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­Given - x is a positive integer divisible by 10, means x is multiple of 10. x=10k
Therefore x^x = (10k)^10k.
x can be 10, 20, 30......
Therefore, 10^10 or 20^20 ... will always have hundreds digit 0 (as 10^3 = 1000 therefore any power greater than 2 will always have hundreds digit is 0)
Answer - A­
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­Ques: If x is a positive integer divisible by 10, what is the hundreds digit of x^(x)
Answer: A (0)
Explaination:
minimum possible positive integer which is divisible by 10 is 10 itself.
             10^(10) has hundred digit 0
2nd possible +ve integer is 20
             20^(20);    [2^20]*[10^20]; again hundred digit is 0
             (here an integer is multiplied with 10th to the power very high value, you will always have minimum 20 zero in this case)
.
.
your 10 to the power will keep on increasing

 
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Bunuel
­If \(x\) is a positive integer divisible by 10, what is the hundreds digit of \(x^x\)?

A. 0
B. 1
C. 2
D. 3
E. 4­
­

The minimum value of \(x = 10\)
­
\(x^1 = 10\)
\(x^2 = 100\)
\(x^3 = 1000\)
\(x^4 = 10000\)
..
.
Hence the hundreds digit of \(x^x\) = 0

Option A
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