Difference between revisions of "G285 MC10B Problems/Problem 1"

Revision as of 19:33, 20 June 2021

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

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

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

2021 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

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 ==Solution==
-We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lfloor 43.5 \rfloor \implies \boxed{\textbf{(B)}\ 43}</cmath>
+We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}</cmath>
-{{MC10B box|year=2021|ab=B|num-b=1|after=2}}
+{{AMC10 box|year=2021|ab=B|num-b=1|after=2}}