Difference between revisions of "2011 AMC 8 Problems/Problem 15"
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That is one <math>1</math> followed by ten <math>0</math>'s, which is <math>\boxed{\textbf{(D)}\ 11}</math> digits. | That is one <math>1</math> followed by ten <math>0</math>'s, which is <math>\boxed{\textbf{(D)}\ 11}</math> digits. | ||
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| + | ==Solution 2== | ||
| + | |||
| + | <cmath>4^5 has four digits while 5^{10} has 7 digits. This means that </cmath>4^5 \cdot 5^{10} has a total of 7+4, 11 digits. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=14|num-a=16}} | {{AMC8 box|year=2011|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:05, 21 November 2022
Problem
How many digits are in the product 
?
Video Solution
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
![\[4^5 has four digits while 5^{10} has 7 digits. This means that\]](http://latex.artofproblemsolving.com/7/5/6/756e8b113291f33da8b8d3a0068dbcc5bcd130d6.png)
4^5 \cdot 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. 
