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If p is the product of the integers from 1 to 30, inclusive, what 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, 20:31
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/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?


You should take only the quotient of the division, that is 30/9 = 3 and 30/27 = 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 16 Jan 2018, 17:29
Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrep

Quote:
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 co-relate 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 the integers from 1 to 30, inclusive, what is [#permalink]

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New post 16 Jan 2018, 19:27
adkikani wrote:
Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrep

Quote:
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 co-relate 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 the integers from 1 to 30, inclusive, what is [#permalink]

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New post 17 Jan 2018, 05:36
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adkikani wrote:
Bunuel chetan2u niks18 VeritasPrepKarishma JeffTargetTestPrep

Quote:
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 co-relate 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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Re: If p is the product of the integers from 1 to 30, inclusive, what is   [#permalink] 17 Jan 2018, 05:36

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