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# Find two numbers whose sum and GCD are given in C++

We have the sum and gcd of two numbers a and b. We have to find both numbers a and b. If that is not possible, return -1. Suppose the sum is 6 and gcd is 2, then the numbers are 4 and 2.

The approach is like, as the GCD is given, then it is known that the numbers will be multiples of it. Now there following steps

If we choose the first number as GCD, then the second one will be sum − GCD

If the sum of the numbers is chosen in the previous step is the same as the sum, then print both numbers.

Otherwise print -1, as a number does not exist.

## Example

#include <iostream> #include <algorithm> using namespace std; void printTwoNumbers(int s, int g) { if (__gcd(g, s - g) == g && s != g) cout << "first number = " << min(g, s - g) << "\nsecond number = " << s - min(g, s - g) << endl; else cout << -1 << endl; } int main() { int sum = 6; int gcd = 2; printTwoNumbers(sum, gcd); }

## Output

first number = 2 second number = 4

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