Solve this alphametic, where each of the capital letters in bold denotes a different decimal digit from 0 to 9. None of the numbers can contain any leading zero.
3√(HOW)+ 3√(AND) = 3√(WHEN)
(In reply to
re: Integrity? (interesting property) by Daniel)
Hey, Daniel! We seem to have radically different understandings of this problem. By "integrity" I was just pointing out that the cube roots of the three terms were NOT integers (my first go at the problem was to see if there were a solution where the roots were integers, to obviate questions of the precision required for the test sum). By substituting (x,y,z,) for (HOW,AND,WHEN) you do not seem to be approaching "this alphametic" as such. We are explicitly told that each capital letter stands for a decimal digit, so (e.g.) HOW = 100*H + 10*O+W ,u.s.w., and that the assignments to each letter must be consistent throughout (e.g. "H" is the same digit in HOW as in WHEN).
I approached it via brute force generation (seven loops, etc.), but there may be some method via factoring the three numbers. In my solution, (192 = 3*2**6), (375 = 3*5**3) and (2187 = 3**7), though that doesn't immediately suggest to me an approach.
Perhaps the proposer (K Sengupta) can explicate.
re(2): Integrity? (interesting property)
