Discussion on: Daily Challenge #94 - Last Digit

[erezwanderman]

Javascript:

  const last_digit = ([...a]) => {
    while (a.length > 1) {
      [x, y] = a.splice(-2);
      a.push(((x % 10) ** ((y - 1) % 4 + 1)) % 10);
    }
    return a[0] || 1;
  }

[Giang Vincent]

can we have some explain please ?

[erezwanderman]

Yes.
Let's look at a spreadsheet of x^y % 10.

For x, the values repeat every 10, that is, x^y % 10 == (x+10)^y % 10.
For y, the values repeat every 4, that is, x^y % 10 == x^y+4 % 10, except for the case y=0.
Why is it so? I don't know, maybe someone else can explain/prove the math. All I know is that I can use it to make the power calculations much smaller. If I need the last digit of 345435^456, I calculate instead the last digit of (345435%10)^(456-1%4+1)=5^3=125, which is 5.

So that's what I'm doing in the function, extracting the last two numbers out of the array, calculating the right-most digit of their power and pushing it back. I also handle the case of empty list as per the requirements, not sure why they wanted it to be 1.