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
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Achari S Ganesha

I am a secondary school teacher teaching Mathematics up to class 10

Updated on July 22, 2022

Comments

  • Achari S Ganesha
    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
      MJD about 8 years
      The unit place of $x^2$ depends only on the unit place of $x$. How many cases are there for the unit place of $x$?