Unit digit of a square integer

An integer's unit digit must be from 0 to 9 and that will determine the unit digit of its square. Squaring the unit digit shows that 1 and 9 end in 1. 2 and 8 end in 4. 3 and 7 end in 9. 4 and 6 end in 6. 5 ends in 5; 0 ends in 0. So there is no way for a square to end in 2, 3, 7 or 8. That would give us 4 correct answers, however only 8 is among the multiple choice solutions.

The units digit of any perfect square is determined only by the integer's unit digit which squares to that number.

Consider the following table: $\begin{matrix} x & { x }^{ 2 } & Units \\ 1 & 1 & 1 \\ 2 & 4 & 4 \\ 3 & 9 & 9 \\ 4 & 16 & 6 \\ 5 & 25 & 5 \\ 6 & 36 & 6 \\ 7 & 49 & 9 \\ 8 & 64 & 4 \\ 9 & 81 & 1 \\ 10 & 100 & 0 \end{matrix}$ From this we can conclude that the units column of a perfect square must be: $0, 1, 4, 5, 6, 9$ Therefore, out of the options given, the answer must be $\boxed {8}$

Any integer squared has the same last digit as its last digit squared.

e.g. 81^2 = 6561 & 1^2 = 1 - both end in 1.

e.g. 728^2 = 529984 & 8^2 = 64 - both end in 4.

No digit squared ends in 8.

1^=1,2^=4,3^=9,.....9^=81......therefore not 8

The problem is equivalent with finding out all the digits in square that isn't possible in $\bmod$ of $10$ .

For $0$ to $9$ , as these number could represent every digits in $\bmod$ of $10$ , we have $0$ , $1$ , $4$ , $6$ , $6$ , $5$ , $9$ , $4$ , and $1$ , as the result of the square of those numbers in $\bmod$ of $10$ respectively. This means that any of $2$ , $3$ , $7$ , and $8$ could not be the numbers in the last digit of any square.

We are asked that which of the given integer has same integer in it's units place if it is squared. For '6', (6)^2=36, which has 6 in it's unit's place.For '5', (5)^2=25,has 5 in it's units place.For '1' (1)^2=1, has 1 in it's units place.But for '8', (8)^2=64, do not have 8 in it's unit's place.so answer follows.