Difference between revisions of "2011 AMC 8 Problems/Problem 15"

Revision as of 19:17, 25 November 2011

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

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

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

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

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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 ==Solution==
+<cmath>4^5 \cdot 5^{10} = 2^10 \cdot 5^{10} = 10^{10}.</cmath>
+That is one <math>1</math> followed by ten <math>0</math>'s, which is <math>\boxed{\text{(D)}\ 11}</math> digits.
 ==See Also==
 {{AMC8 box|year=2011|num-b=14|num-a=16}}