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Joined: 15 Sep 2003
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If x, y, and z are positive integers and 3x=4y=7z then the [#permalink]
 16 Nov 2003, 18:24
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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Manager
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correct dj...
how did you approach this problem??? i just did guess and check, but it took me forever...
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VP
Joined: 13 Nov 2003
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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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Senior Manager
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Would you mind to go in further detail working
_________________
shubhangi
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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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VP
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thanks for writing it down, stoylar.
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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.
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