Interesting Math Problems

[QillerDaemon]

A+BC=2024=2A+C+1 or A+BC=2A+C+1
You can solve for B and get that B=2.

Substituting now that B=2, we get A+2C=2A+c+1. Or C=A+1.

Now, A+2(A+1) = 3A + 2 = 2024. A must be 674. So C must be 675.

A=674
C=675
B=2

It's a Diophantine equation set. It should not be solvable, three variables over just two equations. But the unstated part of the problem is showing the solutions must be non-zero, and in this case also integer solutions. Otherwise it does actually become unsolvable. Your answer, though, is correct!