Question 550557
<font face="Times New Roman" size="+2">


The prime factorization of 64 is *[tex \Large 2^6]


The prime factorization of 100 is *[tex \Large 2^2\,\cdot\,5^2]


The prime factorization of 125 is *[tex \Large 5^3]


So the least common multiple of 64, 100, and 125 is


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2^6\,\cdot\,5^3]


The prime factorization of 80 is *[tex \Large 2^4\,\cdot\,5], so you would need at least 3 factors of 80 to have a product with the required 3 factors of 5.  And indeed, *[tex \Large 80^2\ =\ 6400], obviously divisible by both 64 and 100, is NOT divisible by 125.  But *[tex \Large 80^3\ =\ 512000] is divisible by 125.  Hence, *[tex \Large n\ =\ 3]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
<div style="text-align:center"><a href="http://outcampaign.org/" target="_blank"><img src="http://cdn.cloudfiles.mosso.com/c116811/scarlet_A.png" border="0" alt="The Out Campaign: Scarlet Letter of Atheism" width="143" height="122" /></a></div>
</font>