# Aptitude - Square Root and Cube Root - Discussion

Discussion Forum : Square Root and Cube Root - General Questions (Q.No. 14)

14.

How many two-digit numbers satisfy this property.: The last digit (unit's digit) of the square of the two-digit number is 8 ?

Answer: Option

Explanation:

A number ending in 8 can never be a perfect square.

Discussion:

8 comments Page 1 of 1.
Jayram said:
7 years ago

If the units place is 1, 2, 3, 4, 5, 6, 7, 8, 9.

Square of all number = 1, 4, 9, 16, 25, 36, 49, 64, 81.

Then, take unit value for all squares =1, 4, 9, 6, 5, 6, 9, 4, 1.

And approx is 2, 3, 7, 8.

Square of all number = 1, 4, 9, 16, 25, 36, 49, 64, 81.

Then, take unit value for all squares =1, 4, 9, 6, 5, 6, 9, 4, 1.

And approx is 2, 3, 7, 8.

Abi said:
8 years ago

2 = 4.

3 = 9.

4 = 16.

5 = 25.

7 = 49.

9 = 81.

10 = 100.

As you can see all squares have 0, 1, 4, 5, 6, 9 in units place.

So if we have 2, 3, 7, 8 in units place it square root is not perfect.

3 = 9.

4 = 16.

5 = 25.

7 = 49.

9 = 81.

10 = 100.

As you can see all squares have 0, 1, 4, 5, 6, 9 in units place.

So if we have 2, 3, 7, 8 in units place it square root is not perfect.

BALU said:
8 years ago

I can't understand so please explain again.

(1)

Aarti said:
9 years ago

Number ending with 8 is not a perfect square.

Veena said:
1 decade ago

Number ending with 0, 1, 4, 5, 6, 9 are the perfect squares rather than these are not perfect squares.

M.V.KRISHNA/PALVANCHA said:
1 decade ago

Hello pawan and kala.

The two digit numbers are from 10 to 99.

The numbers with 8 in units place are.

18, 28, 38, 48, 58, 68, 78, 88, 98.

These numbers do not have a perfect square (s*s).

So answer is none of these.

Hope you understand.

The two digit numbers are from 10 to 99.

The numbers with 8 in units place are.

18, 28, 38, 48, 58, 68, 78, 88, 98.

These numbers do not have a perfect square (s*s).

So answer is none of these.

Hope you understand.

Kala said:
1 decade ago

Please explain.

Pawan goyal said:
1 decade ago

hw this could be possible that a number ending with is not a perfect square

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