A Square and a Cube
Why This Matters
Here is a puzzle from the Ganita Prakash textbook.
A queen leaves her fortune of precious stones inside a safe. The code is hidden behind 100 lockers, numbered 1 to 100. A hundred people line up. Person 1 opens every locker. Person 2 toggles every 2nd locker — closes it if open, opens it if closed. Person 3 toggles every 3rd locker, Person 4 every 4th, and so on, up to Person 100.
At the end, only a few lockers are still open. They reveal the code. Which lockers stay open?
You could act it out, 100 lockers at a time. But there is a beautiful shortcut. A locker ends up open only if it was toggled an odd number of times. And how many times is a locker toggled? Exactly once for each of its factors. Locker 6 is touched by persons 1, 2, 3 and 6 — its four factors.
So the question becomes: which numbers from 1 to 100 have an odd number of factors? The answer turns out to be the square numbers — 1, 4, 9, 16, 25, … — and figuring out why is the start of this whole chapter.
This chapter is about squares and cubes: what they are, the secret patterns hidden inside them, and how to undo them with square roots and cube roots. By the end, the locker puzzle will feel obvious.
The Big Idea
A square number is what you get when you multiply a number by itself — like 5 × 5 = 25. A cube number is what you get when you multiply a number by itself three times — like 5 × 5 × 5 = 125. These two ideas come from real shapes: a square’s area is side × side, and a cube’s volume is side × side × side. Once you see squares and cubes as shapes made of unit pieces, their strange-looking patterns — odd factor counts, special last digits, sums of odd numbers — all start to make sense. And every square or cube can be undone: the square root and the cube root take you back to the side you started with.
Let’s Break It Down
We will go in order. First, square numbers and why they are called “squares”. Then the patterns hidden in perfect squares. Then square roots — how to undo a square. Then cube numbers, and finally cube roots.
Before we begin, let us refresh one idea the whole chapter leans on: factors and prime factorisation.
Square numbers and the area model
Back to the locker puzzle. We said a locker stays open only if it has an odd number of factors. Let us see why squares are the special numbers with an odd factor count.
Factors come in pairs. For 6, the pairs are 1 × 6 and 2 × 3. Each factor has a partner so that the two multiply back to 6. Because the partners are all different, the factors come in neat couples: 6 and 3 — that is 4 factors, an even number.
But look at 9. Its pairs are 1 × 9 and 3 × 3. Here something special happens: 3 partners with itself. We do not count that 3 twice — it is just one factor. So 9 has the factors 1, 3, 9 — that is 3 factors, an odd number.
This only happens when a number is a number times itself — that is, a square. The picture below shows both cases side by side.
So in the locker puzzle, every locker whose number is a square — 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 — is toggled an odd number of times and stays open. There are exactly 10 of them.
Now, why do we even call 1, 4, 9, 16 “squares”? Because they are literally the areas of squares. A square with side 3 units is filled by 3 × 3 = 9 little unit squares. A square with side 4 holds 4 × 4 = 16. The number of unit squares (the area) is side × side. The figure below makes this clear.
We write this with a small raised 2. So 3 × 3 = 3² (read “three squared”), 4 × 4 = 4² = 16, and in general n × n = n². The squares of the natural numbers — 1, 4, 9, 16, 25, … — are called perfect squares.
You can square fractions and decimals too. For example (2.5)² = 2.5 × 2.5 = 6.25, and (³⁄₅)² = ³⁄₅ × ³⁄₅ = ⁹⁄₂₅. Squaring just means “multiply the thing by itself”, whatever the thing is.
Let’s use the factor idea on a slightly bigger number.
Use factor pairs to show that 36 has an odd number of factors, and list them.
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Find every factor pair of 36 — two numbers that multiply to 36. Start small: 1 × 36, 2 × 18, 3 × 12, 4 × 9, and 6 × 6.
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Notice the last pair: 6 × 6. The number 6 partners with itself. We count that 6 only once, not twice.
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List all the factors without repeating: 1, 2, 3, 4, 6, 9, 12, 18, 36. Count them — there are 9 factors.
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9 is an odd number, so 36 has an odd number of factors. This happens because 36 = 6 × 6 is a perfect square, so one factor (6) pairs with itself. In the locker puzzle, locker 36 would stay open.
In the locker puzzle, the queen's word-clue says the passcode is the first five locker numbers that were toggled exactly twice. Which five lockers are those?
