Difference between revisions of "2011 AMC 8 Problems/Problem 15"
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<math> \textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(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== | + | ==Video Solution by OmegaLearn== |
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https://youtu.be/rQUwNC0gqdg?t=440 | https://youtu.be/rQUwNC0gqdg?t=440 | ||
Revision as of 16:27, 31 December 2022
Problem
How many digits are in the product 
?
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=440
Solution
![\[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]](http://latex.artofproblemsolving.com/3/9/2/392430c74acdb7141b3e5030628d63655e9dbfbb.png)
That is one 
followed by ten 
's, which is 
digits.
Solution 2
4^5 has 4 digits while 5^10 has 7 digits. This means that 4^5 x 5^10 has a total of 7+4, 11 digits.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
