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If p is the product of the integers from 1 to 30, inclusive, wha

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If p is the product of the integers from 1 to 30, inclusive, wha [#permalink] New post 11 Jul 2010, 01:36
okay, here is the thing !

I was solving 12 O.G book then this question pop up and I had no shortcut for it ! I mean it took me long to do it !

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

I mean if I keep multiplaying each value in order to hit the right one, it will take a lot of time where in GMAT you must spend 1 minute for each a question, is there any shortcut for this type of question ?

thanks
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Re: Can I please know the simple method to solve this problem ? [#permalink] New post 21 Jul 2010, 12:47
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sstudy wrote:
Can anybody please tell me how to solve this one ?

Que :

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

Thanks,
sstudy


I think it should be 3^k not 3k else the highest number i.e. 18 in the option should be the answer
Now lets take 3^k

power 3 in 30! is 30/3 + 30/3^2 ..... till 3^k < 30

=> power of 3 in 30! = 30/3 + 30/3^2 + 30/27 = 10 + 3 + 1 = 14 ( ignore decimals think logically why)

Hence the answer is 14

The general formula is

\frac{n}{k}+ \frac{n}{k^2} +\frac{n}{k^3} till n>k^m

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Re: Can I please know the simple method to solve this problem ? [#permalink] New post 21 Jul 2010, 12:49
Thank you very much for your immediate reply and solution..
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Exponents, factors question [#permalink] New post 04 Aug 2010, 03:10
Can anyone demonstrate the following?

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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Factors, exponents problem [#permalink] New post 07 Aug 2010, 20:18
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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Re: Factors, exponents problem [#permalink] New post 07 Aug 2010, 20:47
p = 1x2x3x4x5x6x7x8x9x10x ....x30


There is a 3 after every 2 numbers. Hence total nos of 3s = 10

Hence K = 10?

Think should be A?
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Re: OG12 - PS [#permalink] New post 08 Aug 2010, 01:45
thks. think i forgot to count the double threes and the triple threes.
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Intergers [#permalink] New post 12 Oct 2010, 03:27
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)20

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Re: Intergers [#permalink] New post 12 Oct 2010, 03:30
This question should be under Quant. Moderators please move it.

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Re: Intergers [#permalink] New post 12 Oct 2010, 06:07
:-O come on, it's CR forum buddy
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Re: Intergers [#permalink] New post 12 Oct 2010, 06:15
anyway let's solve it:
p: 3*6*9*12*15*18*21*24*27*30

P: (3^1)(3*2)(3^2)(3*5)(3^2*2)(3*7)(3*8)(3^3)(3*10)
add up the powers
14
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Product of the Integers [#permalink] New post 12 Oct 2010, 22:57
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)20

I need some great solution to solve this problem quickly.

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Sentence Correction: sentence-correction-strategies-and-notes-91218.html

Arithmatic & Algebra: arithmatic-algebra-93678.html

Helpful Geometry formula sheet: best-geometry-93676.html


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Re: Product of the Integers [#permalink] New post 12 Oct 2010, 23:17
monirjewel wrote:
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)20

I need some great solution to solve this problem quickly.


The answer
(30/3)+(30/9)+(30/27)=10+3+1=14
Answer - (c)

The logic
You need to count the factors of 3 in the product.
So just count all the numbers divisible by 3, which will be (30/3) {3,6,9,...,30}
Now all the numbers divisible by 3^2 or 9 will have 2 factors of 3 so need to count them once more, this will be (30/9) {9,18,27}
And then finally the numbers divisible by 3^3, need to be counted once more, which is (30/27) {27}

No higher powers of 3 will divide numbers <=30, so we are done

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Re: Product of the Integers   [#permalink] 12 Oct 2010, 23:17
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