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If m is the product of all integers from 1 to 40, inclusive

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If m is the product of all integers from 1 to 40, inclusive [#permalink] New post 09 Feb 2011, 14:11
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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
[Reveal] Spoiler: OA
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Re: Number property [#permalink] New post 06 Dec 2012, 19:33
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buymovieposters wrote:
i like this approach, but could you clarify where you wrote "you don't need to find the two's"

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.


For a detailed discussion, check out: http://www.veritasprep.com/blog/2011/06 ... actorials/

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.
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Re: Number property [#permalink] New post 09 Feb 2011, 14:35
Expert's post
MRHDK1 wrote:
Hey guys,

I need some help cracking this question, as I simply don't understand the explanation;

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


Check this: everything-about-factorials-on-the-gmat-85592.html other examples: facorial-ps-105746.html#p827453

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!.

Answer: C.

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Re: Number property [#permalink] New post 09 Feb 2011, 14:39
Thanks Bunuel - now I totally get it...! :)
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Re: Number property [#permalink] New post 09 Feb 2011, 15:53
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.
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Re: Number property [#permalink] New post 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.
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Re: Number property [#permalink] New post 15 Feb 2011, 14:07
This is just an amazing way to solve it. so simple its crazy.
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Re: Number property [#permalink] New post 30 Nov 2012, 15:45
i like this approach, but could you clarify where you wrote "you don't need to find the two's"

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.
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Re: Number property [#permalink] New post 26 Dec 2012, 04:26
buymovieposters wrote:
i like this approach, but could you clarify where you wrote "you don't need to find the two's"

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.


We can also understand it as follows:

The products that give rise to 0's.

(2 * 5) * (12 * 15) * (22 * 25) * (32 *35) * 10 * 20 *30 * 40
= (2 * 5) * (12 * 15) * (22 * 5 *5) * (32 *35) * 10 * 20 *30 * (8 * 5)
= (2 * 5) * (12 * 15) * (22 * 5 ) * (32 *35) * 10 * 20 *30 * (8 * 5* 5)

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.
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Re: Number property [#permalink] New post 06 Nov 2013, 09:53
Bunuel wrote:
MRHDK1 wrote:
Hey guys,

I need some help cracking this question, as I simply don't understand the explanation;

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


Check this: everything-about-factorials-on-the-gmat-85592.html other examples: facorial-ps-105746.html#p827453

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!.

Answer: C.

Similar questions:
if-n-is-the-product-of-integers-from-1-to-20-inclusive-106289.html
product-of-sequence-101187.html
number-properties-question-87191.html
og12-ps-96176.html
ps-103218.html


Why divided 5 though?
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Re: Number property [#permalink] New post 06 Nov 2013, 18:41
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accincognito wrote:
Why divided 5 though?


Check out this link: http://www.veritasprep.com/blog/2011/06 ... actorials/

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.
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Re: Number property   [#permalink] 06 Nov 2013, 18:41
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