For every factor of 10, we need a 2 and a 5.niraj_a wrote:If 10^n is a factor of the product of the first 24 integers, what is the greatest possible value of n?
A) 7
B) 6
C) 5
D) 4
E) 3
i guesstimated 5 = n because 24*20*10*5*2 acount for 4 zeros anyway.
2 won't be the limiting factor (since every even number has at least 1 factor of 2), so let's focus on the 5s.
Looking at the relevant numbers in 24!:
5*10*15*20
means that we'll have 4 5s on our list. Therefore, the maximum number of 10s we can get is also going to be 4.













