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

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09 Aug 2009, 14:02

I can see that any of these numbers is ok. so I just pick 18.

P = 30! = 30x29x28x......x4x3x2x1

if 3K is a factor of p, it means that we can divide p by 3K and the result is an integer. there is a "3" in the expression of p, and all numbers shown are also there.

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

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09 Aug 2009, 23:22

Agree with Economist. But from the list, the best answer would be 18. Since all the prime factors of 18*3(2*3*3*3) would be in the product number. OA pls! _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

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

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11 Aug 2009, 04:04

1

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tarek99 wrote:

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18

please show the fastest way to solve this. thanks

Well ..I think question should read as 3^k ....i.e, 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?

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

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11 Aug 2009, 04:32

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age wrote:

tarek99 wrote:

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18

please show the fastest way to solve this. thanks

Well ..I think question should read as 3^k ....i.e, 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?

Then C ..14 will be correct..

I agree. If C is the OA, then the questions should read as 3^k giving 14 as the answer.

Although if its not 3^k, then another interpretation of 3k could be 300 + k instead of 3*k, because 3*k makes the question silly. This means that when k=10, we get 310 and not 3*10 = 30

Thus proceeding with the "300 + k" assumption, we get: P = 30!

Option A: 310 => 310 = 2 * 5 * 31 => 31 is a prime and is not present in 30! Thus 310 is not a factor of P => A is Eliminated

Option B: 312 => 312 = 2^3 * 3 * 13 Thus 312 is a factor of P => B is the correct answer !

Option C: 314 => 314 = 2 * 157 => 157 is a prime and is not present in 30! Thus 314 is not a factor of P => C is Eliminated

Option D: 316 => 316 = 2^2 * 79 => 79 is a prime and is not present in 30! Thus 316 is not a factor of P => D is Eliminated

Option E: 318 => 310 = 2 * 3 * 53 => 53 is a prime and is not present in 30! Thus 318 is not a factor of P => E is Eliminated

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

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29 Sep 2014, 07:27

tarek99 wrote:

If p is the product of the integers from 1 to 30, inclusive, what is the greatest integer k for which 3k is a factor of p? A. 10 B. 12 C. 14 D. 16 E. 18

please show the fastest way to solve this. thanks

Can we change 3k to 3^K . A minor correction to make this question correct.

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

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.

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