You Will Know Your Answer is Correct.

$2^{2018} + 15 \times 2^{2014} + 2^{2012} + 2^{3} + 2^{2} + 2^{1} +2^{0}$ = $2^{6} \times 2^{2012} + 15 \times 2^{2} \times 2^{2012} + 2^{2012} + 8 + 4 + 2 + 1$ = $64 \times 2^{2012} + 60 \times 2^{2012} + 2^{2012} + 15$ = $(64 + 60 + 1) \times 2^{3} \ times 2^{2009} + 15$ = $125 \times 8 \times 2^{2009} + 15$ = $1000 \times 2^[2009] + 15$ Disregarding the $+ 15$ , we realize that the last 3 digits will all be $000$ . However, for the thousands column, we need to find the last digit of $2^{2009}$ . We realize that there is a pattern for the final digit, as $2^{1} = 2$ , $2^{2} = 4$ , $2^{3} = 8$ , $2^{4} = 16$ , $2^{5} = 32$ Meaning that there is a pattern which goes 2, 4, 8, and 6. $2009/4 = 502R1$ Meaning that $2^{2009}$ Ends in $2$ . $2 \times 1000 + 15$ = $2015$

It suffices to think about $2^{2014}$ ... the other summands can then be dealt with easily. We will first work modulo 625, keeping in mind that $\phi(625)=500.$ Now $2^{2012}=2^{500\times4+12}\equiv{2^{12}}=4096 \pmod{625}$ Since everything is divisible by $2^4=16$ , the above congruency is also valid mod 10000. Multiplying with 2 a few times (and once with 15), we find that $2^{2018}\equiv2144$ and $15\times2^{2014}\equiv5760\pmod{10000}$ , and the whole expression is congruent to $\boxed{2015}\pmod{10000}$ ... and we do indeed know that our result is right.