The number ends with 2 its perfect squre or non perfect square
Answers
Answer:
Non-perfect square it is.
Step-by-step explanation:
Let's see what would be ending digit for the perfect squares of numbers from 0 to 9.
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0^2= 0.
1^2= 1.
2^2= 4.
3^2= 9.
4^2= 16 (ending digit is 6).
5^2= 25 (ending digit is 5).
6^2= 36 (ending digit is 6).
7^2= 49 (ending digit is 9).
8^2= 64 (ending digit is 4).
9^2= 81 (ending digit is 1).
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Hence, it is clear that any integer number, when squared, can give us the ending digit as 0, 1, 4, 5, 6 or 9 only.
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Therefore, ending digit as 2, 3, 7 or 8 is not possible, for the squares of integers (perfect squares).
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For any big number, the ending digit of the number can be only between 0 and 9.
So, the big number's square will also contain ending digit as 0, 1, 4, 5, 6 or 9 only.