Difference between revisions of "2021 GMC 10B Problems/Problem 21"
(Created page with "==Problem== Find the remainder when <math>3^{1624}+7^{1604}</math> is divided by <math>1000</math>. <math>\textbf{(A)} ~122 \qquad\textbf{(B)} ~322 \qquad\textbf{(C)} ~482 \...") |
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\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Note that <math>3^{24} = 9^{12} = (1-10)^{12}</math>, we can apply the binomial theorem to give | |
<cmath>3^{24}\equiv (1-10)^{12}\equiv 1-12\cdot10 + 66\cdot10^2 \equiv481 \pmod{1000}.</cmath> | <cmath>3^{24}\equiv (1-10)^{12}\equiv 1-12\cdot10 + 66\cdot10^2 \equiv481 \pmod{1000}.</cmath> | ||
− | Since we can | + | Since we can compute <math>7^4 \pmod {1000}</math> rather easily, we can finish the problem from here |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
3^{1624}+7^{1604} &\equiv 481 + 7^4 &\pmod{1000} \ | 3^{1624}+7^{1604} &\equiv 481 + 7^4 &\pmod{1000} \ |
Revision as of 11:49, 11 April 2022
Problem
Find the remainder when is divided by .
Solution
Since , we have
Note that , we can apply the binomial theorem to give
Since we can compute rather easily, we can finish the problem from here ~pineconee