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

Revision as of 20:32, 27 October 2020

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
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

@@ Line 1: / Line 1: @@
+==Problem==
 How many digits are in the product <math>4^5 \cdot 5^{10}</math>?
-<math> \text{(A) } 8 \qquad\text{(B) } 9 \qquad\text{(C) } 10 \qquad\text{(D) } 11 \qquad\text{(E) } 12 </math>
+<math> \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12 </math>
+==Video Solution==
+https://youtu.be/rQUwNC0gqdg?t=440
 ==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{\textbf{(D)}\ 11}</math> digits.
 ==See Also==
 {{AMC8 box|year=2011|num-b=14|num-a=16}}
+{{MAA Notice}}