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G285 MC10B Problems/Problem 1

Revision as of 19:35, 20 June 2021 by Geometry285 (talk | contribs)

Problem

Find $\left \lceil {\frac{3!+4!+5!+6!}{2+3+4+5+6}} \right \rceil$

$\textbf{(A)}\ 42\qquad\textbf{(B)}\ 43\qquad\textbf{(C)}\ 44\qquad\textbf{(D)}\ 45\qquad\textbf{(E)}\ 46$

Solution

We have \[\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}\]

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