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If d is a positive integer and f is the product of the first
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If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d? (1) 10^d is a factor of f (2) d>6
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Originally posted by enigma123 on 28 Jan 2012, 18:13.
Last edited by Bunuel on 30 Oct 2012, 01:42, edited 2 times in total.
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If d is a positive integer and f is the product of the first
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28 Jan 2012, 18:18
If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?(1) 10^d is a factor of f > \(k*10^d=30!\). First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) > \(\frac{30}{5}+\frac{30}{25}=6+1=7\) > \(30!\) has \(7\) zeros. \(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) > it tells us only that max possible value of \(d\) is \(7\). Not sufficient. Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we cannot determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient. (2) d>6 Not Sufficient. (1)+(2) From (2) \(d>6\) and from (1) \(d_{max}=7\) > \(d=7\). Answer: C. Hope it helps.
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Re: +ve integer D
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28 Jan 2012, 18:19
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. For more on this concept check Everything about Factorials on the GMAT: everythingaboutfactorialsonthegmat85592.html
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Re: If d is a positive integer and f is the product of the first
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29 Jan 2012, 15:50
Hi Bunuel thanks  all makes sense apart from the concept of trailing zeros. Am I right in saying this is how you said there will be 7 zero's. 30/5 + 30/25 = 6 + 1 (quotient) = 7. Where I am not clear is have you simply divided 30/25? I hope I am making myself clear, if I am not then please let me know.
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Re: If d is a positive integer and f is the product of the first
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Re: If d is a positive integer and f is the product of the first
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04 Jun 2013, 21:17
Hi. I think there is a solution without knowing the trailing zeros formula. Of course I. alone will not be enough (d= 10 or d=100 do the trick) and II. d>6 is vague. Now, to evaluate I and II together, like you know that 10 = 2*5, and 10^x = (2*5)^x = 2^x*5^x, if 10^d is a factor of f, like f = 1*2*3*4*5...*30, you need to see how many 2s and 5ves you can get. You have plenty of 2s, so lets focus on the 5ves. You actually get 7 fives between 1 and 30 (one in 5,10,15,20,30 and two in 25). So basically d could be any number between 1 and 7. Like II is "d>6", you know that d = 7.
(my 1st post, sorry for style... just trying to help, I suck at knowing formulas, although they save you time. (I learned a lot from this forum, Bunuel is my Guru!))



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Re: If d is a positive integer and f is the product of the first
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04 Jun 2013, 22:25
I'm having trouble understanding "the product of the first 30 positive integers"



