.g. 2 between 1 and 4, 4 between 4 and 9).

Largest square below 1000: $31^2=$ 961.

36² = 1296; 2n numbers lie between n² and (n + 1)²; counts 10, 4, 3, 3, 2, 2, 2, 2, 2, 1; largest square below 1000 is 961.

Any two consecutive triangular numbers add up to a square: $1+3=4$, $3+6=9$, $6+10=16$, next $10+15=25=5^2$. In the dot picture, a square of side n splits along a diagonal into the triangles with sides n and n − 1.

Consecutive triangular numbers add to a square, e.g. 10 + 15 = 25.

$1156=2\times 2\times 17\times 17=(2\times 17{)}^2=34^2$: a perfect square, $\sqrt{1156}=34$.

$2800=2\times 2\times 2\times 2\times 5\times 5\times 7$: the 7 has no partner, so it is not a perfect square.

1156 = 34² is a square; 2800 is not (the factor 7 is unpaired).

2032 and 2048 end in 2 and 8, and 1027 ends in 7: none of these can be a square. $1089=33^2$.

Not perfect squares: (i), (ii) and (iii).

(i), (ii), (iii); 1089 = 33².

A square ends in 4 when the number ends in 2 or 8: 108² and 292² (11664 and 85264). 64² and 36² end in 6.

108² and 292²

$126^2=125^2+125+126=15625+251$ (adding the 126th odd number, 251). So (iv): $126^2=15876$.

(iv) 15625 + 251 = 15876

$441=3\times 3\times 7\times 7=21^2$, so the side is 21 m.

21 m

It must be a multiple of LCM(4, 9, 10) $=180=2^2\times 3^2\times 5$. To make a square, the 5 needs a partner: $180\times 5=900=30^2$.

900

$9408=2^6\times 3\times 7^2$. Only the 3 is unpaired, so multiply by 3: $28224=(2^3\times 3\times 7{)}^2=168^2$, so the square root is 168.

Multiply by 3; √28224 = 168.

Between $n^2$ and $(n+1{)}^2$ there are $2n$ numbers: (i) 32 (ii) 198.

(i) 32 (ii) 198

The third number is the product of the first two, and the answer is one more than it:

$4^2+5^2+20^2=$ 21² (16 + 25 + 400 = 441)

$9^2+10^2+$ 90² $=$ 91² (81 + 100 + 8100 = 8281)

21²; 90² and 91²

The picture is a 9 × 9 arrangement of blocks (straight grids and tilted diamonds), and each block has 5 × 5 = 25 tiny squares.

Total $=81\times 25=$ 2025 $=3\times 3\times 3\times 3\times 5\times 5=3^4\times 5^2$ (which is $45^2$).

2025 = 3⁴ × 5² (= 45²)

$2\times 2\times 2=$ 8 cubes; $3\times 3\times 3=$ 27 cubes.

Patterns: cubes can end in any digit 0–9; a number ending in 1, 4, 5, 6, 9 or 0 has a cube ending in the same digit, while 2 ↔ 8 and 3 ↔ 7 swap; odd numbers have odd cubes and even numbers even cubes.

8 and 27; cubes can end in any digit (2 → 8, 8 → 2, 3 → 7, 7 → 3, the rest keep their digit).

1 digit: 1, 8 (2); 2 digits: 27, 64 (2); 3 digits: 125, 216, 343, 512, 729 (5).

No. If a cube ends in 0, its cube root ends in 0 (is a multiple of 10), so the cube is a multiple of 1000 and ends in at least three zeros. The number of zeros at the end of a cube is always a multiple of 3.

2, 2 and 5 cubes; no, a cube ends in 0, 3, 6, … zeros.

$4104=2^3+16^3=9^3+15^3$ (8 + 4096 = 729 + 3375)

$13832=2^3+24^3=18^3+20^3$ (8 + 13824 = 5832 + 8000)

4104 = 2³ + 16³ = 9³ + 15³; 13832 = 2³ + 24³ = 18³ + 20³

In the pattern, the nth row has n odd numbers adding to $n^3$. A row of 10 numbers is the 10th row, so the sum is $10^3=$ 1000.

10³ = 1000

(i) $64=2^6=(2\times 2{)}^3$: 4 (ii) $512=2^9=(2^3{)}^3$: 8 (iii) $729=3^6=(3^2{)}^3$: 9

(i) 4 (ii) 8 (iii) 9

For cubes the differences become constant (all 6) at the third level, just as for squares they become constant (2) at the second level.

Level 1: 7, 19, 37, 61, 91; level 2: 12, 18, 24, 30; level 3: all 6.

$27000=27\times 1000=3^3\times 10^3$: 30.

$10648=2^3\times 11^3$: 22.

30 and 22

$1323=3\times 3\times 3\times 7\times 7$. The 7s need one more: multiply by 7 to get $9261=21^3$.

7 (1323 × 7 = 9261 = 21³)

(i) False: odd × odd × odd is odd (e.g. $3^3=27$).

(ii) False: $2^3=8$, $12^3=1728$.

(iii) False: the smallest is $10^3=1000$, already 4 digits.

(iv) False: the largest is $99^3=970299$, only 6 digits ($100^3$ has 7).

(v) False: e.g. $8$ has the factors 1, 2, 4, 8 (four). Only cubes that are also squares, like 1, 64, 729, have an odd number of factors.

All five are false.

Look at the last digit (it fixes the root's last digit) and the thousands part (which two cubes it lies between):

• 1331: ends in 1 → root ends in 1; 1 thousand lies between $1^3$ and $2^3$ → 11.
• 4913: ends in 3 → root ends in 7; 4 lies between $1^3$ and $2^3$ → 17.
• 12167: ends in 7 → root ends in 3; 12 lies between $2^3$ and $3^3$ → 23.
• 32768: ends in 8 → root ends in 2; 32 lies between $3^3$ and $4^3$ → 32.

11, 17, 23 and 32

Differences of consecutive powers grow as the numbers grow, and cubes grow much faster than squares: $67^3-66^3=3\times 67\times 66+1=13267$, $43^3-42^3=5419$, $67^2-66^2=133$, $43^2-42^2=85$.

The greatest is (i) 67³ − 66³.

(i) 67³ − 66³ (= 13267)

1 to 17: 16, 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8, 17

(Check: 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, 16, 25, 16, 9, 25.)

A computer search shows this is the only way, apart from writing it backwards. The reason: 16 and 17 can each have only one neighbour (16 + 9 = 25 and 17 + 8 = 25 are the only square sums possible for them), so they must be at the two ends, and then each step is forced.

1 to 32 in a circle: 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, 14, 2, 23, 26, 10, 15 (and back to 1: 15 + 1 = 16).

16-9-7-2-14-11-5-4-12-13-3-6-10-15-1-8-17 (unique up to reversal, since 16 and 17 each have only one possible neighbour); a circle for 1–32 is given above.