3 thoughts on “Factoring Complex Numbers- Open Middle Problem”

I thought of flogging it out, but even assuming integers 2 to 9 this gave something like 8!/3! combinations. Then I analysed the product (ac – bd) + (ad + bc)I for odd and even. This reduced the choices to about 36, and the result then fell out.
This is a HARD SUM ! Result 5 2 3 6

I thought of flogging it out, but even assuming integers 2 to 9 this gave something like 8!/3! combinations. Then I analysed the product (ac – bd) + (ad + bc)I for odd and even. This reduced the choices to about 36, and the result then fell out.

This is a HARD SUM ! Result 5 2 3 6

I am glad I could stretch your mind a little on this, but you are short 3 combinations.