For all positive integers n, the expression n! denotes the product of the first n positive integers. When 5! is expressed as an iteger, what is the ones digit?
2007-09-23 13:16:21 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
0
Since it includes the factors 2 and 5, it has to be divisible by 10.
2007-09-23 13:26:05 · answer #1 · answered by Computer Guy 7 · 0⤊ 0⤋
0. Because 5! = 5*4*3*2*1 = 120
Notice that the last digit is going to be 0 for any higher values of n too, because having the "5" and "2" in the factors means it's always going to be a factor of 10, and all factors of 10 end in 0.
2007-09-23 20:20:54 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Yes, it would be 0. 5 Factorial = 5*4*3*2*1
5*4*3*2*1
20*3*2*1
60*2*1
120*1
120
The last digit would be 0.
2007-09-23 20:25:16 · answer #3 · answered by Me.. 4 · 0⤊ 0⤋