The Third To Last Digit

$2016^{2016} \equiv 16^{2016}$ (mod $1000$ )

Using Chinese Remainder Theorem

$x \equiv 16^{2016} \equiv 0$ (mod $8$ )

$x \equiv 16^{2016} \equiv 16^{16} \equiv 6^8 \equiv 36^4 \equiv 46^2 \equiv 116$ (mod $125$ ) (Using Euler's Theorem $16^{\phi\left(125\right)} \equiv 16^{100}$ (mod $125$ ) )

Since $8$ and $125$ are co-prime, there exists $x$ (mod $1000$ ) satisfying these conditions.

Now $x = 8n$ , for some $n$ and $x = 116 + 125m$ , for some $m$ .

Now we get a linear Diophantine equation $8n - 125m = 116$ .

Using extended Euclidean algorithm

$125 = 15 \times 8 + 5$

$8 = 5 + 3$

$5 = 3 + 2$

$3 = 2 + 1$

$3 - 2 = 1$

$2 \times 3 - 5 = 1$

$2 \times 8 - 3 \times 5 = 1$

$47 \times 8 - 3 \times 125 = 1$

$116 \times 47 \times 8 - 3 \times 116 \times 125 = 116$

$5452 \times 47 - 348 \times 125 = 116$

Therefore $n = 5452$ , and thus $x = 5452 \times 8 \equiv 616 \equiv 16^{2016} \equiv 2016^{2016}$ (mod $1000$ )

And from above we get that the 3rd rightmost digit is $\boxed{6}$ .

To avoid a tedious application of the Chinese Remainder Theorem, we will use a little trick. $\frac{2016^{2016}}{8}=252\times2016^{2015}\equiv 2\times 16^{15}\equiv 77 \pmod{125}$ . Multiplying through by 8, we find that $2016^{2016}\equiv 8\times 77= 616 \pmod{1000}$ . The answer is $\boxed{6}$

We simply have to take modulo 1000.

$2016^{2016} = 616 \quad mod(1000)$

Hence the last 3 digits of $2016^{2016}$ is 616. Hence the digit in the $100^{th}$ place is 6.