Physics, asked by soninishit, 5 months ago

Using first, third, sixth and seventh digits of the numbers 93426185
one time each how many
mumbers 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 then answer zero​

Answers

Answered by angshu12
1

Answer:

0

Explanation:

The 1 st ,3rd ,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 answ

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

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