A Square and A Cube
Why do only SOME numbers have neat square shapes? Solve the famous 100-locker puzzle, then master squares, square roots, cubes and cube roots — with every exercise solved.
The 100-locker puzzle
A locker gets toggled once for every number that divides its locker number. Locker 12 gets toggled by persons 1, 2, 3, 4, 6, 12 — that's 6 times (even), so it ends closed. A locker ends OPEN only if it's toggled an odd number of times.
Divisors normally come in pairs: for 12, pair up (1,12), (2,6), (3,4) — 3 pairs = 6 divisors, an even count. But for 16: (1,16), (2,8), (4,4) — wait, 4 pairs with itself! Since 4×4=16, the number 4 doesn't get a partner, so 16 has an odd number of divisors (1,2,4,8,16 = 5 divisors). This only happens when the locker number is a perfect square — that's the whole secret of this chapter!
A locker gets touched exactly twice only if its number has exactly 2 divisors — and that's the definition of a prime number (only 1 and itself as factors)!
The passcode is 2-3-5-7-11 — the first five prime numbers!
Square numbers
A square number (or perfect square) is what you get when you multiply a whole number by itself: n² = n × n. For example, 7² = 7 × 7 = 49.
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| n² | 1 | 4 | 9 | 16 | 25 | 36 | 49 | 64 | 81 | 100 |
A perfect square can only end in 0, 1, 4, 5, 6, or 9 — NEVER in 2, 3, 7, or 8! Quick check: 38² ends in 4 (could be a square), but does that guarantee it? No — this rule only rules OUT numbers, it can't confirm one is a square on its own.
If a number has 3 zeros at the end (like 2000), its square will have exactly 6 zeros (double!). A perfect square can never have an odd number of trailing zeros.
🔢 The sum-of-odd-numbers pattern
Since 35² = 1225, what's 36²? Just add the next odd number! The 36th odd number is 2(36)-1 = 71.
🔺 Squares hide inside triangular numbers too
1+3=2², 3+6=3², 6+10=4². The pattern: T_n + T_{n+1} = (n+1)². Try it yourself: 15+21=?
15 and 21 are neighbouring triangular numbers (1, 3, 6, 10, 15, 21, …), so the rule T_n + T_{n+1} = (n+1)² says their sum should land exactly on a square. Just add them and check:
Is 36 a perfect square? Yes — 6×6=36, so 15+21=36=6². The pattern holds again!
The book shows a growing picture: a small square of dots gets an "L" of extra dots wrapped around two sides to make the next bigger square. In dot-counts: 1+3=2², then that 2×2 square gets an L of 5 dots added (3+2=5) to make 3×3=3², then an L of 7 dots (5+2=7) makes 4×4=4².
Your turn: on squared paper, draw the next box — a 5×5 square of dots — by wrapping one more L-shape of dots around the 4×4 square. How many NEW dots does that L need, and what's the running total?
Finding square roots
The square root undoes squaring: if 7²=49, then √49=7. Every positive perfect square actually has TWO square roots (+7 and -7) — but in this chapter we'll focus on the positive one.
Method 1 · Successive subtraction
Keep subtracting consecutive odd numbers (1, 3, 5, 7…) from your number. If you land EXACTLY on 0, it's a perfect square — and the count of subtractions is the square root!
Method 2 · Prime factorization (pair up the factors)
Break 324 into prime factors, then pair up identical primes:
Every prime paired up perfectly! Take one from each pair and multiply: 2 × 3 × 3 = 18.
156 = 2×2×3×13. The 2's pair up nicely, but 3 and 13 are both alone — no partner! That leftover means 156 is NOT a perfect square.
🎮 Try it: pair up the prime factors
Method 3 · Estimation
Bracket between known squares first: 40²=1600, 50²=2500, so the root is between 40 and 50. Narrow further: 45²=2025 is too big, so it's between 40 and 45. Try 44: 44²=1936.
Cubes and cube roots
A cube number is a number multiplied by itself THREE times: n³ = n × n × n. Unlike squares, cubes can end in ANY digit — even 2, 3, 7, or 8 (e.g. 2³=8, 12³=1728).
| n | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| n³ | 1 | 8 | 27 | 64 | 125 | 216 |
Mathematician Srinivasa Ramanujan once said 1729 seemed like a "dull" number to a friend visiting him in hospital — but Ramanujan instantly replied it's actually fascinating: it's the SMALLEST number that can be written as the sum of two cubes in two different ways:
Just like with 1729, the trick is to hunt for TWO different pairs of cubes that add to the same total. There's no shortcut except trying nearby cubes and checking — let's verify both pairs by actually working out each cube:
4104 = 2³+16³ = 9³+15³, and 13832 = 2³+24³ = 18³+20³ — both numbers really do split into two different cube-pairs, exactly like 1729 does!
