Imperative style solution of the GCD problem

gcd.js

function gcd(numbers) {
    let results = {};
    
    for (let i = 0; i < numbers.length; i++) {
        let divisor = numbers[i];
        for (let j = 0; j < numbers.length; j++) {
            let number = numbers[j];
            if (number == divisor) continue;

            let result = number % divisor;
            if (result === 0) {
                if (results[divisor] === undefined) results[divisor] = 1;
                else results[divisor] += 1;
            }
        }
    }

    let greatest;
    let maxTimes = 0;
    for (let divisor in results) {
        let times = results[divisor];
        if (times > maxTimes) {
            greatest = divisor;
            maxTimes = times;
        }
    }

    return greatest;
}

function testGCD() {
    let nbs;
    let rst;

    console.log(`1st test >>`);
    nbs = [2, 5, 6, 8, 10];
    rst = gcd(nbs);
    console.log(`[${nbs}] == 2 [${rst == 2}]`);

    console.log(`\n2nd test >>`);
    nbs = [1, 2, 5, 8, 10, 12, 15];
    rst = gcd(nbs);
    console.log(`[${nbs}] == 1 [${rst == 1}]`);

    console.log(`\n3rd test >>`);
    nbs = [2, 3, 5, 6, 9, 10, 12, 15, 18];
    rst = gcd(nbs);
    console.log(`[${nbs}] == 3 [${rst == 3}]`);
}

testGCD();

Expected result of this code should be like...

$ node gcd.js
1st test >>
[2,5,6,8,10] == 2 [true]

2nd test >>
[1,2,5,8,10,12,15] == 1 [true]

3rd test >>
[2,3,5,6,9,10,12,15,18] == 3 [true]