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>
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==Video Solution by OmegaLearn==
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https://youtu.be/rQUwNC0gqdg?t=440
  
 
==Solution==
 
==Solution==
  
<cmath>4^5 \cdot 5^{10} = 2^10 \cdot 5^{10} = 10^{10}.</cmath>
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<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==
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<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.
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==Video Solution 1 by SpreadTheMathLove==
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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}}
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{{MAA Notice}}

Latest revision as of 14:09, 17 December 2023

Problem

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

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

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=440

Solution

\[4^5 \cdot 5^{10} = 2^{10} \cdot 5^{10} = 10^{10}.\]

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

Solution 2

$4^5$ has $4$ digits while $5^{10}$ has $7$ digits. This means that $4^5 \cdot 5^{10}$ has a total of $7+4=\boxed{\textbf{(D)} 11}$ digits.

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=mYn6tNxrWBU

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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