🔺 Odd numbers add up to cubes too!
Each cube n³ uses exactly n consecutive odd numbers. So 10³ uses exactly 10 consecutive odd numbers. Try summing 10 odd numbers starting from 91: 91+93+…+109 — it equals exactly 1000 = 10³, with no long addition required!
Finding cube roots by prime factorization
This time, group primes into triples (groups of 3) instead of pairs:
Both groups are perfect triples! Take one from each: 3 × 5 = 15.
500 = 2 × 2 × 5 × 5 × 5. The 5's form a perfect triple, but the 2's only have 2 copies, not 3 — incomplete group, so 500 is not a perfect cube.
Ancient Babylonians (around 1700 BCE) wrote the first known lists of squares and cubes on clay tablets! In Sanskrit, varga means "square" and ghana means "cube" — and the word for "root" comes from mula, meaning the root of a plant. That word travelled through Arabic (jidhr) and Latin (radix) all the way to the word "radical" we still use in maths today!
Every question from the book
Start with the fast check from Section 2: a perfect square can only end in 0, 1, 4, 5, 6 or 9. Look at the units digit of each:
2032, 2048, and 1027 are eliminated immediately by the units-digit rule. But ending in 9 only means 1089 might be a square — it doesn't prove it. So check by finding its square root: 33×33 = 1089. ✓
2032, 2048, and 1027 are NOT perfect squares. 1089 actually IS one — it's 33²!
You don't need to square the whole number — a square's units digit only depends on the units digit of the base number being squared. Look at each base's last digit and square just that:
So only bases ending in 8 or 2 will give a square ending in 4. Let's confirm with the full squares:
108² and 292² end in 4.
126² = 125² + 2(125) + 1 (using (a+1)²=a²+2a+1):
A square garden means side × side = area, so we need the square root of 441. Bracket it between known squares first: 20²=400 and 21²=441 — that's an exact match, no estimating needed! (You could also confirm with prime factorisation: 441 = 3×3×7×7 = (3×7)².)
The side is 21 metres (check: 21×21=441 ✓).
First find the LCM of 4, 9, 10 = 180 = 2² × 3² × 5. The 5 has no partner, so multiply by another 5:
Step 1 — break 9408 down into prime factors by dividing repeatedly by the smallest prime that fits:
So 9408 = 2×2×2×2×2×2×3×7×7 = 2⁶ × 3 × 7².
Step 2 — pair them up: the six 2's make three full pairs (2²)³, and the two 7's make one pair. But the lone 3 has no partner!
Step 3 — to complete the pair, multiply by one more 3:
Multiply by 3 to get 28224 = 168² (check: 168 × 168 = 28224 ✓).
The direct way: work out both squares, then count the whole numbers strictly in between (subtract 1 from the gap, since both endpoints are excluded):
Notice something? 32 is exactly 2×16, and 198 is exactly 2×99. That's because expanding (n+1)² = n² + 2n + 1 shows the two squares always differ by exactly 2n+1 — so subtracting 1 for the endpoints leaves exactly 2n numbers strictly between them.
32 numbers lie between 16² and 17²; 198 numbers lie between 99² and 100². Shortcut: between n² and (n+1)² there are always exactly 2n numbers!
Spot the rule: each row uses n, n+1, and n×(n+1), and they always sum to [n×(n+1)+1]². Check row 1: n=1 gives 1, 2, 1×2=2, and 1×2+1=3 — matches 1²+2²+2²=3²! ✓
Answers: (21)² and (90)² = (91)² — here n=9, so n×(n+1)=90 and n×(n+1)+1=91.
The picture is a checkerboard of 9 columns × 8 rows = 72 tiles (some upright grids, some tilted into diamonds — but each tile, tilted or not, is a 6×6 grid = 36 tiny squares).
2592 = 2⁵ × 3⁴ (prime factorisation: 2592 = 2×2×2×2×2×3×3×3×3). Bonus: since every prime's count is even×odd mixed (5 twos, 4 threes), 2592 is not a perfect square itself — the lone extra 2 has no partner!
∛27000 — break into prime factors and group into triples of 3:
Take one factor from each triple: 2 × 3 × 5 = 30.
∛10648 — same method, divide out primes repeatedly:
∛27000 = 30 (check: 30×30×30=27000 ✓) and ∛10648 = 22 (check: 22×22×22=10648 ✓).
