Difference between revisions of "G285 MC10B Problems/Problem 1"
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| + | ==Problem== | ||
| + | Find <math>\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil</math> | ||
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| + | <math>\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46</math> | ||
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==Solution== | ==Solution== | ||
| + | We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lfloor 43.5 \rfloor \implies \boxed{\textbf{(B)}\ 43}</cmath> | ||
| − | {{ | + | {{MC10B box|year=2021|ab=B|num-b=1|after=2}} |
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![\[\frac{6+24+120+720}{20} = \frac{87}{2} = \lfloor 43.5 \rfloor \implies \boxed{\textbf{(B)}\ 43}\]](http://latex.artofproblemsolving.com/3/6/6/3667a558a9836092246e08a763767945f6a39b37.png)
