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by madsport » Mon Sep 15, 2008 11:39 am
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

The answer is D

I have no clue how to solve this one
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by alescau » Mon Sep 15, 2008 11:58 am
since the coefficients are prime numbers, nothing simplifies.
to make the equality stand, x=4*7, y=3*7, z=3*4. add them up and get 61.
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by madsport » Mon Sep 15, 2008 1:33 pm
how did you come up with this arrangement --->x=4*7, y=3*7, z=3*4?
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by parallel_chase » Mon Sep 15, 2008 1:39 pm
madsport wrote:how did you come up with this arrangement --->x=4*7, y=3*7, z=3*4?
3x=4y=7z

LCM = 84

x = 28
y = 21
z = 12

28+21+12 = 61
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by 4meonly » Tue Sep 16, 2008 1:16 am
3x=4y=7z

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

x + y + z = x + 3x/4 + 3x/7 = 61x/28
to make 61x/28 an integer minimum value of x is 28, because 61 is prime number
so 28 + 30*28/4 + 3*28/7 = 61

Answer D
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