Thread: How do you recreate a long multiplication problem using numbers from 0 to 9 each exactly twice?

There are 20 blank digits in a long multiplication problem. The actual problem is a 3 digit number multiplied by another 3 digit number. The product is a 5 digit number. Each digit is a number from 0 to 9. Each number from 0 to 9 appears twice.

To start, there are 900 x 900 = 810000 possible problems,
because the first digit of each factor cannot be 0, leaving 1-9.
Many of them are eliminated right away
because they will repeat digits in the two factors.

For example, you don't have to bother checking
111 x anything or any other 3 digit repeat.
Also if you have 123 x ABC, then you cannot have
more than one 1,2, or 3 in ABC.

If the problem is ABC x DEF, then F is not 1,
since then ABC repeats immediately, and the
last digit of the final product also C.

Neither factor can end in 0, since then the final
product ends in 0 too.

You could perhaps continue an analysis like that
eliminating great swaths of possible numbers.

Meanwhile, a computer search turns up
179 * 224 = 716 + 358_ + 358__ = 40096
which is the only solution.