SOLUTION: Find the smallest possible value of "n" so that 80^n is divisible by 64,100 and 125.
Algebra.Com
Question 550557: Find the smallest possible value of "n" so that 80^n is divisible by 64,100 and 125.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The prime factorization of 64 is 
The prime factorization of 100 is 
The prime factorization of 125 is 
So the least common multiple of 64, 100, and 125 is
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, 
John
My calculator said it, I believe it, that settles it
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