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# m and m!

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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.\$

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

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