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Re: 3^k is a factor of p [#permalink]
23 Aug 2012, 19:27
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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
Values which we are looking for are 3,6,9,12,.. all multiples till 30 now everything will give you atleast 1 power of 3 but there are values which will give more than 1 power of 3 and those values will be multiples of 9 9 -> will give 2 powers 18-> will give 2 powers 27 -> will give 3 powers NUmber of single powers of 3 = (30-3)/3 +1 - 1(for 9) - 1(for 18) -1(for 27) = 7 so total powers = 2(for 9) + 2(for 18) + 3(for 27) + 7 = 14
Re: 3^k is a factor of p [#permalink]
23 Aug 2012, 21:07
Hi nkdotgupta,
Thanks for your reply but this is the explaation as provided in the OG. I wanted to know if there are other ways of approaching the problem since this method might be cumbersome we encounter larger numbers.
nktdotgupta 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
Values which we are looking for are 3,6,9,12,.. all multiples till 30 now everything will give you atleast 1 power of 3 but there are values which will give more than 1 power of 3 and those values will be multiples of 9 9 -> will give 2 powers 18-> will give 2 powers 27 -> will give 3 powers NUmber of single powers of 3 = (30-3)/3 +1 - 1(for 9) - 1(for 18) -1(for 27) = 7 so total powers = 2(for 9) + 2(for 18) + 3(for 27) + 7 = 14
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!.
Back 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.
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!.
Back 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 1.
Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
22 Oct 2013, 12:20
Correct me if I am mistaken but understanding number properties seems to be in line with these type of problems.
Going back to basics - these type of problems seem to relate to the factor foundation rule - if a is a factor of b and b is a factor of c then a is a factor of c.
In these problems you are trying to find the greatest degree of a common prime- in this case- of 3 in 30!.
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!.
Back 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 1.
Answer: C.
Hope it's clear.
Hi Bunuel,
This explanation makes complete sense but i'm curious as to why another method doesn't work.
I first found the number of integers (30-1+1) = 30 Average of 30+1/2 = 15.5
30 * 15.5 = 465
Factoring 465 only gave me one "3" and therefore I knew it was wrong but I had wasted valuable time. I realized that what I had done doesn't account for the product but rather the sumof the consecutive integers. Correct?
If the question was changed just a bit to say that "p" is the sum of the integers from 1 - 30 then my method would've worked and the answer would've been 1. Is that correct?
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!.
Back 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 1.
Answer: C.
Hope it's clear.
Hi Bunuel,
This explanation makes complete sense but i'm curious as to why another method doesn't work.
I first found the number of integers (30-1+1) = 30 Average of 30+1/2 = 15.5
30 * 15.5 = 465
Factoring 465 only gave me one "3" and therefore I knew it was wrong but I had wasted valuable time. I realized that what I had done doesn't account for the product but rather the sumof the consecutive integers. Correct?
If the question was changed just a bit to say that "p" is the sum of the integers from 1 - 30 then my method would've worked and the answer would've been 1. Is that correct?
Yes, you are correct. If it were "p is the SUM of integers from 1 to 30, inclusive", then the answer would be 1. _________________
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!.
Back 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 1.
Answer: C.
Hope it's clear.
Br
Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?
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!.
Back 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 1.
Answer: C.
Hope it's clear.
Br
Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?
Yes, it's a typo. Edited. Thank you. _________________
Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
14 Aug 2015, 21:30
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
you have to do successive division by 3. The given product is 30! divide 30 by 3; the quotient is 10 divide 10 by 3; the quotient is 3 divide 3 by 3; the quotient is 1 add all the quotients, 10+3+1=14
Brent Hanneson - Founder of GMAT Prep Now, a free & comprehensive GMAT course with: - over 500 videos (35 hours of instruction) - over 800 practice questions - 2 full-length practice tests and other bonus offers - http://www.gmatprepnow.com/
Re: If p is the product of integers from 1 to 30, inclusive [#permalink]
30 Oct 2015, 09:58
Looking at the possible answers I understood what I should be looking for.
But was I the only one confused with the question, since "a factor is an integer that divides another integer without leaving a remainder"? What I mean is: having in mind the definition of a "factor", I could be looking for the biggest "factor", resulting from 3^k, and not the highest power of 3 in 30 (or the number of 3 in the prime factorization of p).
This is probably a dumb question, but I'm probably missing something that you guys are not and I want to be sure that I see it from now on.
gmatclubot
Re: If p is the product of integers from 1 to 30, inclusive
[#permalink]
30 Oct 2015, 09:58
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