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

125000 has 3 trailing zeros;

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: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

As for your question, forget 40!. Let's look at 10!

What is the maximum value of p such that 10^p is a factor of 10! To make a 10, you need a 2 and a 5. In 10!, will you have more 2s or more 5s?

1*2*3*4*5*6*7*8*9*10 - Every alternate number has at least one 2 in it. Every fifth number has at least one 5 in it. You certainly have more 2s than 5s. If you have eight 2s and two 5s, how many 10s can you make? You certainly need a 5 to make a 10. Since you have only two 5s, you can make only two 10s. It doesn't matter how many 2s are extra. You cannot make a 10 with only 2s. _________________

Re: Number property [#permalink]
09 Feb 2011, 18:18

Expert's post

abmyers wrote:

The factors of 10 are 1, 2, 5, and 10.

To find what k is, find how many times you can make 10 with the numbers in 1-40.

First, find the 10's as they are going to be most restrictive. Only 10, 20, 30 and 40 have 10 as a factor. 4 total.

Next find 5's. 5, 15, 25 and 35 each have a 5, with 25 having 2. 5 total.

You dont need to find the twos as you know there are more than 5's.

Add the two you get 9.

I would suggest that you refrain from actually counting the 5s/10s. Not only can it be tedious with bigger numbers, but it is also prone to errors. With smaller numbers, I don't think it will matter very much what you do. For some theory on these kind of questions, check out this link: [url] if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html#p834162[/url]

Once you understand such problems, you wouldn't care how big or small the numbers are... You would be able to solve such questions easily. _________________

The first four terms give rise to four 0's, the next three to three 0's and the last one to two 0's and the total is 9.

Note: You may also multiply 5 with 4 and/or 6 and/or 8 as they contain a 2. Similarly for 15, 25 and 35. But the number of 0's is restricted by the number of 5's. _________________

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

125000 has 3 trailing zeros;

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: If m is the product of all integers from 1 to 40, inclusive, what is the greatest integer p for which 10^p is a factor of m? (A) 7 (B) 8 (C) 9 (D) 10 (E) 11

Given: \(m=40!\). So we should find the # of trailing zeros in 40!, as it'll be the greatest value of p for which 40!/10^p will be an integer.

40! has \(\frac{40}{5}+\frac{40}{5^2}=8+1=9\) trailing zeros, which means that 40! ends with 9 zeros so p=9 is the greatest integer for which 10^p (10^9) is a factor of 40!.

It explains this method in detail. Also, you divide by 5 because to make a 10, you need a 2 and a 5. The number of 5s will be fewer than the number of 2s hence all you need to do is find the number of 5s you have. You can make that many 10s. _________________

Re: If m is the product of all integers from 1 to 40, inclusive [#permalink]
14 Nov 2014, 07:23

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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