A locker is toggled once for each factor it has. Being toggled exactly twice means having exactly two factors — which is the definition of a prime number (factors 1 and itself). The first five primes are 2, 3, 5, 7, 11. So the passcode is 2-3-5-7-11.
Patterns hidden in perfect squares
If you write out the first squares — 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, … — they hide some neat patterns. These patterns are not just curiosities; they are quick tests you can use in exams.
Pattern 1 — the last digit. Look at the units digit (the last digit) of each square: 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, 1, … Every perfect square ends in 0, 1, 4, 5, 6 or 9. None ever ends in 2, 3, 7 or 8.
Why? The last digit of a square depends only on the last digit of the number you squared. Square the digits 0 through 9 and look at their last digits: 0→0, 1→1, 2→4, 3→9, 4→6, 5→5, 6→6, 7→9, 8→4, 9→1. The results only ever end in 0, 1, 4, 5, 6 or 9. The digits 2, 3, 7, 8 simply never appear. So they can never be the last digit of a square.
This gives a one-glance test. The number 4,67,853 ends in 3, so it cannot be a perfect square. But be careful — the test only works one way.
If a number ends in 0, 1, 4, 5, 6 or 9, then it must be a perfect square.
The reverse rule — squares end only in those digits — is true and easy to remember, so it feels natural to flip it around and assume any number ending that way is a square.
The last digit can only tell you when a number is NOT a square. Ending in an allowed digit does not prove it is one. For example 26 ends in 6 but is not a square; 11 ends in 1 but is not a square. You must still check by another method.
Pattern 2 — zeros at the end. Square a number that ends in zeros and watch the zeros double. 10² = 100 (one zero → two zeros). 200² = 40000 (two zeros → four zeros). 700² = 490000 (two zeros → four zeros). So a square always has an even number of zeros at the end. A number ending in an odd number of zeros (like 1000, three zeros) can never be a perfect square.
Pattern 3 — parity. The square of an even number is even (2² = 4, 4² = 16). The square of an odd number is odd (3² = 9, 5² = 25). So a number and its square always have the same parity (both odd or both even).
Pattern 4 — squares are sums of odd numbers. This is the prettiest one. Look at the gaps between consecutive squares:
4 − 1 = 3
9 − 4 = 5
16 − 9 = 7
25 − 16 = 9
The gaps are the odd numbers 3, 5, 7, 9, … So if you add up odd numbers starting from 1, you keep landing exactly on the next square:
1 = 1 = 1²
1 + 3 = 4 = 2²
1 + 3 + 5 = 9 = 3²
1 + 3 + 5 + 7 = 16 = 4²
1 + 3 + 5 + 7 + 9 = 25 = 5²
This is not a coincidence — it has a lovely picture proof. Build a square out of dots, one inverted-L layer at a time. Each new L wraps around the previous square and adds the next odd number of dots. The figure below shows it.
This pattern is also a test: a number is a perfect square exactly when you can reach 0 by subtracting 1, 3, 5, 7, … in turn. Take 25: 25 − 1 = 24, 24 − 3 = 21, 21 − 5 = 16, 16 − 7 = 9, 9 − 9 = 0. We subtracted 5 odd numbers and hit 0, so 25 = 5². Try 38 and you cross past 0 without ever landing on it, so 38 is not a perfect square.
The pattern also lets you jump from one square to the next. The n-th odd number is 2n − 1 (1st is 2×1−1 = 1, 6th is 2×6−1 = 11, and so on). So to get from one square to the next, add the right odd number.
Given that 35² = 1225, find 36² without multiplying 36 by 36.
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Every square is the sum of consecutive odd numbers starting from 1. Since 35² = 1225, that 1225 is the sum of the first 35 odd numbers.
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To get to 36², we just need to add the 36th odd number to 1225. Use the rule: the n-th odd number is 2n − 1.
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The 36th odd number is 2 × 36 − 1 = 72 − 1 = 71.
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Add it on: 1225 + 71 = 1296. So 36² = 1296, found with a single small addition instead of a long multiplication.
There is also a link to triangular numbers — 1, 3, 6, 10, 15, … (dots arranged in growing triangles). Add any two neighbouring triangular numbers and you get a perfect square: 1 + 3 = 4, 3 + 6 = 9, 6 + 10 = 16. Two triangles slot together into a square.
Square roots — undoing a square
Squaring takes a side and gives an area. The square root does the reverse: it takes the area and gives back the side.
