The other day my friend impressed me with her knowledge of perfect squares. She had all the squares up to 30 committed to memory, whereas I struggle to calculate 16^{2}! As she listed the squares, I started to notice a pattern:

20^{2} |
400 |

21^{2} |
441 |

22^{2} |
484 |

23^{2} |
529 |

24^{2} |
576 |

Can you see it? The last digit of the answer can be found by squaring the last digit of the question. For example, 2^{2} = 4, so the last digit of 22^{2} must be 4. This trick works for even large numbers. You may not know what 623^{2} is, but I guarantee you the last digit is 9.^{1}

However, I found another pattern that allowed me to calculate squares in my head. Here is how it works (using 22).

First, subtract 10: 22 – 10 = 12

Second, multiply the number by 4: 12 x 4 = 48

Third, add a zero: 48 → 480

Fourth, add the square of the last digit of the question: 480 + 4 = 484

Cool eh?^{2} After practicing a few times I was able to perform this process in a few seconds. However, we need to be careful as we extend the strategy. Let’s do 29:

First, subtract 10: 29 – 10 = 19

Second, multiply the number by 4: 19 x 4 = 76

Third, add a zero: 76 → 760

Fourth, add the square of the last digit of the question: 760 + 81 = 841

In this case, 19 x 4 is a bit trickier to do mentally. In addition, 760 + 81 requires a mental carry, which takes some practice. After thinking about it for a while, I realized that this pattern only applies to the numbers 20-30. However, there is a way to extend the method to the numbers 10-20. Consider the following example 17:

First, subtract 10: 17 – 10 = 7

Second, multiply the number by 4: 7 x 4 = 28

Third, add a zero: 28 → 280

Fourth, add the square of the **difference from 20 **(20 – 17 = 3): 280 + 9 = 289

To recap, here is my method for calculating squares to 30.

- For the squares from 1-10, simply remember your multiplication facts.
- For the squares from 10-20, use the process, remembering to add the square of the difference from 20.
- For the squares from 20-30, use the process, remembering to add the square of the last digit.
^{3}

My friend criticized my method; calling it overly complicated. She said that memorizing the squares was easy and I should do the same. While I agree that memorizing the squares is faster and more reliable than working through my, somewhat tedious method, I quite enjoyed finding the pattern. I hope the pattern will aid in your memorization. I know it has aided in mine.

^{1}The reason that the trick works can be explained as follows.

Break the number up into two pieces (the last digit and the remaining number): 623 = 620 + 3

Square the number: (620 + 3)^{2} = (620 + 3)(620 + 3)

Expand the brackets using FOIL: (620)(620) + (620)(3) + (3)(620) + (3)(3)

Calculate each term: 384400 + 1860 + 1860 + 9

Observe that every term has a 0 in the 1’s place expect the last term. This will always be the case. Hence, the last digit of the final answer is always determined by the last term of FOIL. The last term of FOIL is calculated by square the last digit of the original question (3×3).

^{2}Similar to the note above, the pattern works because of the distributive property. Consider 22^{2}. The multiply by 4 and add a zero step can be combined by simply multiplying by 40.

(22 – 10) x 40 + 2^{2}

= 12 x 40 + 2^{2} (simplifying inside the brackets)

= 10 x 40 + 2 x 40 + 2^{2 } (splitting off 2 groups of 40)

= 20 x 20 + 2 x 40 + 2^{2 } (rewriting the first multiplication)

= 20 x 20 + 20 x 4 + 2^{2 } (rewriting the second multiplication)

= 20 x 20 + 20 x 2 + 20 x 2 + 2^{2 } (splitting off 2 groups of 20)

= (20 + 2)^{2} (backwards foil)

= (22)^{2} (simplifying inside the brackets)

^{3}I leave it as an exercise for the reader to show why the trick works from 10 – 20.

How about squares in the 40’s and 50’s?

There are separate patterns to find the first two digits and the last two digits…