Difference between revisions of "2000 PMWC Problems/Problem I4"

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==Solution==
 
==Solution==
Given that <math>A^4=75600\times B</math>. If <math>A</math> and <math>B</math> are positive integers, find the smallest value of <math>B</math>.
 
 
 
 
If <math>A</math> and <math>B</math> are positive integers, then <math>B</math> must be the smallest positive integer that, when multiplied by <math>75600</math>, yields a perfect fourth power. The prime factorization of <math>75600</math> is <math>2^4 * 3^3 * 5^2 * 7^1</math>, so the smallest value of <math>B</math> is <math>2^0 * 3^1 * 5^2 * 7^3 = 25,725</math>.
 
If <math>A</math> and <math>B</math> are positive integers, then <math>B</math> must be the smallest positive integer that, when multiplied by <math>75600</math>, yields a perfect fourth power. The prime factorization of <math>75600</math> is <math>2^4 * 3^3 * 5^2 * 7^1</math>, so the smallest value of <math>B</math> is <math>2^0 * 3^1 * 5^2 * 7^3 = 25,725</math>.
  
 
==See Also==
 
==See Also==
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Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems

Latest revision as of 11:03, 23 December 2019

Problem

Given that $A^4=75600\times B$. If $A$ and $B$ are positive integers, find the smallest value of $B$.

Solution

If $A$ and $B$ are positive integers, then $B$ must be the smallest positive integer that, when multiplied by $75600$, yields a perfect fourth power. The prime factorization of $75600$ is $2^4 * 3^3 * 5^2 * 7^1$, so the smallest value of $B$ is $2^0 * 3^1 * 5^2 * 7^3 = 25,725$.

See Also

Back to test: https://artofproblemsolving.com/wiki/index.php/2000_PMWC_Problems