Answer is 47. I found the answer by trial and error such that a = 2, b =3, c = 42. can someone help me learn a quick solution.
my thought:
7bc + 7ac + 7ab = 6 abc
assume a <= b <= c, then 21bc >= 6 abc; meaning a<=3
if a= 1, 7bc + 7c + 7b = 6bc (since a,b,c have to be positive, this cannot be true)
if a =2, 7bc +14c + 14b = 12 bc
b + c = 5/14(bc)
from here, i know that bc has to be the multiple of 14 which means bc can be 14,28,42.....126 .... using a long trial and error method, i deduce that a =2, b =3 and c =42.
I know that there has to be a faster way to solve this. Can someone help! thanks!