(I) ABCD - EABF = GDGA : - - (II) HII + EGKD = EFCK ---------------------- (III) IH * EDH = HABF (IV) (V) (VI) Solution: In (IV) we have two numbers having only digits H and I. If we can identify these the puzzle is done. (III) I * E <= H Therefore 0 < I < H and 0 < E < H. Now I cannot be 1 otherwise D=H in (IV). So I >= 2. As IH * HII < 10000 we have only one possible combinations left as (I,H)=(2,4) is too large. I H IH HII IH*HII 2 3 23 322 7406 2 4 24 422 10128 The rest is easily filled in. 7406 - 1749 = 5657 : - - 322 + 1586 = 1908 ---------------------- 23 * 163 = 3749 References: Hoer Zu, No 51, 2003-12-12