+0  
 
0
59
1
avatar+75 

If $m$ is the smallest positive integer such that $m!$ is a multiple of $4125,$ and $n$ is the smallest positive integer such that $n!$ is a multiple of $281600,$ then find $n-m.$

 Jan 5, 2019

Best Answer 

 #1
avatar
+1

4,125 = 3 * 5^3 * 11 

Since the smallest m must have 5^3 and 11 as factors, then the smallest m = 15, because:

15! / 4,125 =317,011,968 -- The quotient and m! is multiple of 4125.

Similarly,  281,600 = 2^10 * 5^2 * 11, then the smallest n must have these 3 factors as well. And 12 is the smallest n, because: 12! / 281,600 =1,701 -- The quotient and n! is multiple of 281,600

n - m =12 - 15 = - 3

 Jan 5, 2019
 #1
avatar
+1
Best Answer

4,125 = 3 * 5^3 * 11 

Since the smallest m must have 5^3 and 11 as factors, then the smallest m = 15, because:

15! / 4,125 =317,011,968 -- The quotient and m! is multiple of 4125.

Similarly,  281,600 = 2^10 * 5^2 * 11, then the smallest n must have these 3 factors as well. And 12 is the smallest n, because: 12! / 281,600 =1,701 -- The quotient and n! is multiple of 281,600

n - m =12 - 15 = - 3

Guest Jan 5, 2019

22 Online Users

avatar
avatar
avatar