# Finding two numbers that produce equal to the sum of rest in JavaScript

## Suppose following is the problem:

We have a sequence of numbers starting from 1 and upto any arbitrary number, let's call it num. We have to pick two such numbers from the sequence (let's call them m and n), such that:

sum(1 to num) - (m + n) = m * n

And finally, we should return an array of groups of all such numbers.

For example −

If the input is −

const num = 10;

Then the output should be −

const output = [
[7, 6]
];

because sum(1 to 10) = 55

and,

55 - (6 + 7) = 6 * 7 = 42

## Example

The code for this will be −

Live Demo

const num = 10;
const pickNumbers = num => {
const sum = (num) * (num + 1) * (.5);
const results = [];
for (let n = 1; n <= num; n++) {
let first = sum - n;
let second = n + 1;
if (first % second === 0) {
let m = first / second;
if (m < num && m !== n && results.every(group => group + group
!== m + n)){
results.push([m, n]);
}
}
}
return results;
}
console.log(pickNumbers(10));

## Output

And the output in the console will be −

[
[7, 6]
]