219999...9978*4=879999...9912, making A=2, B=1, Q=7, W=8, X=9.
SOLUTION DETAILS
Let's find X first by taking an X within the row of Xs. 4X+c=10c+X, where c is the carryover in the result of multiplying the following X by 4. 3X=9c, X=3c, and c must be between 0 and 3. Therefore X=0, 3, 6, or 9.
Now look at W. 4W=10p+A, where p is a carryover. And 4A+q=W where q is a carryover. Both p and q are less than 4. Multiply the last equation by 4: 16A+4q=4W=10p+A, so 15A=10p-4q. Therefore, we need to find values of p and q that are multiples of 15.
| |
0 |
4 |
8 |
12 |
| 0 |
0 |
-4 |
-8 |
-12 |
| 10 |
10 |
6 |
2 |
-2 |
| 20 |
20 |
16 |
12 |
8 |
| 30 |
30 |
26 |
22 |
18 |
The first column contains 10p values and the first row 4q values. The other cells are the result of subtracting the column header from the row header. Only two cells are divisible by 15: 0 and 30, implying (p,q)=(0,0) or (3,0). We can reject (0,0) because it's likely to lead to the trivial solution where the numbers are zero.
So we go for p=3, q=0, making A=2. This tells us that 4W=32 and W=8. And this means that there is no carryover for 4B, therefore B must be less than 3. We can also see that 4Q+3=10c+B, because 4W=32 making the carryover 3. We know B can only be 0, 1 or 2. So 4Q=10c-3, 10c-2, or 10c-1. This has to be divisible by 4, so we can eliminate 10c-3 and 10c-1 which are odd numbers. So B=1. That leaves 4Q=10c-2, and Q=2, c=1 or Q=7, c=3. X is therefore 3 or 9. To recap, we are left with 2133...328*4?=?8233...312 or 2199...978*4?=?8799...912. Only the latter satisfies the arithmetic.