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by sanju09 » Mon Apr 26, 2010 3:53 am
Let x, y, and z be three positive integers satisfying y = 3 x, z = 4 x, x + y + z = 3 k, k is an integer. Which of the following is the smallest value of k for which x, y, and z are all even?
(A) 8
(B) 10
(C) 12
(D) 16
(E) 24
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by harshavardhanc » Mon Apr 26, 2010 4:12 am
sanju09 wrote:Let x, y, and z be three positive integers satisfying y = 3 x, z = 4 x, x + y + z = 3 k, k is an integer. Which of the following is the smallest value of k for which x, y, and z are all even?
(A) 8
(B) 10
(C) 12
(D) 16
(E) 24
IMO D
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Harsha
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by Nilakshi1001 » Mon Apr 26, 2010 4:23 am
We will use the traditional substitution method. We will put the various values of k, and find out.

x+y+z=3k
=>x+3x+4x=3k
=>8x=3k

(a) k=8

8x=3k=3*8
x=3, rejected.

Similarly, we will do for b, c, d and e. Only for k=16, we get positive values. Thus the answer is D.
Last edited by Nilakshi1001 on Mon Apr 26, 2010 4:39 am, edited 1 time in total.
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by liferocks » Mon Apr 26, 2010 4:36 am
from the question we get
x+y+z=3k
or x+3x+4x=3k
or 8x=3k
now LCM of 8 and 3 is 24 so either side is equal to 24 or its multiple
if 8x=24 ,x=3 so not correct
if 8x=24*2, x=6

hence the lowest value of k for which x,y and z are integer is 24*2/3 or 16
Ans D
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