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

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

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New post 29 Oct 2007, 20:04
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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
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Re: PS [#permalink]

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New post 29 Oct 2007, 20:09
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
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 [#permalink]

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New post 29 Oct 2007, 20:17
I thought so too...that is not the OA, however...
Director
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Re: PS [#permalink]

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New post 29 Oct 2007, 20:24
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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Re: PS [#permalink]

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New post 29 Oct 2007, 20:43
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, 20:43
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if x, y, and z are positive integers and 3x=4y=7z, then the

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