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Find how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by \(25 = 5^2\) . So, the answer is 24. The correct answer is B.

I have absolutely no Idea what they are telling me... Can someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?

Find how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by \(25 = 5^2\) . So, the answer is 24. The correct answer is B.

I have absolutely no Idea what they are telling me... Can someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?

Thanks!

Trailing zeros: Trailing zeros are a sequence of 0's in the decimal representation (or more generally, in any positional representation) of a number, after which no other digits follow.

Fro example, 125000 has 3 trailing zeros (125000);

The number of trailing zeros in the decimal representation of n!, the factorial of a non-negative integer n, can be determined with this formula:

\(\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k}\), where k must be chosen such that 5^(k+1)>n

It's more simple if you look at an example:

How many zeros are in the end (after which no other digits follow) of 32!? \(\frac{32}{5}+\frac{32}{5^2}=6+1=7\) (denominator must be less than 32, \(5^2=25\) is less)

So there are 7 zeros in the end of 32!

The formula actually counts the number of factors 5 in n!, but since there are at least as many factors 2, this is equivalent to the number of factors 10, each of which gives one more trailing zero.

BACK TO THE ORIGINAL QUESTION:

According to above 100! has \(\frac{100}{5}+\frac{100}{25}=20+4=24\) trailing zeros.

Answer: B.

For more on this issues check Factorials and Number Theory links in my signature.

Find how many times the factor 5 is contained in 100!. That is, we have to find the largest such that 100! is divisible by . There are 20 multiples of 5 in the first hundred but 25, 50, 75, and 100 have to be counted twice because they are divisible by \(25 = 5^2\) . So, the answer is 24. The correct answer is B.

I have absolutely no Idea what they are telling me... Can someone please post a simple explanation for the rationale behind the explanation, or (even better) provide an alternative simple approach?

Thanks!

100! = 1 x 2 x 3 x ..... x 100

10 = 5 x 2

2s are in abundance however there is limited supply of 5s

How many multiples of 5 are there from 1 to 100- One way is counting other way is 100/5 = 20 How many multiples of 25 are there which contain an extra five = 100/25 = 4 There is no point going forward as the next power of 5 is 125 which is greater than 100. That does it: 20 + 4 = 24
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It asks for the number of zeros 100! ends with. If a number ends with zero it has to be a multiple of 10. So essentially we have to find out how many times in 100! does multiplication by 10 happen. But multiplication by 10 also happens when multiplication by both 5 and 2 happen. So it is better to see the factors of the number and then find the number of times multiplication by all those factors happen.

Consider the simpler example. 4*5*6. How many zeros does it end with?

We see 2 occurs twice in 4 which is 2*2, and once in 6 which is 2*3 and 5 occurs only once. So 5 is a limiting factor. Since 5 occurs only once, the number of times multiplication by both 5 and 2 happen or in other words the number of times multiplication by 10 happens or the number of zeros the number ends with is only 1.

In the case of 100!, 5 occurs in 5, 10, 15, 20 and so on up to 100 i.e, 20 times. But remember 5 occurs twice in 25 which is 5*5 , twice in 50 which is 5*5*2 and similarly twice each in 75 and 100. So it actually occurs 24 times.

2 occurs a lot more times and so 5 is the limiting factor.

Applying the same logic as in the simpler example, the number of zeros 100! ends with is 24.
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Re: How many zeros does 100! end with? [#permalink]

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Re: How many zeros does 100! end with? [#permalink]

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10 Dec 2015, 19:33

Hi Guys,

Am still abit confused about the whole concept. Some clarification would be appreciated.

With the formula, I see that the trailing zeros for the following are: 1) For 32!, (32/5) + (32/25) = 7 2) For 25!, (25/5) + (25/25) = 6 3) For 10!, (10/5) = 2 4) For 5!, (5/5) = 1

However, as i plug the numbers in and count on the calculator, the answer is different 1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros 2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros 3) For 10!, 3,628,800. A total of 2 trailing zeros 4) For 5!, 120. A total of 1 trailing zeros

Why is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.

Am still abit confused about the whole concept. Some clarification would be appreciated.

With the formula, I see that the trailing zeros for the following are: 1) For 32!, (32/5) + (32/25) = 7 2) For 25!, (25/5) + (25/25) = 6 3) For 10!, (10/5) = 2 4) For 5!, (5/5) = 1

However, as i plug the numbers in and count on the calculator, the answer is different 1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros 2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros 3) For 10!, 3,628,800. A total of 2 trailing zeros 4) For 5!, 120. A total of 1 trailing zeros

Why is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.

Am still abit confused about the whole concept. Some clarification would be appreciated.

With the formula, I see that the trailing zeros for the following are: 1) For 32!, (32/5) + (32/25) = 7 2) For 25!, (25/5) + (25/25) = 6 3) For 10!, (10/5) = 2 4) For 5!, (5/5) = 1

However, as i plug the numbers in and count on the calculator, the answer is different 1) For 32!, 263,130,836,933,693,000,000,000,000,000,000,000. A total of 21 trailing zeros 2) For 25!, 15,511,210,043,331,000,000,000,000. A total of 12 trailing zeros 3) For 10!, 3,628,800. A total of 2 trailing zeros 4) For 5!, 120. A total of 1 trailing zeros

Why is it that the first 2 doesnt comply whereas the last 2 does. I mean I understand the answer is correct here but am just trying to get my head around this.

I think there is some problem with the calculator that you are using

32! = 263130836933693530167218012160000000 25! = 15511210043330985984000000 Both of these values comply with our understanding.