Question

Using first, third, sixth and seventh digits of the number $93426185$, one time each, how many numbers of two digits are possible which are perfect squares of a number. What Is that number? If two such numbers are possible then answer $11$, if no such number is possible than answer zero.

• A. $0$
• B. $7$
• C. $9$
• D. $11$

Answer 1

Answer: (A)

$0$

The answer is option $A$

The $1^{st},3^{rd},6^{th},7^{th}$ digits of $93426185$ are $5,1,4,3$ respectively

To be form a perfect square with these digits, the number needs to end either with $5$ or $1$ or $4$

Possible two digit numbers without repetition ending with the above digits are

$>15,45,35,31,41,51,14,34,54$

We can see none of them as perfect squares

Since no such number is possible, the answer is $0$