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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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14 Mar 2016, 01:14



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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18 May 2016, 11:27
ajju2688 wrote: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p ?
(A) 10 (B) 12 (C) 14 (D) 16 (E) 18 Solution: We are given that p is the product of the integers from 1 to 30 inclusive, which is the same as 30!. We know that 30! must be divisible by all the numbers from 30 to 1, inclusive. In this question, however, we only care about the numbers in the factorial that are divisible by 3. All of the other numbers are irrelevant. For example, 20 is not a multiple of 3, and because 3 does not evenly divide into 20, there’s no need to consider it. Only the multiples of 3 matter for this problem and they are: 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3. What we’re really being asked is how many prime factors of 3 are contained in the product of 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3. However, instead of actually calculating the product and then doing the prime factorization, we can simply count the number of prime factors of 3 in each number from our set of multiples of 3: 30, 27, 24, 21, 18, 15, 12, 9, 6, and 3. 30 has 1 prime factor of 3. 27 has 3 prime factors of 3. 24 has 1 prime factor of 3. 21 has 1 prime factor of 3. 18 has 2 prime factor of 3. 15 has 1 prime factor of 3. 12 has 1 prime factor of 3. 9 has 2 prime factors of 3. 6 has 1 prime factor of 3. 3 has 1 prime factor of 3. When we sum the total number of prime factors of 3 in our list we get 14. It follows that 14 is the greatest value K for which 3^k divides evenly into 30!, and thus k = 14. Answer: C Alternate SolutionThere’s actually a shortcut that can be used. The question is what is the largest possible value of k such that 30!/3^k is an integer. To use the shortcut, divide the divisor 3 into 30, which is simply the numerator without the factorial. This division is 30/3 = 10. We then divide 3 into the resulting quotient of 10, and ignore any remainder. This division is 10/3 = 3. We again divide 3 into the resulting quotient of 3. This division is 3/3 = 1. Since our quotient of 1 is smaller than the divisor of 3, we can stop here. The final step is to add together all the quotients: 10 + 3 + 1 = 14. Thus, the largest value of k is 14. One caveat to this shortcut is that it only works when the denominator is a prime number. For example, had the denominator been 9, this rule could not have been used.
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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10 Aug 2016, 18:13
Attached is a visual that should help.
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Screen Shot 20160810 at 7.09.59 PM.png [ 108.13 KiB  Viewed 999 times ]
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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14 Oct 2016, 17:28
I was wondering, shouldn't the answer be k=15 in reality? It was not specified that the integer k must be positive, hence 3^0=1 would also be one of the factors? Please correct me if I'm wrong.



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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15 Oct 2016, 01:25



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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15 Oct 2016, 01:34
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ajju2688 wrote: If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3^k is a factor of p ?
(A) 10 (B) 12 (C) 14 (D) 16 (E) 18 Product of the integers from 1 to 30 = 30! So p = 30! Quote: what is the greatest integer k for which 3^k is a factor of p To find the greatest integer of a factor keep dividing it... 30/3 = 10 10/3 = 3 3/1 = 1 So, Maximum power of 3^k is ( 10 + 3 + 1 ) = 14
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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05 Mar 2017, 12:28
Bunuel wrote: ajju2688 wrote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? a) 10 b) 12 c) 14 d) 16 e) 18 This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming? Finding the number of powers of a prime number k, in the n!.The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\) For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!. Check for more: http://gmatclub.com/forum/everythingab ... 85592.html and http://gmatclub.com/forum/mathnumbertheory88376.htmlBack to the original question:If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18 Given \(p=30!\). Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 14. Answer: C. Hope it's clear. Could you clarify how you solved for 10 ,3 and 1 for this step of checking the highest power of 3 in 30!? One method I used was simply doing a sum of the # of factors of 3. For example, 3,6,9,12 have a total of 5 factors of three (9 has 3X3 which is 2 factors). I did this for all multiples of 3 up until 30 and did a sum of all. Posted from my mobile device Posted from my mobile device



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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01 Apr 2017, 04:53
p= 30! 30! is the product of all the numbers from 1 till 30 BUT we have to just focus on the multiples of 3 because we have to find the power of 3 in the final answer of this factorial. 3,6(2*3),9(3*3),12(2^2*3),15(3*5),18(2*3^2),21(3*7),24(2^3*3),27(3^4),30(2*3*5) Counting the number of 3's k= 14
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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25 Apr 2017, 15:56
What if the question only asked this:
What is the product of integers from 1 to 30, inclusive?
What is the strategy?



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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27 Apr 2017, 10:21
But how do you solve p=30! if they wanted a numerical answer??



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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27 Apr 2017, 19:12
GMAT usually never asks to calculate big nos. as the exam is not testing our calculation but our analytical thinking. It just tests how fast and efficiently we can find an answer to a solution in the given time frame. Its a test for people who are or will be business folks. So GMAT is aware that there are calculators available to business folks. I am not an expert but according to me, you should not worry about calculating such big nos. Even if you check on calculator, the answer is 2.6525286e+32. Too something !! Also do check the nos. in the official guide, and see if you can find out questions that ask factorials of any big no. If such nos. are asked, there would be another values given to simplify the expression. I hope this helps........
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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28 May 2017, 06:29
Hello, To respond to GMAT01 I didn't understand as well and wanted to ask, but while typing this I looked back at the previous post on factorials It mentions you have to take the quotient of the division, which is why 30/3^2 = 3 and 30/3^3 = 1



