Difference between revisions of "2021 April MIMC 10 Problems/Problem 17"

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==Solution==
 
==Solution==
To be Released on April 26th.
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<cmath>\sum_{k=0}^{60} {60 \choose k}</cmath> can be expressed as <math>2^{60}</math>, and <math>{60 \choose 0}</math> is equal to <math>1</math>. Therefore, we can simplify the original expression into <math>2^{60}-1+2^{59}-1+...+2^3-1-2^{10}=2^{60}+2^{59}+...+2^{3}+2^3-58-1024=2^{61}-(8+58+1024)=2^{61}-1090</math>. The expression that the answer wants would be <math>2^2\cdot(2\cdot 61+2\cdot2+1090)=4\cdot 1216=\fbox{\textbf{(D)} 4864}</math>.
<cmath>\sum_{k=0}^{60} {60 \choose k}</cmath> can be expressed as <math>2^{60}</math>, and <math>60 \choose 0</math> is equal to <math>1</math>. Therefore, we can simplify the original expression into <math>2^{60}-1+2^{59}-1+...+2^3-1-2^{10}=2^{60}+2^{59}+...+2^{3}+2^3-58-1024=2^{61}-(8+58+1024)=2^{61}-1090</math>. The expression that the answer wants would be <math>2^2\cdot(2\cdot 61+2\cdot2+1090)=4\cdot 1216=\fbox{\textbf{(D)} 4864}</math>.
 

Latest revision as of 13:50, 26 April 2021

The following expression \[\sum_{k=1}^{60} {60 \choose k}+\sum_{k=1}^{59} {59 \choose k}+\sum_{k=1}^{58} {58 \choose k}+\sum_{k=1}^{57} {57 \choose k}+\sum_{k=1}^{56} {56 \choose k}+\sum_{k=1}^{55} {55 \choose k}+\sum_{k=1}^{54} {54 \choose k}+...+\sum_{k=1}^{3} {3 \choose k}-2^{10}\] can be expressed as $x^{y}-z$ which both $x$ and $y$ are relatively prime positive integers. Find $2^{x}(xy+2x+z)$.

$\textbf{(A)} ~4632 \qquad\textbf{(B)} ~4844 \qquad\textbf{(C)} ~4860\qquad\textbf{(D)} ~4864 \qquad\textbf{(E)} ~8960$

Solution

\[\sum_{k=0}^{60} {60 \choose k}\] can be expressed as $2^{60}$, and ${60 \choose 0}$ is equal to $1$. Therefore, we can simplify the original expression into $2^{60}-1+2^{59}-1+...+2^3-1-2^{10}=2^{60}+2^{59}+...+2^{3}+2^3-58-1024=2^{61}-(8+58+1024)=2^{61}-1090$. The expression that the answer wants would be $2^2\cdot(2\cdot 61+2\cdot2+1090)=4\cdot 1216=\fbox{\textbf{(D)} 4864}$.

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