Suppose a square garden has an area of 49 square metres. What is the length of its side? We need a number that, times itself, gives 49. Since 7 × 7 = 49, the side is 7 metres. We say 7 is the square root of 49, and write √49 = 7.
In general, if y = x², then x is a square root of y, written x = √y.
One subtlety: both 7 × 7 = 49 and (−7) × (−7) = 49 (a negative times a negative is positive). So a perfect square actually has two square roots, one positive and one negative: √49 = +7 or −7. In this chapter we will only use the positive square root.
So how do we test whether a big number like 324 is a perfect square, and find its root if it is? The most reliable method uses prime factorisation.
The idea: a perfect square is a number times itself. So in its prime factorisation, the primes must split into two identical groups. Whatever is in one group, multiplied out, is the square root. The figure below shows this for 324, and also shows a number (156) that fails the test.
Let’s run this test on a fresh number, step by step.
Is 1156 a perfect square? If so, find its square root using prime factorisation.
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Find the prime factorisation of 1156. Divide by the smallest primes: 1156 ÷ 2 = 578, 578 ÷ 2 = 289. Now 289 is not even; try the next primes. 289 ÷ 17 = 17. So 1156 = 2 × 2 × 17 × 17.
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Try to make two identical groups. Each group needs the same primes. Take one 2 and one 17 into each group: (2 × 17) and (2 × 17).
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Both groups are the same, so 1156 = (2 × 17) × (2 × 17) = 34 × 34. It is a perfect square.
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The square root is one whole group: 2 × 17 = 34. So √1156 = 34. (Check: 34 × 34 = 1156. ✓)
What if the number is not a perfect square, but you still want a rough idea of its root? Then you estimate by trapping it between two squares you know.
Let’s estimate the square root of a number that isn’t a perfect square.
Estimate √250 — find the whole number it is closest to.
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Trap 250 between two squares you know. 15² = 225 and 16² = 256. Since 225 < 250 < 256, the root is between 15 and 16: so 15 < √250 < 16.
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Decide which end it is nearer to. Compare the gaps: 250 − 225 = 25, but 256 − 250 = only 6. So 250 is much closer to 256.
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Since 250 is so close to 16² = 256, its square root is very close to 16 (just a little less than 16).
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So √250 ≈ 16 (slightly under 16). We could not get an exact whole number because 250 is not a perfect square, but we pinned it down quickly.
Cube numbers — multiplying three times
Squares come from flat squares. Cubes come from solid cubes. A cube is a solid box where all edges are equal and all corners are right angles — like a dice or an ice cube.
How many tiny 1 cm cubes fit inside a cube of side 2 cm? Picture it as layers. The bottom layer is a 2 × 2 square of small cubes = 4 cubes. There are 2 such layers stacked up. So 2 × 2 × 2 = 8 small cubes. For a cube of side 3, each layer is 3 × 3 = 9 cubes, and there are 3 layers: 3 × 3 × 3 = 27. The figure below shows the idea.
We write a number times itself three times with a small raised 3. So 5 × 5 × 5 = 5³ (read “five cubed”) = 125, and in general n × n × n = n³. The numbers 1, 8, 27, 64, 125, … are called perfect cubes.
Is 9 a cube? No — 2³ = 8 and 3³ = 27, and there is nothing in between, so no number from 10 to 26 is a cube either. Cubes grow fast, so there are far fewer of them. Just like squares, you can cube fractions and negatives: (−6)³ = −6 × −6 × −6 = −216 (two negatives make a positive, then one more negative makes the answer negative).
Cubes hide patterns too. Add consecutive odd numbers in growing blocks and you get the cubes:
1 = 1³
3 + 5 = 8 = 2³
7 + 9 + 11 = 27 = 3³
13 + 15 + 17 + 19 = 64 = 4³
Block n uses n odd numbers, and they always add to n³.
There is a famous cube story. The mathematician Srinivasa Ramanujan was ill in hospital, and his friend G. H. Hardy came to visit. Hardy mentioned that his taxi was numbered 1729, a “rather dull number”. Ramanujan instantly replied that it was very interesting — it is the smallest number that can be written as the sum of two cubes in two different ways:
1729 = 1³ + 12³
1729 = 9³ + 10³
Because of this, 1729 is now called the Hardy-Ramanujan number, and such numbers are called taxicab numbers.
