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# If x is the product of the positive integers from 1 to 8, in

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Intern
Joined: 29 Nov 2009
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If x is the product of the positive integers from 1 to 8, in  [#permalink]

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20 Jun 2010, 22:52
2
3
00:00

Difficulty:

15% (low)

Question Stats:

78% (01:05) correct 22% (01:07) wrong based on 209 sessions

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If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i*3^k*5^m*7^p$$, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-x-is-the- ... 46157.html
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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21 Jun 2010, 01:15
2
divakarbio7 wrote:
If x is the product of the positive integers from 1 to 8,
inclusive, and if i, k, m, and p are positive integers
such that x = 2i3k5m7p, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

I am not understanding the reasoning behind this problem.

It should be "if $$x=2^i*3^k*5^m*7^p$$".

$$x=8!=2*3*4*5*6*7*8=2*3*(2^2)*5*(2*3)*7*(2^3)=2^7*3^2*5^1*7^1=2^i*3^k*5^m*7^p$$ --> $$i=7$$, $$k=2$$, $$m=1$$, and $$p=1$$ --> $$i+k+m+p=7+2+1+1=11$$.

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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30 Nov 2014, 16:54
Here the idea is to go expressing each factor from one to eight, decomposed into its primes:

1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 = 1 * 2 * 3 *(2*2) * 5 +(2*3) * 7 *(2*2*2) = 2 ^ 7 * 3 ^ 2 * 5 ^ 1 * 7 ^ 1

Then

i = 7

k = 2

m = 1

p = 1

i + k + m + p = 11

tutoring courses GMAT part math homeworkers in Chile hurtado claudio gmatchile@yahoo.com http://www.gmatchile.cl
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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30 Nov 2014, 21:59
1
Incidentally 2,3,5,7 are prime factors of x = 8!

$$x = 2^i * 3^k * 5^m * 7^p$$ , we require to find (i + k + m + p)

There is no need to find individual values of i,k,m,p

Just write all prime numbers in there powers & add

$$x = 2^1 * 3^1 * (2^2) * 5^1 * (2^1 3^1) * 7^1 * (2^3)$$

i+k+m+p = 1+1+2+1+1+1+1+3 = 11

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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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03 Dec 2014, 07:27
Following the logic of
PareshGmat

There would be no need to write products between cousins, it is enough to write directly i + k + m + p = 11, that if you can do it in the CAT, obviously, but for people who can't do it directly would require the previous steps, to have crystal clarity and understand how to deal with similar situations, on the understanding that this is the purpose of some learning.

tutoring courses GMAT part math homeworkers in Chile hurtado claudio gmatchile@yahoo.com http://www.gmatchile.cl

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Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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16 Jan 2018, 06:59
divakarbio7 wrote:
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that $$x = 2^i*3^k*5^m*7^p$$, then i + k + m + p =

(A) 4
(B) 7
(C) 8
(D) 11
(E) 12

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-x-is-the- ... 46157.html
_________________
Re: If x is the product of the positive integers from 1 to 8, in &nbs [#permalink] 16 Jan 2018, 06:59
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