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

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Answer 1 · 2018-02-01T22:33:32

The minimum possible value of #x+y+z# is #61#

Note that any two of #3#, #4# and #7# have no common factor greater than #1#.

Hence the smallest positive integer values of #x#, #y#, #z# correspond to:

#3x=4y=7z=3 * 4 * 7 = 84#

...yielding:

#{ (x=28), (y=21), (z=12) :}#

So:

#x+y+z=28+21=12 = 61#