\(\begin{array} {lr} & ABCD \\ \times & 9 \\ \hline & DCBA \end{array}\)

Find the four-digit number ABCD.

Guest May 24, 2020

#1**0 **

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

_{.}

Guest May 24, 2020