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If m is the product of all integers from 1 to 40, inclusive
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09 Feb 2011, 15:11
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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
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Re: Number property
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09 Feb 2011, 15:35
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: everythingaboutfactorialsonthegmat85592.html other examples: facorialps105746.html#p8274531. 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 nonnegative 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)>nIt'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: ifnistheproductofintegersfrom1to20inclusive106289.htmlproductofsequence101187.htmlnumberpropertiesquestion87191.htmlog12ps96176.htmlps103218.html
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Re: Number property
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09 Feb 2011, 15:39
Thanks Bunuel  now I totally get it...!



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Re: Number property
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09 Feb 2011, 16:53
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 140.
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
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09 Feb 2011, 19:18
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 140.
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] ifnistheproductofintegersfrom1to20inclusive106289.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
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15 Feb 2011, 15:07
This is just an amazing way to solve it. so simple its crazy.
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Re: Number property
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30 Nov 2012, 16:45
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 140.
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
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06 Dec 2012, 20:33
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 140.
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
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26 Dec 2012, 05:26
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 140.
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
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06 Nov 2013, 10:53
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: everythingaboutfactorialsonthegmat85592.html other examples: facorialps105746.html#p8274531. 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 nonnegative 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)>nIt'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: ifnistheproductofintegersfrom1to20inclusive106289.htmlproductofsequence101187.htmlnumberpropertiesquestion87191.htmlog12ps96176.htmlps103218.htmlWhy divided 5 though?
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Re: Number property
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06 Nov 2013, 19:41
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: If m is the product of all integers from 1 to 40, inclusive
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14 Nov 2014, 09:08



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Re: If m is the product of all integers from 1 to 40, inclusive
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10 Jan 2016, 08:11
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?
10^p is a factor of 40! i.e., how many zeros(trailing zero's) are at the end of integer if 40! is expanded in full form.
then it becomes very simple all of a sudden as we use
The number of trailing zeros in the decimal representation of n!, the factorial of a nonnegative 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
\(\frac{40}{5}+\frac{40}{5^2}+\frac{40}{5^3}\)......................
8+1+0=9
Thus C is the correct answer.



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Re: If m is the product of all integers from 1 to 40, inclusive
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28 Jun 2016, 04:35
I know this is an old post but Bunuel I understand your answer and I did remember to use the formula too however I didnt know what to do with 40/25. To take that as a 2 or a 1. Is it that we have to take an INT value or do we round it up? Just stuck there Thanks Parth 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 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 nonnegative 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)>nIt'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: If m is the product of all integers from 1 to 40, inclusive
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28 Jun 2016, 05:15



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Re: If m is the product of all integers from 1 to 40, inclusive
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13 Aug 2016, 15:46
Hey Bunuel, You mentioned the value of K should be picked by 5^(K+1)>n...although in verbal explanation for the question, you said 5^(K+1) should be LESS than n, in this case 40. Im guessing the sign in the equation for K should be <n?



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Re: If m is the product of all integers from 1 to 40, inclusive
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17 Mar 2018, 11:56
Hi All, We're asked to find largest value of P so that 10^P divides into the product of the first 40 positive integers. This question comes down to "prime factorization" and finding all the "5s" and "10s" in the first 40 positive integers. 10 = (5)(2)…so we're looking for all of the 10s that can be "made" in this big product. Let's look at all the multiples of 5…. 5, 10, 15, 20, 25, 30, 35, 40 So we have some obvious 10s 10 20 = (2)(10) 30 = (3)(10) 40 = (4)(10) We can "make" some more 10s with the numbers that end in 5…. (5)(any even) = multiple of 10 (15)(any even) = multiple of 10 (35)(any even) = multiple of 10 I've separated the "25" for a reason; it creates an EXTRA 0 when multiplied by a multiple of 4… (25)(4) = 100 = 10^2 So we have….4 + 3 + 2 = nine 10s…..P = 9 Final Answer: GMAT assassins aren't born, they're made, Rich
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