Difference between revisions of "2021 GMC 10B Problems/Problem 21"

Revision as of 12:49, 11 April 2022

Problem

Find the remainder when $3^{1624}+7^{1604}$ is divided by $1000$ .

$\textbf{(A)} ~122 \qquad\textbf{(B)} ~322 \qquad\textbf{(C)} ~482 \qquad\textbf{(D)} ~882 \qquad\textbf{(E)} ~922$

Solution

Since $\varphi(1000) = 400$ , we have $\begin{align*} 3^{1624} + 7^{1604} &\equiv 3^{400\cdot4+24} + 7^{400\cdot4+4} &\pmod{1000} \\ &\equiv 3^{24} + 7^{4} &\pmod{1000} \end{align*}$

Note that $3^{24} = 9^{12} = (1-10)^{12}$ , we can apply the binomial theorem to give $3^{24}\equiv (1-10)^{12}\equiv 1-12\cdot10 + 66\cdot10^2 \equiv481 \pmod{1000}.$

Since we can compute $7^4 \pmod {1000}$ rather easily, we can finish the problem from here $\begin{align*} 3^{1624}+7^{1604} &\equiv 481 + 7^4 &\pmod{1000} \\ &\equiv 481 + 401 &\pmod{1000} \\ &\equiv \boxed{\textbf{(D)}~882} &\pmod{1000}. \end{align*}$ ~pineconee

@@ Line 11: / Line 11: @@
 \end{align*}</cmath>
-Since <math>3^{24} = 9^{12} = (1-10)^{12}</math>, we can apply the binomial theorem to give
+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>
-Since we can bash <math>7^4 \pmod {1000}</math> rather easily, we can finish the problem from here
+Since we can compute <math>7^4 \pmod {1000}</math> rather easily, we can finish the problem from here
 <cmath>\begin{align*}
 ^{1624}+7^{1604} &\equiv 481 + 7^4 &\pmod{1000} \\

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Difference between revisions of "2021 GMC 10B Problems/Problem 21"

Revision as of 12:49, 11 April 2022

Problem

Solution