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

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

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New post 20 Jun 2010, 23:52
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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
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Re: If x is the product of the positive integers from 1 to 8, in [#permalink]

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New post 21 Jun 2010, 02:15
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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.

Please help :(


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 \(p=1\) --> \(i+k+m+p=7+2+1+1=11\).

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

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New post 30 Nov 2014, 17:54
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

Correct answer D


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

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New post 30 Nov 2014, 22:59
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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

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

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New post 03 Dec 2014, 08:27
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.

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

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New post 16 Jan 2018, 07:59
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
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: If x is the product of the positive integers from 1 to 8, in   [#permalink] 16 Jan 2018, 07:59
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