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

Revision as of 20:35, 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 $\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}$

Revision as of 20:33, 20 June 2021 (view source) Geometry285 (talk \| contribs) m ← Older edit		Revision as of 20:35, 20 June 2021 (view source) Geometry285 (talk \| contribs) m Newer edit →
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	We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}</cmath>		We have <cmath>\frac{6+24+120+720}{20} = \frac{87}{2} = \lceil 43.5 \rceil \implies \boxed{\textbf{(C)}\ 44}</cmath>

−	{{~~AMC10~~ box\|year=2021\|ab=B\|num-b=1\|after=2}}	+	{{MC10B box\|year=2021\|ab=B\|num-b=1\|after=Problem 2}}