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If 3^k is a divisor of the product of all even integers between 2

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If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 07 Feb 2016, 18:08
5
13
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A
B
C
D
E

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  45% (medium)

Question Stats:

65% (01:49) correct 35% (01:43) wrong based on 229 sessions

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If \(3^k\) is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?

A. 4
B. 6
C. 10
D. 13
E. 14
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Re: If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 07 Feb 2016, 18:32
4
1
TeamGMATIFY wrote:
If \(3^k\) is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?

A. 4
B. 6
C. 10
D. 13
E. 14


Hi,
since the Q tells us \(3^k\) is a divisor of the product of all even integers between 2 and 30 (inclusive)
the Q basically asks us power of 3 in the product..

lets see the even multiples of 3 from 2 to 30..
6*12*18*24*30..
3^5(2*4*6*8*10)..
or 3^6*2*4*2*8*10..
ans 6
B
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Re: If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 17 Feb 2016, 16:39
5
6
The way i went about it was as follows

Product of even integers between 2 and 30 are listed below:-
=2*4*6*8.........28*30 -this can also be written as 2*1 * 2*2 * 2*3.....2*14 *2*15
2^15 *(1*2*3*4.....14*15) => 2^15* 15!

3^k would not be a divisor of 2^15, so we should find out how many times 3^k would go into 15!

15/ 3 + 15/3^2 ---> 5 +1 = 6
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Re: If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 17 Feb 2016, 16:42
4
we can simplify it

Product of even integers between 2 and 30 are listed below:-
=2*4*6*8.........28*30 -this can also be written as 2*1 * 2*2 * 2*3.....2*14 *2*15
2^15 *(1*2*3*4.....14*15) => 2^15* 15!

3^k would not be a divisor of 2^15, so we should find out how many times 3^k would go into 15!

15/ 3 + 15/3^2 ---> 5 +1 = 6
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Re: If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 23 Mar 2017, 08:25
2
TeamGMATIFY wrote:
If \(3^k\) is a divisor of the product of all even integers between 2 and 30 (inclusive), what is the maximum value of k?

A. 4
B. 6
C. 10
D. 13
E. 14


Set of all even integers between 2 and 30 (inclusive) = { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 22 , 24 , 26 , 28 , 30 }

Or, Set of all even integers between 2 and 30 (inclusive) = 2 ( 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 )

Or, Set of all even integers between 2 and 30 (inclusive) = 2*15!

Now, check the highest power of \(3^k\) in 2*15!

Highest power of \(3^k\) in 2*15! is = \(\frac{15}{3} => 5\)

Or, \(\frac{5}{3} = 1\)

So, the maximum value of k in \(3^k\) is 6 ( ie, 5 + 1 )

Thus, answer must be (B) 6
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Re: If 3^k is a divisor of the product of all even integers between 2  [#permalink]

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New post 01 Dec 2018, 11:12
3^k is a divisor of the product of all even integers between 2 and 30
Looking for any even integers that are divisible by 3 from 2 -> 30:
6 ; 18 ; 24 ; 30
Factoring these number to product of primes
6 = 2*3
18 = 2*3*3
24=2*2*2*3
30= 2*3*5
It looks like there are 5 "3" that can go into 3^k. But we also have an option of 0.
So there are 6 options for 3 that will go to 3^k.
=> k = 6 => B
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Re: If 3^k is a divisor of the product of all even integers between 2   [#permalink] 01 Dec 2018, 11:12
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