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

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

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New post 20 Jun 2010, 22:59
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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
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If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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

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New post 24 Jun 2014, 02:16
21
4
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

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

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New post 21 Jun 2010, 01:23
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37
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/everything-ab ... 85592.html and http://gmatclub.com/forum/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.

Answer: C.

Hope it's clear.
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Re: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 23 Jun 2010, 12:38
3
15. No need to actually find the product of 1 to 30. Just look at the numbers that have factors of 3.

3 = 3
6 = 3 * 2
9 = 3 * 3
12 = 3 * 2 * 2
15 = 3 * 5
18 = 3 * 3 * 2
21 = 3 * 7
24 = 3 * 2 * 2 * 2
27 = 3 * 3 * 3
30 = 3 * 2 * 5

There are 14 3's, so the largest factor of p for 3k is 3*14.
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Re: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 19 Jun 2012, 05:13
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peter2000 wrote:
p is product of integers from 1 to 30 inclusive, what is the greatest integer k for which3^k is a factor of p?

A)10
B)12
C)14
D)16
E)18

I understand the way the GMAT guide does the problem, I was wondering if there was a quicker way?


Check out this post for an explanation of the method you can use to solve such questions:
http://www.veritasprep.com/blog/2011/06 ... actorials/
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Re: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 23 Aug 2012, 19:27
8
1
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: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 25 May 2014, 08:54
1
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: 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.

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

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New post 25 May 2014, 10:00
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



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.

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

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New post 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

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A.png [ 10.77 KiB | Viewed 35870 times ]


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

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New post 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 the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 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 the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 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 the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 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 Solution

There’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 the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 10 Aug 2016, 18:13
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Attached is a visual that should help.
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Re: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 16 Aug 2016, 14:56
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divakarbio7 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)

Answer:

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

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New post 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 the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 15 Oct 2016, 01:25
1
liseh wrote:
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.


No. 3^15 is not a factor of p = 30! while 3^14 is.
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Re: If p is the product of the integers from 1 to 30, inclusive, what is  [#permalink]

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New post 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:

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

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New post 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/everything-ab ... 85592.html and http://gmatclub.com/forum/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.

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?
GMAT Club Bot
Re: If p is the product of the integers from 1 to 30, inclusive, what is &nbs [#permalink] 14 Jan 2018, 15:20

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