A perfect square always ends in 0, 1, 4, 5, 6 or 9. Can you say the same about perfect cubes? Look at the cubes 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000.
No. Cubes can end in any digit. Look at the last digits of 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000: they are 1, 8, 7, 4, 5, 6, 3, 2, 9, 0 — every digit from 0 to 9 appears. Unlike squares, cubes are not restricted: 2, 3, 7 and 8 are all possible last digits of a cube.
Cube roots — undoing a cube
Just as the square root undoes squaring, the cube root undoes cubing. Since 8 = 2³, we say 2 is the cube root of 8 and write ³√8 = 2.
In general, if y = x³, then x = ³√y.
To find a cube root, we use prime factorisation again — but with one change. For a square, we split the primes into two identical groups. For a cube, we split them into three identical groups. The figure below shows this for 3375, and shows 500 failing the test.
Why threes? Because cubing a number repeats every prime factor three times. If 15 = 3 × 5, then 15³ = (3 × 5)³ = 3 × 3 × 3 × 5 × 5 × 5 — each prime appears exactly three times. So a number is a perfect cube exactly when each of its primes appears a number of times that is a multiple of 3, letting them form complete triples.
Let’s find a cube root from scratch.
Find the cube root of 1728 using prime factorisation.
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Prime factorise 1728. Keep dividing by 2: 1728 ÷ 2 = 864, ÷ 2 = 432, ÷ 2 = 216, ÷ 2 = 108, ÷ 2 = 54, ÷ 2 = 27. That is six 2s. Then 27 = 3 × 3 × 3. So 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3.
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Group the primes into triples: (2 × 2 × 2), (2 × 2 × 2), and (3 × 3 × 3). The six 2s make two triples of 2; the three 3s make one triple of 3.
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For a cube root, build one factor by taking one prime from each triple. There are two triples of 2, so take a 2 for each, and there is one triple of 3, so take a 3. That gives 2 × 2 × 3.
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Multiply: 2 × 2 × 3 = 12. So ³√1728 = 12. (Check: 12³ = 12 × 12 × 12 = 1728. ✓)
There is also a quick “guessing” trick for nice cube roots, using the last digit. The cubes of 0–9 end in 0,1,8,7,4,5,6,3,2,9 — and crucially each last digit is unique. So the last digit of a cube tells you the last digit of its cube root. For 1331: it ends in 1, and the only digit whose cube ends in 1 is 1, so the root ends in 1. Since 1331 is a 4-digit number between 10³ = 1000 and 20³ = 8000, the root is between 10 and 20. Ending in 1 → the root is 11. Indeed 11³ = 1331.
Common Mistakes
These trip up students every year. Spotting them now keeps the marks safe.
The square root of a number times itself, like √(7 × 7), needs a calculation — you must work out 49 first, then find its root.
Square root usually feels like a hard step that 'must be computed', so doing the multiplication first feels like the safe, proper way to start.
Squaring and square-rooting are opposite operations that cancel each other. √(7 × 7) = √(7²) = 7 straight away. There is no need to compute 49 and undo it. In general √(n²) = n.
A square root and a cube root work the same way — split the prime factors into two equal groups for both.
Both are 'roots' found by prime factorisation, and the square-root method of pairing is learnt first, so the brain reuses it for cubes by habit.
The number of groups must match the power. For a square root, split the primes into TWO identical groups. For a cube root, split them into THREE identical groups. One whole group is the root in each case.
Squaring a number makes it bigger and cubing makes it even bigger, so squares are always smaller than the original number and roots are always smaller too — wait, so a square root must be larger than the number.
With ordinary whole numbers bigger than 1, squaring does make things larger, so it is tempting to flip that and assume the inverse, the square root, must make things larger again.
The square root makes a number SMALLER (for numbers above 1), because it undoes the growth that squaring caused. √49 = 7, and 7 is smaller than 49. The root takes you back to the side, which is smaller than the area.
Quick Check
Answer each one, then read the explanation that appears.
Which of these numbers can you say at once is NOT a perfect square, just by its last digit?
Perfect squares can only end in 0, 1, 4, 5, 6 or 9. The number 2048 ends in 8, which is impossible for a square, so it is definitely not a perfect square. (1024 = 32², 1089 = 33² and 1156 = 34² are all genuine squares — their last digits 4, 9 and 6 are allowed.)
How many of the numbers 1 to 100 stay open in the locker puzzle?
Only lockers with an odd number of factors stay open, and those are exactly the perfect squares. The squares from 1 to 100 are 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100 — that is 10 lockers.
