Olympiad papers rarely ask you to calculate a square root. They ask you to spot one — from the last digit, from the digit count, or from a factor pattern. That is the skill this topic builds.
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Three tests will reject most non-squares in a second, without any working.
🔍 The three quick tests
Last digit. A perfect square can only end in 0, 1, 4, 5, 6 or 9. Anything ending in 2, 3, 7 or 8 is never a square.
Trailing zeros. A square ends in an even number of zeros. So 1000 is not a square, but 1600 may be.
Prime factors. In a perfect square every prime appears an even number of times; in a perfect cube, a multiple of three times.
🔢 Last digits worth knowing
| Ends in | Square ends in | Cube ends in |
|---|---|---|
| 2 | 4 | 8 |
| 3 | 9 | 7 |
| 4 | 6 | 4 |
| 7 | 9 | 3 |
| 8 | 4 | 2 |
| 9 | 1 | 9 |
Cubes are friendlier than squares. The last digit of a cube tells you the last digit of its cube root exactly: 2 and 8 swap, 3 and 7 swap, and every other digit stays the same. So the cube root of 4913 must end in 7, and since 17³ = 4913, the answer is 17.
Three identities turn long sums into one-line answers.
📈 Three patterns
Sum of the first n odd numbers = n². So 1 + 3 + 5 + … + 19 = 10² = 100.
Between n² and (n+1)² there are exactly 2n non-squares. Between 36 and 49 there are 2 × 6 = 12.
n² − (n−1)² = 2n − 1. Consecutive squares differ by the odd numbers in order.
The (a ± b)² shortcut. 97² = (100 − 3)² = 10000 − 600 + 9 = 9409. 103² = 10000 + 600 + 9 = 10609. Under Olympiad time this beats long multiplication every time.
√(a + b) is not √a + √b. √(9 + 16) = √25 = 5, but √9 + √16 = 3 + 4 = 7. This wrong step is deliberately offered as a distractor in Olympiad papers.
Prime factorisation gives the root exactly, and it also answers the "smallest number to multiply by" questions.
🌳 The method
Break the number into prime factors: 1296 = 2⁴ × 3⁴.
For a square root, halve each power: 2² × 3² = 4 × 9 = 36.
For a cube root, take a third of each power: ∛1728 = ∛(2⁶ × 3³) = 2² × 3 = 12.
To make a number a perfect square, multiply by whatever primes appear an odd number of times.
The classic question: what is the smallest number by which 2352 must be multiplied to give a perfect square? 2352 = 2⁴ × 3 × 7². The 3 is alone, so multiply by 3 — giving 7056 = 84².
12 Olympiad-level questions. Look for the pattern before you reach for long division. 🍀
