challenge for programmable calculators

[Albert Chan]

(12-24-2013 05:43 PM)cruff Wrote:  \[abc-1 = \frac{9(11a+b)}{a+b+c}\]

I don't follow how above (from post#32) can conclude *only* a + b + c = 9

(abc - 1)(a + b + c) ≡ 0 (mod 9)

→ (a + b + c) ≡ 0 (mod 9)

If a+b+c=18, min(abc*(a+b+c)) = (9*8*1)(9+8+1) = 1296 > 999, thus a+b+c = 9

7 combinations, 2 solutions: 117, 126, 135, 144, 225, 234, 333

But, more checks are needed:

→ abc ≡ 1 (mod 9)
9 combinations, 0 solution: 125, 147, 188, 227, 248, 444, 455, 578, 777

→ abc ≡ 1 (mod 3, but not mod 9) and (a + b + c) = (6 or 12 or 15)
3 combinations, 0 solution: 114, 177, 447