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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 $n-m.$

sudsw12 Jan 5, 2019

#1**+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**

Guest Jan 5, 2019

#1**+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