Question

Which of the following cannot be a digit at the units place of a perfect square ?​

Answer 1

Answer:

2,3,7,8

Step-by-step explanation:

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

11²=121

12²=144

and the pattern repeates so on..

The only numbers that are there in the units place are 1,4,5,6,9,0.

Hence the numbers that aren't at the units place are 2,3,7,8

Hence the answer is 2,3,7,8

Answer 2

Answer:

2,3,7 and 8

Step-by-step explanation:

$1^{2}$ = 1             (digit at the units place is 1)

$2^{2}$ = 4            (digit at the units place is 4)

$3^{2}$ = 9            (digit at the units place is 9)

$4^{2}$ = 16           (digit at the units place is 6)

$5^{2}$ = 25          (digit at the units place is 5)

$6^{2}$ = 36          (digit at the units place is 6)

$7^{2}$ = 49          (digit at the units place is 9)

$8^{2}$ = 64          (digit at the units place is 4)

$9^{2}$ = 81           (digit at the units place is 1)

$10^{2}$= 100        (digit at the units place is 0)

we know that all the numbers will end with 1,2,3,4,5,6,7,8,9 or 0.

so the square of these numbers will always end with 0,1,4,5,6 and 9 (examples given above). So, the only numbers that cannot be a digit at the units place of a perfect square is 2,3,7 and 8.

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