prove that any number that is a square must have one of the following for its units digit: 0,1,4,5,6,9.

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- Feb 26th 2009, 02:18 PM #1

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- Feb 26th 2009, 03:34 PM #2

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- Feb 26th 2009, 04:04 PM #3
Consider a two digit number with unit digit as a and tens digit as b

So a and b both are positive interger less than 10

So number can be written as 10b + a

Now $\displaystyle (10b + a)^2 = 100b^2 + 20ab + a^2$

Now $\displaystyle 100b^2$ and $\displaystyle 20ab$ both have 0 at unit place. So unit digit of $\displaystyle 100b^2 + 20ab + a^2$ will be decided by $\displaystyle a^2$

And for a being single digit integer, you only get 0,1,4,5,6,9.at units place of $\displaystyle a^2.$

You can extend this proof for three digit number 100c + 10b + a.

Same logic will be applied in this case