Originally posted by joe shmo
Is there a way to solve
a + b + c = a*b*c
To arrive definitively at those results?
one equation with three unknowns is not definitely defined.
what you can do is to apply some logic on the digits:
* no "0" involved since the product involving zero is zero, but 000 is not a valid three digit number.
* not more than one digit may be 1 since 1*1*n=n but 1+1+n=n+2
from the first two: no digit bigger than 3 since the smallest including 4 is 124 with 1+2+4=7 and 1*2*4=8, the products increase faster than the sums.
so 123 and its permutations are left over 1+2+306 1+2*3=6, all valid solutions are:
123 132 213 231 312 321 as given above.