Difference between revisions of "2021 April MIMC 10 Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | <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 ![\[\sum_{k=1}^{60} {60 \choose k}+\sum_{k=1}^{59} {59 \choose k}+\sum_{k=1}^{58} {58 \choose k}+\sum_{k=1}^{57} {57 \choose k}+\sum_{k=1}^{56} {56 \choose k}+\sum_{k=1}^{55} {55 \choose k}+\sum_{k=1}^{54} {54 \choose k}+...+\sum_{k=1}^{3} {3 \choose k}-2^{10}\]](http://latex.artofproblemsolving.com/f/3/e/f3ec40504a0dd4366dbbd4b910eb14ab633ae037.png)
can be expressed as 
which both 
and 
are relatively prime positive integers. Find 
.
Solution
![\[\sum_{k=0}^{60} {60 \choose k}\]](http://latex.artofproblemsolving.com/c/6/2/c62fc86f29e4ce7b68ca2c65d47854fd0603c4cc.png)
can be expressed as 
, and 
is equal to 
. Therefore, we can simplify the original expression into 
. The expression that the answer wants would be 
.
