3 Digit Factorials

Number Theory Level 4

If the number of trailing zeros in n ! n! is 3 3 more than the number of trailing zeros of ( n − 1 ) ! (n-1)! , then how many three digit values can n n take?

9 8 7 6 5

Ravneet Singh by Ravneet Singh , 30, India

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1 solution

Ravneet Singh
Aug 3, 2015

n ! \large n! will have 3 more zeros in its decimal expansion than ( n − 1 ) ! \large (n-1)! exactly when n \large n is divisible by 5 3 = 125. 5^{3}=125. There are seven 3 digit multiples of 125 since 8.125 = 1000 8.125=1000 . But we can't include 625 because there is a difference of 4 zeros between 624 ! a n d 625 ! 624! and 625! because 625 = 5 4 625 = 5^{4} and hence only values are 125 ! , 250 ! , 375 ! , 500 ! , 750 ! , 875 ! \large 125!,250!,375!,500!,750!,875! . Total values 6 \huge 6

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