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If p is the product of integers from 1 to 30, inclusive

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If p is the product of integers from 1 to 30, inclusive [#permalink]

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23 Aug 2012, 10:18
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 12 Dec 2012, 04:36, edited 2 times in total.
Edited the question.
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Re: 3^k is a factor of p [#permalink]

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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

Hope it helps!
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Re: 3^k is a factor of p [#permalink]

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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

Hope it helps!
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Posts: 43916
If p is the product of integers from 1 to 30, inclusive [#permalink]

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24 Aug 2012, 00:55
30
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Expert's post
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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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

Hope it's clear.
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Re: 3^k is a factor of p [#permalink]

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24 Aug 2012, 01:26
1
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BOOKMARKED
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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

Hope it's clear.

Thanks Bunuel for the crystal clear explanation!
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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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!.

Is this correct?
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Re: 3^k is a factor of p [#permalink]

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25 May 2014, 08:54
1
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BOOKMARKED
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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

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?
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Joined: 02 Sep 2009
Posts: 43916
Re: 3^k is a factor of p [#permalink]

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25 May 2014, 10:00
Expert's post
1
This post was
BOOKMARKED
russ9 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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

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.
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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24 Jun 2014, 02:16
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Just wrote down the multiples of 3 & the count of 3's in them

3 >> 1

6 >> 1

9 >> 2

12 > 1

15 > 1

18 > 2

21 > 1

24 > 1

27 > 3

30 > 1

Total = 14

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

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11 Sep 2014, 11:17
2
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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!

to find the greatest integer K i.e 3^k

30/3 +30/9 +30/27 = 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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08 Mar 2015, 19:50
1
KUDOS
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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

Hope it's clear.

Br

Bunuel..."So the highest power of 3 in 30! is 1."...You mean 14 right?
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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08 Mar 2015, 19:53
Avinashs87 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

[Reveal] Spoiler:
c

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: everything-about-factorials-on-the-gmat-85592.html and math-number-theory-88376.html

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.

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.
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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09 Mar 2015, 00:29
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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

Simplest and the easy way
Plz find the attached picture

Attachments

A.png [ 10.77 KiB | Viewed 23326 times ]

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

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14 Aug 2015, 21:30
1
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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

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

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22 Oct 2015, 09:18
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

This question is really asking us to determine the number of 3's "hiding" in the prime factorization of p.

p = (1)(2)(3)(4)(5)(6)(7)(8)(9) . . . (27)(28)(29)(30)
= (1)(2)(3)(4)(5)(2)(3)(7)(8)(3)(3)(10)(11)(3)(4)(13)(14)(3)(5)(16)(17)(3)(3)(2)(19)(20)(3)(7)(22)(23)(3)(8)(25)(26). . . (3)(3)(3)(28)(29)(3)(10)
= (3)^14(other non-3 stuff)

Cheers,
Brent
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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22 Oct 2015, 21:21
fayea 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

What is the quickest way and how?

You need to find out the number of 3's in 30!

There is a direct formula for this:

$$[30/3] + [30/3^2] + [30/3^3]$$ = 10 + 3 + 1 = 14
Option C

Here [x] = integral part of the number
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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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.
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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30 Oct 2015, 11:03
GuidoVdS wrote:
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.

The answer to your question is that you need to sometimes use the options provided in a PS question to guide you in order to understand the way/method to apply. Once you look at the options, the only way the statement you have mentioned above in red can be interpreted is by finding that power 'k' of 3, such that 3^k divides or is a factor of a number = 1*2*3*4*5*6...*30

Lets say you get an answer of 9 as the "biggest factor", this can also be mentioned as 3^2, giving you the value of k=2.

The only way a 'factor' can be a factor of another integer is by having the same prime factors as in the integer and having the same powers of these prime factors.

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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30 Oct 2015, 20:03
1
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GuidoVdS wrote:
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.

The question is asking for the biggest power of 3^k.
This can be found out by finding the number of 3's is the series.

Does this help?
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Re: If p is the product of integers from 1 to 30, inclusive [#permalink]

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10 Dec 2015, 14:33
Bunuel, you're a savage! Thanks for the explanation.
Re: If p is the product of integers from 1 to 30, inclusive   [#permalink] 10 Dec 2015, 14:33

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