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# If x, y, and z are positive integers and 3x=4y=7z, then the

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Intern
Joined: 26 Jul 2006
Posts: 5
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If x, y, and z are positive integers and 3x=4y=7z, then the [#permalink]  06 Sep 2006, 15:37
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!
VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 42 [0], given: 0

(D) 61

Given 3x = 4y = 7z

LCM(3,4,7) = 84
=> x = 28, y = 21, z = 12

Sum = 61.
VP
Joined: 29 Dec 2005
Posts: 1349
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Kudos [?]: 32 [0], given: 0

Re: very interesting #'s properties q [#permalink]  06 Sep 2006, 17:10
chrismeiyu wrote:
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!

since 3, 4 and 7 has no common factors, the only way to have 3x=4y=7z is 3 (4x7) = 4 (3x7) = 7 (3x4). so x = 28, y= 21, and z = 12.

therefore x + y + z = 28+21+12 = 61.
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Joined: 11 May 2006
Posts: 260
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Kudos [?]: 6 [0], given: 0

Re: very interesting #'s properties q [#permalink]  06 Sep 2006, 21:33
chrismeiyu wrote:
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!

you are almost there
now x + y + z will be a +ve integer since x,y, and z each are +ve integers.

so (61/28)x will be a integer => x is a multiple of 28.
now (61/28)x will be minimum when x = 28
hence x+y+z = 61
Re: very interesting #'s properties q   [#permalink] 06 Sep 2006, 21:33
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