reduction of complexity numbers procedure

I’m looking for a solution to my problem

if I have numbers

var first = 14:1
var next = 13:8

therefore, the console should give a result

console.log(first_result)  // 141
console.log(next_result)  // 141

and I want the numbers counted like 141 in the result to be

simply if there is an example

13:8 if both digits are the last, 3 and 8 are bigger than ten, then turn the tens and turn and insert into the previous number and leave the rest at the end

so 13:8 becomes 141

>Solution :

If you are starting with strings, then you just simply split the string on : to get your 2 numbers.

To get the last digit, you can simply use x % 10. Then just add the numbers and see what happens.

let value = '13:8',
    nums = value.split(':').map(Number),
    last_digits = nums.map(x => x % 10),
    // or last_digits.reduce((x,y) => x+y, 0)
    sum = last_digits[0] + last_digits[1],
    result;

if (sum > 10) {
    nums[0]++;
    nums[1] = sum - 10;  // or, probably sum % 10
}

result = nums.join('');
console.log(result);