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67% (01:42) correct
33% (00:01) wrong based on 9 sessions

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 _________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Last edited by snipertrader on 14 Aug 2009, 04:44, edited 1 time in total.

Nice One sniper. But there is a problem with the question. I did take a look at it previously but dont have the link with me right now. Accroding to me, best answer here would be 18.

Reason : 1*2*3*4....*29*30. Here, k could be any of those numbers i.e. 1 to 30. Accordingly, the greatest number k would be 30. However, since that is not in the option, I would go with 18,the highest one among the options.

Sniper, if you dont mind dear, could you tell me the source of the question? _________________

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

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

i think the question should be 3^k

ie the question is asking ( how many multiples of 3 r there)

Nice One sniper. But there is a problem with the question. I did take a look at it previously but dont have the link with me right now. Accroding to me, best answer here would be 18.

Reason : 1*2*3*4....*29*30. Here, k could be any of those numbers i.e. 1 to 30. Accordingly, the greatest number k would be 30. However, since that is not in the option, I would go with 18,the highest one among the options.

Sniper, if you dont mind dear, could you tell me the source of the question?

Hi Nikhil

The link that you were referring to is here: factor-82137.html _________________

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

i think the question should be 3^k

ie the question is asking ( how many multiples of 3 r there)

30-1 / 3 +1 = 10.............A

I think you have mistakenly used the formula for finding the multiples of 3 in a range of consecutive nos......

The answer for the number of powers of 3 in 30! is 14 => 30/3 + 30/(3^2) + 30/(3^3) + 30/(3^4) + ........ => 10 + 3 + 1 + 0 + ...... => 14 _________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Source - the infamous banned gmat sets. Sorry if it is a repeat question.

bhanushalinikhil wrote:

Nice One sniper. But there is a problem with the question. I did take a look at it previously but dont have the link with me right now. Accroding to me, best answer here would be 18.

Reason : 1*2*3*4....*29*30. Here, k could be any of those numbers i.e. 1 to 30. Accordingly, the greatest number k would be 30. However, since that is not in the option, I would go with 18,the highest one among the options.

Sniper, if you dont mind dear, could you tell me the source of the question?

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Can u explain further.. i am not able to follow.. is there a generic formula for this ?

samrus98 wrote:

The answer for the number of powers of 3 in 30! is 14 => 30/3 + 30/(3^2) + 30/(3^3) + 30/(3^4) + ........ => 10 + 3 + 1 + 0 + ...... => 14

_________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

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

i think the question should be 3^k

ie the question is asking ( how many multiples of 3 r there)

30-1 / 3 +1 = 10.............A

I think you have mistakenly used the formula for finding the multiples of 3 in a range of consecutive nos......

The answer for the number of powers of 3 in 30! is 14 => 30/3 + 30/(3^2) + 30/(3^3) + 30/(3^4) + ........ => 10 + 3 + 1 + 0 + ...... => 14

Can u explain further.. i am not able to follow.. is there a generic formula for this ?

samrus98 wrote:

The answer for the number of powers of 3 in 30! is 14 => 30/3 + 30/(3^2) + 30/(3^3) + 30/(3^4) + ........ => 10 + 3 + 1 + 0 + ...... => 14

Yes, there is a formula for calculating the number of powers of a prime number, say p, in the factorial of a number, say n. The formula is: n/p + n/p^2 + n/p^3 .......and so on till n > p^m

Eg: How many powers of 2 exist in 54!? => 54/2 + 54/4 + 54/8 + 54/16 + 54/32 + 54/64 => 27 + 13 + 6 + 3 + 1 + 0 => 50

samrus98 - I think you need to start gmat quant tutoring!

Looks like all the work for IIM CATs will come in handy here! PS: Do you have a list of formulas just like your SC notes.

Much appreciated _________________

If you have made mistakes, there is always another chance for you. You may have a fresh start any moment you choose, for this thing we call "failure" is not the falling down, but the staying down.

Thanks samrus98 for the formula.but if the no is small then i will prefer to do it this way. 30!=1*2*3*4*5*3*2*7*8*3*3*10*11*3*4*13*14*3*5*16*17*3*3*2*19*20*3*7*22*23*3*4*25*26*3*3*3*28*29*3*10

so no of 3=14

I think this method is useful when no is not prime.

Thanks samrus98 for the formula.but if the no is small then i will prefer to do it this way. 30!=1*2*3*4*5*3*2*7*8*3*3*10*11*3*4*13*14*3*5*16*17*3*3*2*19*20*3*7*22*23*3*4*25*26*3*3*3*28*29*3*10

so no of 3=14

I think this method is useful when no is not prime.

The formula is applicable even when the number is not prime, but with a small change in the methodology.

Eg. How many powers of 36 are there in 30!? 36 = 2^2 * 3^2

Now we know that powers of 2 would definitely be more than powers of 3, lets calculate. Powers of 2 in 30! => 30/2 + 30/4 + 30/8 + 30/16 + 32/16 => 15 + 7 + 3 + 1 + 0 => 26

For 36 we need a pair of 2 and 3 each and we have 7 pairs of of 3 and 13 pairs of 2. Thus we have min(7, 13) = 7 powers of 36 in 30!

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