M04-26

Official Solution:


Is \(\frac{7^7}{7^x}\) an integer?

The expression \(\frac{7^7}{7^x} = 7^{7-x}\) will result in an integer if \(x\) is an integer and \(x\) is less than or equal to 7. For example, if \(x\) takes values of 7, 6, 5, ..., 0, -1, ... and so on, then \(7^{7-x}\) will always be an integer. However, for values of \(x\) such as 8, 9, 10, ... or any non-integer, such as 3.5, or -2.5, \(7^{7-x}\) will not result in an integer.

(1) \(0 \le x \le 7\).

If \(x\) is an integer from this range, \(\frac{7^7}{7^x}\) will be an integer but if \(x\) is a non-integer from this range, \(\frac{7^7}{7^x}\) will NOT be an integer.

(2) \(|x| = x^2\)

Square \(|x| = x^2\);

\(x^2 = x^4\);

\(x^2(x^2 - 1) = 0\)

\(x^2(x - 1)(x + 1) = 0\);

\(x = 0\), 1, or -1. For each of those values \(\frac{7^7}{7^x}\) IS an integer. So, we have an YES answer.


Answer: B