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Re: How countless zeros walk 100! finish with?<#permalink>
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16 Nov 2018, 03:51
Find how numerous times the element 5 is had in 100!. That is, we have to find the largest such that 100! is divisible through . There are 20 multiples that 5 in the an initial hundred however 25, 50, 75, and 100 need to be count twice because they are divisible by (25 = 5^2) . So, the answer is 24.The correct answer is B.

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I have actually absolutely no Idea what they space telling me...Can who please write-up a basic explanation because that the reason behind the explanation, or (even better) provide an alternative straightforward approach?Thanks!
I take it 100! in excel, and also I"ve obtained this = 9.3326E+157So, 100! ends v a lot much more 0 보다 24.Did i miss something?Really appreciated because that explanation.
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Find how numerous times the element 5 is contained in 100!. That is, we have actually to uncover the biggest such the 100! is divisible through . There are 20 multiples of 5 in the an initial hundred but 25, 50, 75, and also 100 have to be counting twice because they room divisible by (25 = 5^2) . So, the prize is 24.The exactly answer is B.
I have actually absolutely no Idea what they space telling me...Can who please article a basic explanation for the rationale behind the explanation, or (even better) carry out an alternative basic approach?Thanks!
I take it 100! in excel, and also I"ve obtained this = 9.3326E+157So, 100! ends with a lot an ext 0 보다 24.Did i miss something?Really appreciated because that explanation.
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Find how numerous times the element 5 is consisted of in 100!. The is, we have actually to discover the biggest such the 100! is divisible by . There are 20 multiples that 5 in the an initial hundred however 25, 50, 75, and also 100 have to be counted twice because they are divisible through (25 = 5^2) . So, the answer is 24.The correct answer is B.

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I have actually absolutely no Idea what they space telling me...Can who please article a straightforward explanation for the reason behind the explanation, or (even better) carry out an alternative basic approach?Thanks!
Trailing zeros:Trailing zeros are a succession of 0"s in the decimal depiction (or an ext generally, in any type of positional representation) the a number, after i beg your pardon no other digits follow.Fro example, 125000 has actually 3 trailing zeros (125000);The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative essence n, deserve to be established with this formula:(fracn5+fracn5^2+fracn5^3+...+fracn5^k), wherein k need to be liked such the 5^(k+1)>nIt"s more simple if friend look at an example:How countless zeros are in the finish (after i m sorry no various other digits follow) of 32!?(frac325+frac325^2=6+1=7) (denominator need to be much less than 32, (5^2=25) is less)So there space 7 zeros in the finish of 32!The formula in reality counts the variety of factors 5 in n!, but due to the fact that there space at least as countless factors 2, this is indistinguishable to the variety of factors 10, every of which provides one more trailing zero.BACK to THE initial QUESTION:According to over 100! has (frac1005+frac10025=20+4=24) rolling zeros.Answer: B.For an ext on this issues examine Factorials and also Number Theory links in my signature.Hope it helps.
Hi Bunuel, deserve to you please explain why K = 2 and not k=3 this time as 125>100. (as described in the formula above 5^(K+1)>n