Using prime factors, which grouping correctly gives the square root of 324 (= 2 × 2 × 3 × 3 × 3 × 3)?
For a square root the primes must split into two identical groups. The only way to do that with 2, 2, 3, 3, 3, 3 is (2 × 3 × 3) and (2 × 3 × 3), each equal to 18. So √324 = 18. The other groupings are not two equal groups.
What is the cube root of 27000?
27000 = 27 × 1000 = 3³ × 10³ = (3 × 10)³ = 30³. So ³√27000 = 30. (Quick check by last digit: 27000 ends in 0, so the root ends in 0; and it sits between 10³ and 100³, pointing to 30.)
Practice Problems
Try each one yourself first, then tap to see the full solution.
Easy
Find the length of the side of a square whose area is 441 m².
The side length is √441. Prime factorise 441: 441 ÷ 3 = 147, 147 ÷ 3 = 49, 49 = 7 × 7. So 441 = 3 × 3 × 7 × 7.
Make two identical groups: (3 × 7) and (3 × 7). Both are 21, so 441 = 21 × 21.
The square root is one group: 3 × 7 = 21.
So the side of the square is 21 metres.
Which of these are NOT perfect squares, just by checking the last digit: 2032, 2048, 1027, 1089?
A perfect square can only end in 0, 1, 4, 5, 6 or 9.
- 2032 ends in 2 → not a perfect square.
- 2048 ends in 8 → not a perfect square.
- 1027 ends in 7 → not a perfect square.
- 1089 ends in 9 → allowed, so the test cannot rule it out (in fact 1089 = 33²).
So 2032, 2048 and 1027 are not perfect squares.
Find the cube roots of 64, 512 and 729.
Use the cubes you know: 4³ = 64, 8³ = 512, 9³ = 729.
So:
- ³√64 = 4
- ³√512 = 8
- ³√729 = 9
(You can check 729 by factorising: 729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3)³ = 9³.)
Medium
Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Then find the square root of that product.
Prime factorise 9408: 9408 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7 × 7.
Group into pairs to spot what is unpaired: (2 × 2), (2 × 2), (2 × 2), (7 × 7), and a lonely 3.
For a perfect square, every prime must appear an even number of times (so it can be paired). The 2s are fine (six of them), the 7s are fine (two), but the 3 appears only once. We need one more 3.
So multiply by 3: 9408 × 3 = 28224.
Now 28224 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7. Take one prime from each pair for the root: (2 × 2 × 2) × 3 × 7 = 8 × 21 = 168.
So multiply by 3, and √28224 = 168.
Find the smallest square number that is divisible by each of 4, 9 and 10.
First, the number must be divisible by 4, 9 and 10, so it must be a multiple of their LCM.
LCM of 4, 9, 10: 4 = 2², 9 = 3², 10 = 2 × 5. Take the highest power of each prime: 2² × 3² × 5 = 180.
So the number must be a multiple of 180 = 2² × 3² × 5. But for it to be a perfect square, every prime needs an even power. Here 2 and 3 already have even powers, but 5 appears only once.
Multiply by one more 5: 180 × 5 = 900.
Check: 900 = 2² × 3² × 5² = (2 × 3 × 5)² = 30². It is a perfect square, it is divisible by 4, 9 and 10, and it is the smallest such. Answer: 900.
Without doing the full multiplication, find the sum 91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109.
These are 10 consecutive odd numbers. Recall the cube pattern: a block of n consecutive odd numbers (chosen as the right block) adds up to n³.
Here there are 10 numbers, so the block adds up to 10³ = 1000.
Quick check using pairs: pair the outer terms — 91 + 109 = 200, 93 + 107 = 200, 95 + 105 = 200, 97 + 103 = 200, 99 + 101 = 200. Five pairs of 200 = 1000. ✓
So the sum is 1000.
Challenge
Decide if each statement is true or false, with a reason: (i) The cube of any odd number is even. (ii) There is no perfect cube that ends with 8. (iii) The cube of a 2-digit number may have seven or more digits.
(i) False. Odd × odd = odd, and odd × odd × odd is still odd. For example 3³ = 27 (odd), 5³ = 125 (odd). The cube of an odd number is always odd, never even.
(ii) False. 2³ = 8 ends in 8. Also 12³ = 1728 ends in 8. So perfect cubes can certainly end in 8 (cubes can end in any digit).
