2016 AMC 10B Problem 8

The tens digit depends only on the last two digits, so work modulo $100$.

Since $2015 \equiv 15 \pmod{100}$ and $15^2 = 225 \equiv 25$, we get $2015^{2016} = (2015^2)^{1008} \equiv 25^{1008} \pmod{100}$. A number ending in $25$ stays ending in $25$ when multiplied by $25$ (because $25 \cdot 25 = 625$), so $25^{1008} \equiv 25 \pmod{100}$.

Then
$$2015^{2016} - 2017 \equiv 25 - 17 = 8 \pmod{100},$$
so the number ends in $08$. Its tens digit is $0$.

The answer is $\boxed{\textbf{(A)}\ 0}$.

Questions about a specific digit are questions about a remainder: the tens digit needs only arithmetic mod $100$. Numbers ending in $25$ (like those ending in $76$, $0$, or $1$) are fixed under multiplication mod $100$, so huge exponents collapse instantly. Reduce the base first, then look for a stable last-two-digit pattern.

Compute a few powers: $15, 225, 3375, 50625, \ldots$; the last two digits alternate $15, 25, 75, 25, 75, \ldots$, and even exponents at least $2$ give $25$. Then $25 - 17 = 8$.

• Looking only at the units digit ($5 - 7$ with a borrow) and reporting the borrow's effect wrongly, or assuming all powers of $2015$ end in $15$.
• Forgetting the subtraction and reporting the tens digit of $2015^{2016}$ itself, which is $2$ (not a choice, but a sign of misreading).