How to solve this mathematic puzzle with a program?

Question

Find 2 numbers:
Number A: 3 digits. The last digit is 9.
Number B: 3 digits. The second digit is 6.

A + B = C

C is a 4 digits number, and the second digit is 0.

• • 9
+ • 6 •
-------------
= • 0 • •

All digits 0 to 9 must be placed, and only once time.

Thanks

Answer 1

There are two possible solutions to this problem:
Option 1: A = 289 and B = 764
Option 2: A = 788 and B = 264
In both cases the sum is 1053

Answer 2

Seen like this. (As idk what language will you use)

for (int A = 100; A < 1000; A++)
for (int B = 100; B < 1000; B++) {
if (A % 10 == 9 && (B/10)%10 == 6 && (A+B)/100%10 == 0) print("%d, %d" A, B);
}

(And additional loops to check for 1 to 9

Answer 3

You know that 'a' is a three digit number that ends in a 9, so you can use for(int a=109; a < 1000; a+= 10) in your loop.

For 'b' you could do something like:

for(int i=100; i<1000; i+=100) {
for(int j=0; j<10; j++) {
b = i+60+j;
}
}

(there might be a better way, not sure).

Seems complicated, but spares you thousands of unnecessary comparisons and calculations.

Answer 4

Use loops. check each the number until found the condition you want.

Answer 5

You need to put digits 1234578 in the right order onto the points. So generate all the permutations (7! Altogether), place it one by one, then check the addition.