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?
factorial problem
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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
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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Your reasoning is correct for the problemcartera wrote:50/5+50/5^2=12
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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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.
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.
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