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

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

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16 Nov 2003, 17:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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16 Nov 2003, 18:53
correct dj...

how did you approach this problem??? i just did guess and check, but it took me forever...
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16 Nov 2003, 19:15
derive x, z in y's form. you will get the minimum.
you must confirm by deriving in X's and Z's forms, just to make sure.

with all X it is 64, with all Z's it is 81 (i think), with all Y's 61.
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16 Nov 2003, 22:15
Would you mind to go in further detail working
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shubhangi

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16 Nov 2003, 22:21
the least common multiple is 3*4*7
So, 3*4*7=3x=4y=7z, x=28,y=21, and z=12
28+21+12=61
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16 Nov 2003, 22:43
thanks for writing it down, stoylar.
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17 Nov 2003, 09:34
Say 3x=4y=7z=k (where k is a positive integer)
=>x=k/3,y=k/4,z=k/7

x+y+z= k/3+k/4+k/7= k61/84. Since x+y+z has to be a positive integer, least value of k61/84 will be when k=84 and the least value will be 61.
17 Nov 2003, 09:34
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# If x, y, and z are positive integers and 3x=4y=7z then the

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