factorial problem

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arvindhs2000: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Mon Dec 08, 2008 6:06 pm

factorial problem

by arvindhs2000 » Mon Dec 15, 2008 11:24 pm

I do not know whether this problem has been posted before. This is not a GMAT problem. but I would like to know the answer: the problem goes like this:
How many zeroes are there in 1!x2!x3!....x49!x50! ? here ! stands for factorial of the number?

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Source: Beat The GMAT — Problem Solving | Mon Dec 15, 2008 11:24 pm

4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 am; Thanked: 27 times; Followed by:1 members; GMAT Score:660(

by 4meonly » Tue Dec 16, 2008 1:25 am

any product of 2 and 5 will yield a nuber with 0 as units digit.
So here you should count the number of 5's and 2's
there are a lot of 2's, so it is easer to find the number of 5's. there are less 5's then 2's.

in
1!x2!x3!....x49!x50!
after 5! you have one five in each number, this gives us 46 zeros
after 25! you have two five in each number, this gives us additional 25 zeros
So I think the answer should be 46+25=71 zeros

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cartera: Senior | Next Rank: 100 Posts; Posts: 59; Joined: Tue Jun 10, 2008 6:49 pm; Thanked: 2 times

by cartera » Tue Dec 16, 2008 2:14 am

50/5+50/5^2=12

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4meonly: Legendary Member; Posts: 891; Joined: Sat Aug 16, 2008 4:21 am; Thanked: 27 times; Followed by:1 members; GMAT Score:660(

by 4meonly » Tue Dec 16, 2008 2:17 am

cartera wrote:50/5+50/5^2=12

Your reasoning is correct for the problem
How many zeroes are there in 50!

But we have
How many zeroes are there in 1!x2!x3!....x49!x50!
It is the product of 50 consecutive factorials
1!x2!x3!x4! = 1*1*2*1*2*3*1*2*3*4

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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pm; Location: Pittsburgh, PA; Thanked: 113 times; Followed by:27 members; GMAT Score:710

by dmateer25 » Tue Dec 16, 2008 6:00 am

50! has 12 5’s
45! – 49! each have 10 5’s (50)
40! – 44! each have 9 5’s (45)
35! – 39! each have 8 5’s (40)
30! – 34! Each have 7 5’s (35)
25! – 29! Each have 6 5’s (30)
20! – 24! Each have 4 5’s (20)
15! – 19! Each have 3 5’s (15)
10! – 14! Each have 2 5’s (10)
5! – 9! Each have 1 5 (5)
1! – 4! Have 0 5’s (0)

I get 262 0's.