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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14 Jan 2018, 11:32
Hi All, These types of questions are based on a math concept called "prime factorization", which basically means that any integer greater than 1 is either prime OR the product of a bunch of primes. Here's a simple example: 24 = (2)(2)(2)(3) Now, when it comes to this question, we're asked to multiply all the integers from 1 to 30, inclusive and find the greatest integer K for which 3^K is a factor of this really big number. Here's a simple example with a smaller product: 1 to 6, inclusive… (1)(2)(3)(4)(5)(6) Then numbers 1, 2, 4 and 5 do NOT have any 3's in them, so we can essentially ignore them: 3 = one 3 6 = (2)(3) = one 3 Total = two 3's So 3^2 is the biggest "power of 3" that goes into the product of 1 to 6, inclusive. Using that same idea, we need to find all of the 3's in the product of 1 to 30, inclusive. Here though, you have to be careful, since there are probably MORE 3's than immediately realize: 3 = one 3 6 = one 3 9 = (3)(3) = two 3s 12 = one 3 15 = one 3 18 = (2)(3)(3) = two 3s 21 = one 3 24 = one 3 27 = (3)(3)(3) = three 3s 30 = one 3 Total = 14 3's Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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14 Jan 2018, 15:20
Bunuel wrote: ajju2688 wrote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? a) 10 b) 12 c) 14 d) 16 e) 18 This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming? Finding the number of powers of a prime number k, in the n!.The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\) For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!. Check for more: http://gmatclub.com/forum/everythingab ... 85592.html and http://gmatclub.com/forum/mathnumbertheory88376.htmlBack to the original question:If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18 Given \(p=30!\). Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 14. Answer: C. Hope it's clear. Could you further explain how you arrived at "10+3+1=14" from the equation which you created before it?



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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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14 Jan 2018, 19:31
destinyawaits wrote: Bunuel wrote: ajju2688 wrote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? a) 10 b) 12 c) 14 d) 16 e) 18 This is an OG12 question. Can someone tell me if there is a quick way to solve these kinds of questions since the OG explanation seems to be very time consuming? Finding the number of powers of a prime number k, in the n!.The formula is: \(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}\) ... till \(n>k^x\) For example: what is the power of 2 in 25! (the highest value of m for which 2^m is a factor of 25!) \(\frac{25}{2}+\frac{25}{4}+\frac{25}{8}+\frac{25}{16}=12+6+3+1=22\). So the highest power of 2 in 25! is 22: \(2^{22}*k=25!\), where k is the product of other multiple of 25!. Check for more: http://gmatclub.com/forum/everythingab ... 85592.html and http://gmatclub.com/forum/mathnumbertheory88376.htmlBack to the original question:If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18 Given \(p=30!\). Now, we should check the highest power of 3 in 30!: \(\frac{30}{3}+\frac{30}{3^2}+\frac{30}{3^3}=10+3+1=14\). So the highest power of 3 in 30! is 14. Answer: C. Hope it's clear. Could you further explain how you arrived at "10+3+1=14" from the equation which you created before it? You should take only the quotient of the division, that is 30/9 = 3 and 30/27 = 1.
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If p is the product of integers from 1 to 30, inclusive [#permalink]
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16 Jan 2018, 16:29
Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrepQuote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? Can experts please advise what I am missing in my below approach: I need to find a factor of 3^k such that when 30! is divided by 3^k the remainder is zero. Since k is a positive integer we need to see max value of 3^k ; now 3^k can take 3,9, 27.. I was stuck beyond this point to corelate 30! and max power of 3^k such that remainder is zero. Quote: We know that 30! must be divisible by all the numbers from 30 to 1, inclusive. In this question, however, we only care about the numbers in the factorial that are divisible by 3. Are not we supposed to find no in factorial divisible by 3^k and not 3 ?
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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16 Jan 2018, 18:27
adkikani wrote: Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrepQuote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? Can experts please advise what I am missing in my below approach: I need to find a factor of 3^k such that when 30! is divided by 3^k the remainder is zero. Since k is a positive integer we need to see max value of 3^k ; now 3^k can take 3,9, 27.. I was stuck beyond this point to corelate 30! and max power of 3^k such that remainder is zero. Quote: We know that 30! must be divisible by all the numbers from 30 to 1, inclusive. In this question, however, we only care about the numbers in the factorial that are divisible by 3. Are not we supposed to find no in factorial divisible by 3^k and not 3 ? Hi.. You have to find how many 3s are there in 30!.. 30/3=10 gives us the number which have 3 in it.. 3,6,9...,30 30/9=3.xy~3 gives us the number which are div by 9 as there is extra 3 in these numbers... 9,18,27 30/27=1.. numbers containing 3*3s..27 So our Ans 10+3+1=14
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
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17 Jan 2018, 04:36
adkikani wrote: Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrepQuote: If p is the product of integers from 1 to 30, inclusive, what is the greatest integer k for which \(3^k\) is a factor of p? Can experts please advise what I am missing in my below approach: I need to find a factor of 3^k such that when 30! is divided by 3^k the remainder is zero. Since k is a positive integer we need to see max value of 3^k ; now 3^k can take 3,9, 27.. I was stuck beyond this point to corelate 30! and max power of 3^k such that remainder is zero. Quote: We know that 30! must be divisible by all the numbers from 30 to 1, inclusive. In this question, however, we only care about the numbers in the factorial that are divisible by 3. Are not we supposed to find no in factorial divisible by 3^k and not 3 ? Check out this post for a detailed discussion on this concept: https://www.veritasprep.com/blog/2011/0 ... actorials/
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