2011 AMC 8 Problems/Problem 15: Difference between revisions

Revision as of 00:59, 5 July 2013

How many digits are in the product $4^5 \cdot 5^{10}$ ?

$\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12$

$4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.$

That is one $1$ followed by ten $0$ 's, which is $\boxed{\textbf{(D)}\ 11}$ digits.

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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 ==See Also==
 {{AMC8 box|year=2011|num-b=14|num-a=16}}
+{{MAA Notice}}