(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