Official Solution:If \(357^x*117^y=a\), where \(x\) and \(y\) are positive integers, what is the units digit of \(a\)? (1) \(100 \lt y^2 \lt x^2 \lt 169\).
Since both \(x\) and \(y\) are positive integers, \(x^2\) and \(y^2\) are perfect squares. Now, there are only two perfect squares in the given range: \(121=11^2\) and \(144=12^2\), so \(y=11\) and \(x=12\). Sufficient. (As the cyclicity of the units digit of 7 in integer power is 4, therefore the units digit of \(7^{23}\) is the same as the units digit of \(7^3\), which is 3).
(2) \(x^2-y^2=23\).
From this equation, we have \((x-y)(x+y) = 23\). Since 23 is prime and both \(x\) and \(y\) are positive integers, it follows that \(x-y=1\) and \(x+y=23\). Solving for \(x\) and \(y\), we find \(y=11\) and \(x=12\). Sufficient.
Answer: D