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Re: If d is a positive integer and f is the product of the first
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27 Apr 2015, 05:07
REALY HARD just count the numbers of the number 2 and 5, to see that the max is 7.
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If d is a positive integer and f is the product of the first
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thangvietnam wrote: REALY HARD just count the numbers of the number 2 and 5, to see that the max is 7. I hope the students are clear here about why we only need to consider the number of 5s in the product 30*29*28. . .3*2*1 If not, then please read on. 10 = 2*5 So, to make one 10, we need one 2 and one 5. In the product 30*29*28. . .3*2*1, the number of 2s far exceeds the number of 5s. Therefore, since 5 is the limiting multiplicand here, we only need to consider the number of 5s. Apply this discussionSuppose the question was: If p is the maximum integer such that \(35^p\) is a factor of the product of the first 30 positive integers, what is the value of p?How would you proceed to find the value of p? Regards Japinder
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Re: If d is a positive integer and f is the product of the first
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28 Apr 2015, 00:17
EgmatQuantExpert wrote: thangvietnam wrote: REALY HARD just count the numbers of the number 2 and 5, to see that the max is 7. I hope the students are clear here about why we only need to consider the number of 5s in the product 30*29*28. . .3*2*1 If not, then please read on. 10 = 2*5 So, to make one 10, we need one 2 and one 5. In the product 30*29*28. . .3*2*1, the number of 2s far exceeds the number of 5s. Therefore, since 5 is the limiting multiplicand here, we only need to consider the number of 5s. Apply this discussionSuppose the question was: If p is the maximum integer such that \(35^p\) is a factor of the product of the first 30 integers, what is the value of p?How would you proceed to find the value of p? Regards Japinder Hello EgmatQuantExpert. Really artful question ) We know that \(35\) has two factors \(5\) and \(7\) First impulse is to just take answer from previous question because of presence of \(5\) but we should calculate that number, that has less occurrences So \(5\) in \(30!\) meets \(7\) times but \(7\) in \(30!\) meets \(4\) times. And we can infer that \(30!\) will be divisible by \(35\) only \(4\) times. So \(p = 4\)
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Re: If d is a positive integer and f is the product of the first
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28 Apr 2015, 04:27
Harley1980 wrote: Hello EgmatQuantExpert. Really artful question ) We know that \(35\) has two factors \(5\) and \(7\) First impulse is to just take answer from previous question because of presence of \(5\) but we should calculate that number, that has less occurrences So \(5\) in \(30!\) meets \(7\) times but \(7\) in \(30!\) meets \(4\) times. And we can infer that \(30!\) will be divisible by \(35\) only \(4\) times. So \(p = 4\) Dear Harley1980Spoton analysis and correct answer. Good job done! Best Regards  Japinder
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Re: If d is a positive integer and f is the product of the first
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01 Jun 2015, 17:10
EgmatQuantExpert wrote: thangvietnam wrote: REALY HARD just count the numbers of the number 2 and 5, to see that the max is 7. I hope the students are clear here about why we only need to consider the number of 5s in the product 30*29*28. . .3*2*1 If not, then please read on. 10 = 2*5 So, to make one 10, we need one 2 and one 5. In the product 30*29*28. . .3*2*1, the number of 2s far exceeds the number of 5s. Therefore, since 5 is the limiting multiplicand here, we only need to consider the number of 5s. Apply this discussionSuppose the question was: If p is the maximum integer such that \(35^p\) is a factor of the product of the first 30 positive integers, what is the value of p?How would you proceed to find the value of p? Regards Japinder I think We need to do prime factorization of 35 = 5 x 7 so to make one 35 we need one 5 & one 7. Now, we will calculate how many 5's & 7's are there in the 30!. It will be : 30/5 + 30/5x5 = 7 # of 5's Also, 30/7 = 4 # of 7's. Thus, 7 is the limiting multiplicand here. We have four such pairs of 5 x 7 . Thus the maximum power of 35 will be 4 so as to divide 30! evenly. I really like your step by steo approach to each question. Regards, Ankush.



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Re: If d is a positive integer and f is the product of the first
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30 Dec 2015, 01:04
Bunuel : I understood the solution but what i didn't understand is ..why isn't statement 1 sufficient. We know the trailing zeros are 7and can only be 7 because it would otherwise exceed 30! ..so wouldn't A be enough to tell us that d=7. although statement 2 says d>6 ..and if we combine them it just reconfirms the same thing. can you pls help me fill the gap in my understanding .



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Re: If d is a positive integer and f is the product of the first
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30 Dec 2015, 01:10
puneetkaur wrote: Bunuel : I understood the solution but what i didn't understand is ..why isn't statement 1 sufficient. We know the trailing zeros are 7and can only be 7 because it would otherwise exceed 30! ..so wouldn't A be enough to tell us that d=7. although statement 2 says d>6 ..and if we combine them it just reconfirms the same thing. can you pls help me fill the gap in my understanding . hi, although asked from bunuel, i'll try it for you.. statement 1 is "(1) 10^d is a factor of f " Also rightly said by you there are 7 0s in 30!.. but 30! factor can be 10^2, 10^1, or 10^7.. we only know tht the max value of d is 7.. hope it helped
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Re: If d is a positive integer and f is the product of the first
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29 Aug 2017, 07:46
Bunuel wrote: If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?
(1) 10^d is a factor of f > \(k*10^d=30!\).
First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) > \(\frac{30}{5}+\frac{30}{25}=6+1=7\) > \(30!\) has \(7\) zeros.
\(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) > it tells us only that max possible value of \(d\) is \(7\). Not sufficient.
Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we cannot determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient.
(2) d>6 Not Sufficient.
(1)+(2) From (2) \(d>6\) and from (1) \(d_{max}=7\) > \(d=7\).
Answer: C.
Hope it helps. So the thing is: 10^d is not the only factor, it is one of the factors, that's why we cannot surely say that d=7. But if the question would have said that 10^d is the only factor then, "A" would be the right answer. Am i right?



