stunn3r wrote:
If n is a positive integer greater than 2, n=?
(1) The tens' digit of 11^n is 4.
(2) The hundreds' digit of 5^n is 6.
The units digit of \(11^n\) for any positive integer (n) is always 1. Also, the tens digit of \(11^n\), follows a cyclic order, and assumes all the values from 0-9.
\(11^1 = 11\)
\(11^2 = X21\)
\(11^3 = XX31\)
\(11^4 = XXX41\) and so on
The tens and units digit for \(5^n\), is always 25, for\(n\geq2\). The hundreds digit keeps switching between 1 for n = odd OR as 6 for n = even
From F.S 1, we would have the tens digit as 4 for n = 4, and then again for n = (4+10) = 14. Insufficient.
From F. S 2, we would have the hundreds digit as 6 for n = 4,6,8 and so on. Insufficient.
Even after combing both fact statements, n could be n = 4 or n = 14. Insufficient.
E.
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