Difference between revisions of "2003 AMC 10B Problems/Problem 14"

 
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<math>405</math> is not a perfect power, so the smallest possible value of <math>a+b</math> is <math>405+2=\boxed{\textbf{(D)}\ 407}</math>.
 
<math>405</math> is not a perfect power, so the smallest possible value of <math>a+b</math> is <math>405+2=\boxed{\textbf{(D)}\ 407}</math>.
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==Video Solution by WhyMath==
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https://youtu.be/xhhmWhaOK2E
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~savannahsolver
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2003|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2003|ab=B|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:22, 23 June 2021

Problem

Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$.

$\textbf{(A) } 25 \qquad\textbf{(B) } 34 \qquad\textbf{(C) } 351 \qquad\textbf{(D) } 407 \qquad\textbf{(E) } 900$

Solution

\[3^8\cdot5^2 = (3^4)^2\cdot5^2 = (3^4\cdot5)^2 = 405^2\]

$405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=\boxed{\textbf{(D)}\ 407}$.

Video Solution by WhyMath

https://youtu.be/xhhmWhaOK2E

~savannahsolver

See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 AMC 10 Problems and Solutions

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