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OG12 #110

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OG12 #110 [#permalink] New post 01 Jul 2009, 10:11
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 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

Please show work. Are there any specific formulas to multiply consecutive integers? Thanks.

[Reveal] Spoiler:
OE is C
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Re: OG12 #110 [#permalink] New post 01 Jul 2009, 10:19
Anyone who answers this question, kindly let me know the appropriate book to handle these number games.

I am totally ignorant of adding, multiplying big sets of numbers!!!;(

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Re: OG12 #110 [#permalink] New post 01 Jul 2009, 10:22
This is a typical question, and has a specific way of solving

The question is asking - how many powers of 3 are there in !30

!30 is product of all integers from 1 to 30

Keep dividing 30 by 3 untill you get a zero in the quotient. Then add up all the quotients. Their sum is your answer.

30/3 = 10
10/3 = 3 (dont need to consider the remainders here)
3/3 = 1

10 + 3 + 1 = 14

so there are 14 powers of 3 in !30

Logic -

!30 will have as many multiples of 3 as are there from 1 to 30 (so divide 30 by 3)
!30 will also have as many multiples of 3^2 (ie multiples of 9) as there are from 1 to 30 , but one of those 3s in 9 already gets eliminated in step 1 above. So we divide by only 3 again to get multiples of 9
!30 will also have multiples of 3^3 (27), out of which 3^2 multiples have already been eliminated. Now you need to eliminate only the 3rd power of 3 from the remaining multiples of 27, so again divide by 3.

The quotients of all the above divisions give you the multiples of 3, 9 and 27 in terms of powers of 3. Add them up to get your answer.
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Re: OG12 #110 [#permalink] New post 01 Jul 2009, 10:25
There is no such formula. But there are tricks for this problem.
It is a multiplication problem.
1x2x3x4x5x6........x27x28x29x30
I wrote factors that are divisible by three in bold.
There are 10 numbers in this sequence. That are divisible by 3.
These are 3, 6, 9, ..., 27,30
1 of them can be divided to 3, three times: 27 (3x3x3)
2 of them can be divided to 3, two times : 9, 18 (3x3;2x3x3)
Remaining seven can be divided to 3 only once.
So this multiplication becomes like this.
1x2x3x4x5x(2x3)x7x.......26x(3x3x3)x28x29x(10x3)
So this becomes
1x2x4x5x...28x29x3^{14}xsomething
So the biggest k is 14 ;)
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Re: OG12 #110 [#permalink] New post 01 Jul 2009, 10:35
Hussain15 wrote:
Anyone who answers this question, kindly let me know the appropriate book to handle these number games.

I am totally ignorant of adding, multiplying big sets of numbers!!!;(

Posted from my mobile device Image


You don't need to do lengthy calculations. Gmat doesnt allow calculators, so it will never give you questions involving lengthy calculations. It would be helpful though if you knew reciprocals upto 30 and squares upto 25. Makes calculations easier, though its not very important.
Almost all questions are concept based, so try to practice a variety of questions and always look for the logic behind them, you will be able to handle them better and easier. I learnt this concept from a book on basic mathematics by the TIME institute in India. But really you don't need specific books, just look for the concepts.
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Re: OG12 #110 [#permalink] New post 01 Jul 2009, 11:05
I just listed all of the multiples of 3 up to 30 and counted up all of the 3's that constituted each multiple. Took me 2.5 minutes though, maybe another approach is better
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Re: OG12 #110 [#permalink] New post 02 Jul 2009, 04:01
maliyeci wrote:
There is no such formula. But there are tricks for this problem.
It is a multiplication problem.
1x2x3x4x5x6........x27x28x29x30
I wrote factors that are divisible by three in bold.
There are 10 numbers in this sequence. That are divisible by 3.
These are 3, 6, 9, ..., 27,30
1 of them can be divided to 3, three times: 27 (3x3x3)
2 of them can be divided to 3, two times : 9, 18 (3x3;2x3x3)
Remaining seven can be divided to 3 only once.
So this multiplication becomes like this.
1x2x3x4x5x(2x3)x7x.......26x(3x3x3)x28x29x(10x3)
So this becomes
1x2x4x5x...28x29x3^{14}xsomething
So the biggest k is 14 ;)


This definitely works. Is it the quickest way?
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Re: OG12 #110 [#permalink] New post 02 Jul 2009, 09:27
Quickest way is dividing method. But it shows the logic.
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Re: OG12 #110 [#permalink] New post 08 Jul 2009, 02:34
Questions asks how many powers of 3 are there in 30! (1*2*3..................*30)

Look at the answer choices
A 10
B 12
C 14
D 16
E 18

1.0 We know that its definitely greater than 10
2.0 It is more than 12 , because be have 9 , 18, 27,
3.0 It is less than 15 , why? we have only 3 nines (9*1,9*2 and 9*3)
4.0 so it must be 14

II method : We know that all multiples of 3 ( but not multiples of 9) contribute a 3 to the multiple so there are 7 such numbers so we have 7 3's
Multiples of 9 contribute 2 3's and 27 will contribute 3 3's so we have 7 3's

together we have 7+7 = 14 , 3's
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Re: OG12 #110 [#permalink] New post 08 Jul 2009, 04:19
Go with option C.

We have to find the 3 from Integers 1 to 30 (inclusive), therefore, 3 x (1 to 10)..
this makes 3 ten time..and from 1 to 10, there are 3 integers with 3, those are 3, 6 and 9= 4 more 3's.
so value of K is 14.
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Re: OG12 #110 [#permalink] New post 08 Jul 2009, 10:37
As noted before you are only looking for numbers that are divisible by 3.

I would skip making the complete multiplication and simply make a list of these numbers. You just need to count these numbers ensuring you remember that 9,18 and 27 contain more than one 3 when factoring.

3,6,9, - "9" counts as 2 (3x3) = 4
12,15,18, "18" counts for 2 (3x3x2) = 4
21,24,27, "27: counts as 3 (3x3x3) = 5
30 = 1
4+4+5+1 = 14 - C!
Re: OG12 #110   [#permalink] 08 Jul 2009, 10:37
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