The symbol 3! Means 3 x 2 x 1, which equals 6. Similarly 5! Means 5 x 4 x 3 x 2 x 1, which equals 120, and so on. Suppose that N! ends in exactly 3 zeros after fully multiplying out. What is the smallest value that N can have?

Guest Jun 13, 2019

#1**+2 **

For a number end in \(x\) zeros, it must be divisible by \(10^x\)

So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or \(5^3\cdot2^3\)

To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N.

15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

15, 10 and 5 are multiples of 5. There are also three 2s in 15!.

\(\boxed{15}\) is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros.

power27 Jun 13, 2019

#1**+2 **

Best Answer

For a number end in \(x\) zeros, it must be divisible by \(10^x\)

So for N! to have 3 zeros on the end of it, it must be divisible by 10*10*10 or \(5^3\cdot2^3\)

To get 3 fives when multiplying out N! there must be 3 multiples of 5 less that or equal to N.

15!=15*14*13*12*11*10*9*8*7*6*5*4*3*2*1

15, 10 and 5 are multiples of 5. There are also three 2s in 15!.

\(\boxed{15}\) is the smallest value N can be because 15! will have 3 sets of 2s and 5s to make zeros.

power27 Jun 13, 2019