2016 AMC 10B Problem 8

The tens digit only depends on the last two digits of the number, so we only need to follow the last two digits.

1. The last two digits of $2015$ are $15$. When you multiply, the last two digits of the answer depend only on the last two digits of the numbers you multiply. So we can work with $15$.
2. Square it: $15^2 = 225$, which ends in $25$. So $2015^2$ ends in $25$.
3. Multiply something ending in $25$ by something ending in $25$: $25 \cdot 25 = 625$, which again ends in $25$. So once we reach $25$, multiplying by another $2015^2$ keeps the ending $25$.
4. $2015^{2016} = \left(2015^2\right)^{1008}$ is $2015^2$ multiplied by itself $1008$ times, so it ends in $25$.
5. Now subtract $2017$, which ends in $17$. The ending $25$ minus $17$ gives $08$. (The huge number is much larger than $2017$, so nothing strange happens in the front digits.)
6. The number ends in $\ldots 08$, so its tens digit is $0$.

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

A question about one particular digit is really a question about a remainder. For the tens digit, divide by $100$ and look at the remainder, which is the last two digits. Some endings, like $25$ and $76$, stay the same when you multiply them by themselves, so even a giant power is easy to handle.

Write out the last two digits of the first few powers of $15$:

After the first power the endings take turns: $25$ for even powers, $75$ for odd powers. The exponent $2016$ is even, so $2015^{2016}$ ends in $25$, and $25 - 17 = 08$.

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 $\boxed{\textbf{(A)}\ 0}$.