Difference between revisions of "2011 AMC 8 Problems/Problem 15"
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+ | ==Problem== | ||
How many digits are in the product <math>4^5 \cdot 5^{10}</math>? | How many digits are in the product <math>4^5 \cdot 5^{10}</math>? | ||
<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 by OmegaLearn== | ||
+ | https://youtu.be/rQUwNC0gqdg?t=440 | ||
==Solution== | ==Solution== | ||
− | <cmath>4^5 \cdot 5^{10} = 2^10 \cdot 5^{10} = 10^{10}.</cmath> | + | <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. | 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== | ||
+ | |||
+ | <math>4^5</math> has <math>4</math> digits while <math>5^{10}</math> has <math>7</math> digits. This means that <math>4^5 \cdot 5^{10}</math> has a total of <math>7+4=\boxed{\textbf{(D)} 11}</math> digits. | ||
+ | |||
+ | ==Video Solution 1 by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
==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}} |
Latest revision as of 14:09, 17 December 2023
Contents
Problem
How many digits are in the product ?
Video Solution by OmegaLearn
https://youtu.be/rQUwNC0gqdg?t=440
Solution
That is one followed by ten 's, which is digits.
Solution 2
has digits while has digits. This means that has a total of digits.
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
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.