Squares and Cubes Reference

Knowing squares and cubes by heart saves time on any arithmetic-heavy question. This page lists squares from 1 to 30 and cubes from 1 to 20.

Squares (1 to 30)

Cubes (1 to 20)

Last-digit patterns

The last digit of a square or cube depends only on the last digit of the number, which makes it a fast first check on any answer:

Two things stand out. A perfect square can never end in 2, 3, 7 or 8, so a number like 4,567 is ruled out at a glance. And every digit appears exactly once in the cube column, so the last digit of a perfect cube tells you the last digit of its cube root for certain.

Shortcut: squaring a number ending in 5

For a number written as $n5$, multiply $n$ by the next whole number and write 25 after the result:

It works because $(10n+5)^2 = 100n(n+1) + 25$.

Shortcut: squaring a number near 50

Write the number as $50 \pm a$ and expand. The middle term is always a round hundred, so it stays easy to do in your head:

Worked example: the square root of a perfect square

Find $\sqrt{5476}$, given that it is a whole number.

Solution: The last digit is 6, so from the table the root ends in 4 or 6. Ignoring the last two digits leaves 54, which sits between $7^2 = 49$ and $8^2 = 64$, so the root is 74 or 76. Since $75^2 = 5625$ is already bigger than 5476, it has to be the smaller one:

Worked example: the cube root of a perfect cube

Find $\sqrt[3]{474552}$, given that it is a whole number.

Solution: The last digit is 2, and only a number ending in 8 has a cube ending in 2, so the root ends in 8. Dropping the last three digits leaves 474, which sits between $7^3 = 343$ and $8^3 = 512$, so the tens digit is 7:

This only works when you know the number is a perfect cube. For any other number the method still produces an answer, just a wrong one, with no warning.

Common mistakes

• Taking a square root of a number that isn't a perfect square. 5,477 ends in 7, which no square does, so it has no whole-number square root. Check the last digit before using the method.
• Stopping at one candidate for a square root. A square ending in 6 comes from a number ending in 4 or 6, so there are always two candidates to compare. Cube roots never have this problem.
• Dropping the middle term near 50. $53^2$ is not $2500 + 9$: the $100a$ term, 300 here, carries most of the difference.
• Confusing $n^2$ with $2n$. $3^2 = 9$, not 6. The two agree only at $n = 0$ and $n = 2$.