Theorem

Last digit of a perfect square

Arithmetic • Number System • Perfect Squares

Statement

A perfect square ends in one of

0, 1, 4, 5, 6, 9

only, because those are the last digits of

0^{2}

9^{2}

; a number ending in

2, 3, 7

8

is therefore never a square, while one ending in a permitted digit still has to be checked.

How to use it

1Read the units digit first: a $2$ , $3$ , $7$ or $8$ settles the question with no computation at all.
2For a survivor, bracket it between two nearby squares to decide.
3A square ending in $5$ must end in $25$ , and a square ending in $0$ must end in an even number of zeros.

Which of

4358

6724

and

9803

can be perfect squares?

Step-by-step

1
$4358$ ends in $8$ and $9803$ ends in $3$
Why: Neither digit can end a perfect square, so both are ruled out.
2
$6724$ ends in $4$ , and $8 2^{2} = 6724$
Why: A permitted last digit only shortlists a number; the exact root is what confirms it.

Answer

6724

only

Watch out

The test never proves a number is a square —

24

and

44

end in

4

and neither is one. It rules out and nothing more.

Also known as

units digit of a perfect square, can this be a perfect square, squares end in 0 1 4 5 6 9, last digit test for squares