Can someone help me solve this problem? Thank you!
LOL @vidishas99 We both deleted at the same time!
1 Like
Here I am waiting for your solutions and looking it up on wolframalpha at the same time
2 Likes
Adding again @HoldMyBeer
The number of zeroes indicate the number of times you can factor out 10 from a number
10 = 2 * 5
25! has a lot of 2s in it:
2,4,6,8,10 alone give 8 2s
So 5 is the limiting factor, the number of 5s in 25!:
1 (5) + 1(10)+ 1 (15) + 1(20) + 2(25) = 6
Number of 10s = 6
So number of zeroes at the end of 25! = 6
3 Likes
Oh maaaan, I did figure out 5 but when I saw it had 6, that was such a bummer. Nicely explained!
Thanks a lot!
If you have 125!..then the number of 0s will be… 25 (All multiples of 5 till 125) +5 (Five 5’s for all 25 multiples)+1 (one five extra for 125)
The general formula to find the number of factors of a prime number p in the factorial of a number n,
number of factors of p = [\,\frac{n}{p}]\, + [\,\frac{n}{p^2}]\, + [\,\frac{n}{p^3}]\,+ ...
where [x] refers to the “step of x” or “greatest integer less than or equal to x”
According to greg’s previous solutions, I was supposed to solve this like 25!/ y = integer. However, as there is no value for y. I mean you said that zero’s are indicating 10 but I don’t know why I should choose 5 as the limiting factor.
Because 25! has plenty of 2s
2,4,6,8,10 alone give me 8 2s
Compare that to 5s
There are a total of 6 5s
So you can multiply each of these 5s to 2 and get a 6 10s
Which would give you 6 0s at the end