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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>
  
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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>
 
<cmath>3^{24}\equiv (1-10)^{12}\equiv 1-12\cdot10 + 66\cdot10^2 \equiv481 \pmod{1000}.</cmath>
  
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Since we can bash <math>7^4 \pmod {1000}</math> rather easily, we can finish the problem from here
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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} \\

Latest revision as of 11:50, 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

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