\(\begin{array} {lr} & ABCD \\ \times & 9 \\ \hline & DCBA \end{array}\)
Find the four-digit number ABCD.
A has to be 1 because anything larger would put five digits in the product.
The 1 at the end of the product means that D is 9.
B has to be 0 because anything else would carry over to the A and put five digits in the product.
Since B is 0 and you carried 8 over from the D, C has to be 8.
1 0 8 9
x 9
———–
9 8 0 1
A = 1
B = 0
C = 8
D = 9
I think that works
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