Step 1 — find the prime factorisation of 1323 by dividing repeatedly:
So 1323 = 3×3×3×7×7 = 3³ × 7².
Step 2 — for a perfect cube, every prime needs a group of THREE (not two, like for squares). The three 3's already form a complete triple, but the two 7's are one short of a triple.
Step 3 — multiply by one more 7 to complete the triple of 7's:
Multiply by 7 to get 9261 = 21³ (check: 21×21×21=9261 ✓).
ALL FIVE are False!
(i) 3³=27 is odd, not even — odd³ is always odd.
(ii) 2³=8 and 12³=1728 both end in 8!
(iii) The smallest 2-digit number's cube, 10³=1000, already has 4 digits.
(iv) The largest 2-digit number's cube, 99³=970299, has only 6 digits, never 7+. (v) 2³=8 has factors 1,2,4,8 — that's 4 factors, an even count.
Two clues do all the work here, no factorising needed: (1) the last digit of the cube tells you the last digit of the root (1→1, 8→2, 7→3, 4→4, 5→5, 6→6, 3→7, 2→8, 9→9, 0→0), and (2) the number of digits tells you roughly which "ten" the root sits in, using the table of cubes from Section 4 (10³=1000, 20³=8000, 30³=27000…).
∛1331=11, ∛4913=17, ∛12167=23, ∛32768=32 — all found by combining the last-digit trick with a quick range estimate!
Before crunching numbers, think about size: cube-differences (near 67³ and 43³) involve much bigger numbers than square-differences (near 67² and 43²), so the two cube options are already likely to beat the two square options. Between the cube options, bigger base numbers (67 vs 43) tend to give a bigger gap too. Let's confirm by working out all four:
67³−66³ = 13,267 is the greatest — exactly as predicted: cube-differences beat square-differences, and the larger base (67) beats the smaller one (43).
Practice like the real exam
Section A · MCQ (1 mark each)
Recall the rule from Section 2: a perfect square can only end in 0, 1, 4, 5, 6, or 9. Check each option against that list: 6 is allowed (e.g. 4²=16), 9 is allowed (e.g. 3²=9), but 8 is not in the list — no square number ever ends in 8.
(c) 8
Ask: what number, multiplied by itself three times, gives 125? Test option (b): 5×5×5 = 25×5 = 125 — that matches! (Option (a) 25³ and (c) 15³ are both much bigger than 125, so they can be ruled out immediately.)
(b) 5
Section B · short answer (2 marks each)
It never lands exactly on 0 — 148 is NOT a perfect square. (You can verify this in the Section 3 calculator above!)
Section C · longer answer (3 marks each)
√576 = 24 (check: 24×24=576 ✓)
Section D · 4 marks each
Factorize: 1500 = 2² × 3 × 5³. The 5's already form a perfect triple, but 2² needs one more 2, and the lone 3 needs two more 3's to form a triple:
18 extra sets of blocks needed (in the same ratio as the original factors) to reach 27,000 = 30³.
Section E · case study (4 marks)
(a) √2025 = 45 m (since 45×45=2025) — the hall is a 45 m × 45 m square.
Every prime pairs up perfectly — confirming 2025 is a perfect square, and √2025 = 3×3×5 = 45. ✓ Matches part (a)!
Puzzle Time: Square Pairs! 🧩
Arrange the numbers 1 to 17 (each used exactly once) in a single row so that every pair of neighbours adds up to a perfect square. Grab some paper and try it before you peek at the answer — think about which numbers are the "hardest to place" first (hint: 16 and 17 each have only ONE possible partner, so they HAVE to go at the two ends of the row)!
Check every neighbour pair:
Every single pair is a perfect square (9, 16, or 25)! Notice 16 and 17 had to sit at the two ends — each only pairs with ONE other number (16+9=25 and 8+17=25), so they can't be stuck in the middle with a partner on both sides. That's why there's basically one "shape" of solution (you could also just flip the whole row end-to-end).
Now the harder version: arrange 1 to 32 (each used once) in a circle so that every pair of neighbours — including the pair that wraps back around to close the loop! — adds up to a perfect square.
Every one of the 32 neighbour-pairs (including 15 wrapping back to 1) lands on a perfect square:
A circle is trickier than a row because every number needs TWO working neighbours (not just an inner number) — there's no "end of the line" to hide an awkward number at. This is just one valid loop (read it forwards or backwards from any starting point) — keen puzzlers can try hunting for a genuinely different one!
You did it! 🎉
🏁 Chapter 1 of 7 · Term 1 Maths · Prishita, Class 8