(iii) False. The biggest 2-digit number is 99, and 99³ = 970299, which has only 6 digits. Every 2-digit number cubed gives at most 6 digits, never 7. (The smallest 2-digit cube, 10³ = 1000, has 4 digits, so 2-digit cubes have 4, 5 or 6 digits.)
Which is the greatest: (i) 67³ − 66³, (ii) 43³ − 42³, (iii) 67² − 66², (iv) 43² − 42²? Explain your reasoning without heavy calculation.
First handle the squares using the difference pattern. The gap between consecutive squares n² and (n−1)² is the odd number 2n − 1.
- (iii) 67² − 66² = 2 × 67 − 1 = 133.
- (iv) 43² − 42² = 2 × 43 − 1 = 85.
These differences of squares are small (around a hundred).
Now the cubes. The gap between consecutive cubes n³ − (n−1)³ grows much faster — it works out to 3n² − 3n + 1, which for two-digit n is in the thousands.
- (ii) 43³ − 42³ = 3 × 43² − 3 × 43 + 1 = 3 × 1849 − 129 + 1 = 5547 − 128 = 5419.
- (i) 67³ − 66³ = 3 × 67² − 3 × 67 + 1 = 3 × 4489 − 201 + 1 = 13467 − 200 = 13267.
The cube differences dwarf the square differences, and among the two cubes the one with the larger base (67) gives the bigger gap.
So (i) 67³ − 66³ is the greatest.
Summary
- A number multiplied by itself is a square number; the squares of natural numbers (1, 4, 9, 16, …) are perfect squares. We write n × n as n².
- A number multiplied by itself three times is a cube number; 1, 8, 27, 64, … are perfect cubes. We write n × n × n as n³.
- Only square numbers have an odd number of factors, because in a square one factor pairs with itself.
- Perfect squares end only in 0, 1, 4, 5, 6 or 9, never in 2, 3, 7 or 8, and have an even number of trailing zeros. (A number ending in an allowed digit still need not be a square.)
- The sum of the first n odd numbers is n² (a fact you can see by building a square out of inverted-L dot layers). Cubes are sums of blocks of consecutive odd numbers.
- The square root undoes a square; the cube root undoes a cube. The symbols are √ and ³√.
- A number is a perfect square if its prime factors split into two identical groups, and a perfect cube if they split into three identical groups. One whole group is the root.
What’s Next
You can now spot squares and cubes on sight, find their roots, and use their hidden patterns. That number-sense will keep paying off.
Next, in Chapter 2 — Power Play, you will go beyond squares and cubes to powers in general — what 2⁵ or 10⁸ really mean, the rules for multiplying and dividing powers, and how powers let us write enormous (and tiny) numbers neatly. The squaring and cubing you mastered here are just the first two steps of that bigger story. Onward!
Frequently Asked Questions
Why do only square numbers have an odd number of factors?
Factors come in pairs that multiply to give the number, like 1 and 6, or 2 and 3 for the number 6. In a square number one factor pairs with itself, like 3 and 3 for 9. That repeated factor is counted only once, so the total count is odd. Every non-square has its factors in clean, different pairs, giving an even count.
What digits can a perfect square end in?
A perfect square can only end in 0, 1, 4, 5, 6 or 9. It can never end in 2, 3, 7 or 8. So if a number ends in 2, 3, 7 or 8 you can say at once it is not a perfect square. But ending in an allowed digit does not prove it is a square.
How do you find a square root using prime factorisation?
Break the number into its prime factors. If you can split those primes into two identical groups, the number is a perfect square. The product of the primes in one group is the square root. For example 324 is 2 times 2 times 3 times 3 times 3 times 3, which splits into two groups of 2 times 3 times 3 equals 18, so the square root of 324 is 18.
How is finding a cube root different from finding a square root?
For a square root you split the prime factors into two identical groups and one group is the root. For a cube root you split them into three identical groups instead, and one group is the root. For example 3375 is 3 times 3 times 3 times 5 times 5 times 5, which makes three groups of 3 times 5 equals 15, so the cube root of 3375 is 15.
What is the Hardy-Ramanujan number 1729?
1729 is the smallest number that can be written as the sum of two cubes in two different ways. It is 1 cubed plus 12 cubed, and also 9 cubed plus 10 cubed. Ramanujan spotted this instantly when Hardy called it a dull taxicab number, so it is now called the Hardy-Ramanujan number.