# factorial math problem help!

Anonymous
timer Asked: Feb 6th, 2019
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### Question Description

How many consecutive zeroes are there at the end of 100! (100 factorial). For example, 12! - 479,001,600 has two consecutive zeroes at the end.

portcs
School: UCLA

The factorial of a natural number n is the product of all numbers from 1 up to n. So for instance10! = 10x9x8x7x6x5x4x3x2x1Your question is not concerned with calculating this notion efficiently, but rather with determining the number of trailing zeroes. Consider first the number of trailing zeros in the above example, 10!. It is clear that one zero will be present for each time that 10 is a factor of 10!, so the 10 in the above product directly contributes one trailing zero. However, since 10 = 5*2, you also have to look for factors that will generate 10 indirectly. 10! contains 5 and 2 as factors, and consequently there will be a second trailing zero in this number.Now consider 20! = 20x19x18x17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1Clearly the factors 20 and 10 contribute one trailing zero each; 5 generates a third zero (when multiplied by 2 as before); and 15 generates an additional one, when multiplied for instance by 4. So how can we determine this in a systematic way? It is not difficult. The key point is that for each occurrence of 5 as a factor, there is certainly a matching occurrence of 2, since the expanded product contains more numbers that are multiples of 2 (every even number) than numbers that are multiples of 5. So you can just count how many factors in the factorial are multiples of 5, which is easy. Observe that the above expansion can be written as:20! = (5x4)x19x18x17x16x(5x3)x14x13x12x11x(5x2))x9x8x7x6x5x4x3x2x1In order to determine the number of times that 5 is a factor of n!, you can just calculate the integer division by 5 of the number n whose factorial you are considering. So for instance 10/5=2, so 10! has 2 trailing zeroes, and 20/5 = 4, so 20! has 4 trailing zeroes. So can we determine, in general, the number of trailing zeroes in n! by simply di...

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Anonymous
Thanks, good work

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