document.write( "Question 550557: Find the smallest possible value of \"n\" so that 80^n is divisible by 64,100 and 125. \n" ); document.write( "

Algebra.Com's Answer #358870 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "The prime factorization of 64 is

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\n" ); document.write( "\n" ); document.write( "The prime factorization of 100 is

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\n" ); document.write( "\n" ); document.write( "The prime factorization of 125 is

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\n" ); document.write( "\n" ); document.write( "So the least common multiple of 64, 100, and 125 is\r
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\n" ); document.write( "\n" ); document.write( "The prime factorization of 80 is

, so you would need at least 3 factors of 80 to have a product with the required 3 factors of 5. And indeed,

, obviously divisible by both 64 and 100, is NOT divisible by 125. But

is divisible by 125. Hence,

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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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$\"The$

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