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# factor of p

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14 Aug 2009, 03:20
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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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Last edited by snipertrader on 14 Aug 2009, 04:44, edited 1 time in total.
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14 Aug 2009, 03:33
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?
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14 Aug 2009, 03:59
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
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14 Aug 2009, 04:09
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?

Hi Nikhil

The link that you were referring to is here: factor-82137.html
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14 Aug 2009, 04:13
Quote:
i think the question should be 3^k

In that case, we can say we have 10 multiples of 3 ie 10
One more 3 in 9(3*3)
Two more 3 in 27(3*3*3)

Therefore, 3^13 ie k = 13. But thats not there in the options. What I am missing here?
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14 Aug 2009, 04:14
yezz 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

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
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14 Aug 2009, 04:43
My mistake - it should be 3^k
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Senior Manager
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14 Aug 2009, 04:46
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?

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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.

Senior Manager
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14 Aug 2009, 04:55
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

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14 Aug 2009, 05:29
Ya, even i didn't follow samru's explanation of 30/3+30/(3^2) etc, but here is another way to solve it.

If you separate the multiples of 3 from 1 to 30, you get: 3,6,9,12,15,18,21,24,27,30.

Now, try to write these numbers in terms of the lowest prime factors i.e. 2, 3 or 5 and you get

3 * (2*3) * (3*3) * (2*2*3) * (3*5)*(2*3*3)*(3*7)*(2*2*2*3)*(3*3*3)*(2*3*5)

and count the powers of 3, you get 3^14*2^..*5^...

Hope this helps....
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14 Aug 2009, 05:46
i think the answer is 14
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14 Aug 2009, 05:52
samrus98 wrote:
yezz 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

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

yes samrus, u r defenetly right...my bad
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14 Aug 2009, 07:51
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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

I hope this helps!!
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14 Aug 2009, 18:34

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
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15 Aug 2009, 00:37
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.
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15 Aug 2009, 01:42
hrish88 wrote:
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 check for powers of 2 and 3 separately.

Powers of 3 in 30! => 30/3 + 30/(3^2) + 30/(3^3) + 30/(3^4)
=> 10 + 3 + 1 + 0
=> 14

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!

Hope this helps!
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15 Aug 2009, 01:50
This does help.thanks
Re: factor of p   [#permalink] 15 Aug 2009, 01:50
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