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Re: If d is a positive integer and f is the product of the first
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29 Aug 2017, 09:08
saswatdodo wrote: Bunuel wrote: If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?
(1) 10^d is a factor of f > \(k*10^d=30!\).
First we should find out how many zeros \(30!\) has, it's called trailing zeros. It can be determined by the power of \(5\) in the number \(30!\) > \(\frac{30}{5}+\frac{30}{25}=6+1=7\) > \(30!\) has \(7\) zeros.
\(k*10^d=n*10^7\), (where \(n\) is the product of other multiples of 30!) > it tells us only that max possible value of \(d\) is \(7\). Not sufficient.
Side notes: 30! is some huge number with 7 trailing zeros (ending with 7 zeros). Statement (1) says that \(10^d\) is factor of this number, but \(10^d\) can be 10 (d=1) or 100 (d=2) ... or 10,000,000 (d=7). Basically \(d\) can be any integer from 1 to 7, inclusive (if \(d>7\) then \(10^d\) won't be a factor of 30! as 30! has only 7 zeros in the end). So we cannot determine single numerical value of \(d\) from this statement. Hence this statement is not sufficient.
(2) d>6 Not Sufficient.
(1)+(2) From (2) \(d>6\) and from (1) \(d_{max}=7\) > \(d=7\).
Answer: C.
Hope it helps. So the thing is: 10^d is not the only factor, it is one of the factors, that's why we cannot surely say that d=7. But if the question would have said that 10^d is the only factor then, "A" would be the right answer. Am i right? 1 is the only positive integer which has 1 factor. All other positive integers have more factors. It does not make sense to say that 10^d is the only factor of 30!.
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Re: If d is a positive integer and f is the product of the first
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05 Sep 2017, 17:48
enigma123 wrote: If d is a positive integer and f is the product of the first 30 positive integers, what is the value of d?
(1) 10^d is a factor of f (2) d>6 We are given that d is a positive integer and f = 30!. We need to determine the value of d. Statement One Alone: 10^d is a factor of f Since 10^1 and 10^2 could each divide into 30!, we do not have a unique value for d. Statement one alone is not sufficient to answer the question. Statement Two Alone: d > 6 Since d could be 7, 8, or greater, statement two alone does not allow us to determine a unique value of d. Statements One and Two Together: Using both statements, since we know that d > 6, let’s determine the maximum value d can be given that 10^d divides into 30!. Essentially, we need to determine the maximum number of fivetwo pairs. (Recall that each fivetwo pair creates a factor of 10.) Since there are more twos than fives, let’s determine the number of fives. The factors that are multiples of 5 in 30! are 5, 10, 15, 20, 25 = 5^2, and 30. So, we see there are 7 fives in 30!, and thus the maximum value of d is 7. Since d > 6, d must be 7. Answer: C
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Re: If d is a positive integer and f is the product of the first
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05 Apr 2018, 09:10
The answer is C. Statement 1  With the help of statement 1 we can find maximum values of zeros in 30!. Not sufficient. Statement 2  With this, all we know is d>6. Not sufficient. Statement 1 + 2 = # of trailing zeros in 30! is 7 and d>6. So answer is 7. Sufficient.
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