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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
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B
7
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C
9
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D
11
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Solution

The correct option is B 0
The answer is option A
The 1st,3rd,6th,7th 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

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