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+ 5

# How to solve this mathematic puzzle with a program?

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

5 Answers

+ 6

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

+ 5

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

+ 4

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.

+ 2

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

0

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.