Let the two numbers be represented thus ABCD×4=EFGH.
The table below focusses on A and H. A can only be 1 or 2, otherwise the product would exceed 4 digits, so there are only two rows for A. The columns show the even number candidates for H and the two rows. The body of the table shows the only 10 possible values for D, and each has a label 1-10 in parentheses.
| A⬇️ H➡️ |
2 |
4 |
6 |
8 |
| 1 |
3 (1)
8 (2) |
6 (3) |
4 (4) |
2 (5)
7 (6) |
| 2 |
❌ |
1 (7)
6 (8) |
4 (9) |
7 (10) |
Using these references as starting points we can eventually deduce what the remaining letters can represent. All possible solutions can be determined.
(A,D,H)=
- (1,3,2): E=4⇒B=1 or 2 (already used)❌; E=5 and 4C+1=4{4,6,7,8}+1={17,25,29,33}⇒C=4, but the remaining digits 6 and 8 generate 1643×4=6572 and 1843×4=7372❌; E=6 and 4C+1=4{4,5,7,8}+1⇒C=4, but the remaining digits 5 and 8 generate 1543×4=6172 and 7372❌; E=7 and 4C+1={4,5,6,8}+1={17,21,25,33}⇒C=6, and the remaining digits 4 and 8 generate 1463×4=5852❌ and 1863×4=7452✔️
- (1,8,2): E≠4 (see 1); E=5 and 4C+3=4{3,4,6,7}+3={15,19,27,31}⇒C=6, and the remaining digits 3 and 4 generate 1368×4=5472✔️ (note that these are rearrangements of the digits of the earlier solution (see 1) and 1468×4=5872❌
- (1,6,4): E=5 and 4C+2=4{2,3,7,8}+2={10,14,30,34}❌; E=7 and 4C+2=4{2,3,5,8}+2={10,14,22,34}⇒C=5, leaving digits 3 and 8 and products 5424 and 7424❌
- (1,4,6): E=5 and 4C+1=4{2,3,7,8}+1={9,13,29,33}⇒C=8, leaving digits 2 and 7 and products with duplicated digits❌; E=7 and 4C+1=4{2,3,5,8}+1={9,13,21,33}⇒C=8, leaving digits 2 and 5 and products with duplicated digits❌
- (1,2,8) ❌ (use similar logic in 1-4)
- (1,7,8) ❌ (use similar logic in 1-4)
- (2,1,4)⇒E=8; 4C=4{3,5,6,7}={12,20,24,28}❌
- (2,6,4)⇒E=8; 4C+2=4{1,3,5,7}+2={6,14,22,30}❌
- (2,4,6)⇒E=8; 4C+1=4{1,3,5,7}+1={5,13,21,29}⇒C={1,5}⇒B>2 and E can't be 8 because of carryover❌
- (2,7,8)⇒E=8 (duplication)❌
Two solutions: 1863×4=7452, 1368×4=5472.
