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

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Current Student
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if x, y, and z are positive integers and 3x=4y=7z, then the [#permalink]

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29 Oct 2007, 19:04
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if x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is:

33
40
49
61
84
VP
Joined: 28 Mar 2006
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29 Oct 2007, 19:09
young_gun wrote:
if x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is:

33
40
49
61
84

3x=4y=7z = k (say)

x=k/3 y=k/4 z=k/7

I think it is 84
Current Student
Joined: 31 Aug 2007
Posts: 369
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Kudos [?]: 126 [0], given: 1

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29 Oct 2007, 19:17
I thought so too...that is not the OA, however...
Director
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29 Oct 2007, 19:24
trivikram wrote:
young_gun wrote:
if x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is:

33
40
49
61
84

3x=4y=7z = k (say)

x=k/3 y=k/4 z=k/7

I think it is 84

Should be 61.

x+y+z => x + 3x/4 +3x/7 => x* 61/28
For this to be integer , and for sum to minimum x should be 28.
So sum is 61.
Director
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29 Oct 2007, 19:43
young_gun wrote:
if x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is:

33
40
49
61
84

Yep, D it is.

3x=4y=7z

since 3 is not divisible by either 4 or 7, X must be divisible by both 4 and 7, thus

X min=4*7

the same logic for Y and Z:

Y = 3*7, Z = 3*4

X+Y+Z=28+21+12=61.
Re: PS   [#permalink] 29 Oct 2007, 19:43
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