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A square number can't end with 2, 3, 7, or 8.

Remember that 0^2=0, 1^2=1, 2^2=4, 3^2 =9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64, and 9^2=81.

Looking at some examples of squared number and using the distributive property shows:

Using the distributive property on 72 x 72

72 x 72 = 72 x 70 + 2 x72 = 72 x 70 +70 x 2 +2 x 2

Here, the problem that gives the ones place digit is 2 x 2.

Using the distributive property on 981 x 981

981 x 981 = 981 x 900 + 981 x 80 + 981 x 1 = 981 x 900 +981 x 80 +900 x 1 +80 x 1 +1 x1

Here, the problem that gives the ones place digit is 1 x1.

Using the distributive property on 86 x 86 (I skipped middle steps, but I think you get the point from above problems)

= 86 x 80 +80 x 6 +6 x 6

Again, 6x6 gives the ones place digit.

In any problem, after you break down the problem with the distributive property, you can see that the ones place digit of the number in the original problem squared is what will give the ones place digit in the answer.

As we can see in looking at one digit numbers squared, none of them will end in 2,3, 7 or 8, so no square number will end in 2, 3, 7, or 8.

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Q: Can a square number end with a 2 or 3 or 7 or 8 and why?
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