2011 AMC 8 Problems/Problem 15

Revision as of 11:45, 14 January 2023 by Trex226 (talk | contribs) (Solution 2)

Problem

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

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

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=440

Solution

\[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]

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

Solution 2

$4^5$ has $4$ digits while $5^{10}$ has $7$ digits. This means that $4^5 \cdot 5^{10}$ has a total of $7+4=\boxed{\textbf{(D)} 11}$ digits.

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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