Last digit of a perfect square must be $0, 1, 4, 5, 6,$ or $9$
3,788
Solution 1
Hint: Compute the remainder when you divide the square of each possible digit by 10. Then these are the only possible values for the last digit you can get.
Solution 2
Here is a sketch
You can express any integer in the form $10n\pm r$ with $0\le r \le 5$
Then $(10n\pm r)^2=10\cdot(10n^2\pm 2nr)+r^2$
The final digit of the square will be the final digit of $r^2$, for which there are just six possibilities.
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Achari S Ganesha
I am a secondary school teacher teaching Mathematics up to class 10
Updated on July 22, 2022Comments

Achari S Ganesha over 1 year
We know that a perfect square number can contain 0,1,4,5,6 and 9 in its unit place.
How can I prove that a perfect square number cannot contain 2,3,7 and 8 in its unit place?

MJD about 8 yearsThe unit place of $x^2$ depends only on the unit place of $x$. How many cases are there for the unit place of $x$?
