Squaring is Daring (2)
If you haven't done the earlier Squaring and Daring challenge, do it now.
Now that you have experimented a bit with two-digit numbers, let's start looking at 3-digit numbers. But first we must make an important realization by asking a simple question:
What is the greatest number we are able to get if we start with a 3 digit number?
Well, an input of 999 would give the greatest possible output.
\(9^2\) + \(9^2\) + \(9^2\) = \(81 * 3\) = 243
Now it is up to you to continue. Will we see new "kinds" of sequences with 3-digit, 4-digit, 5-digit, n-digit numbers?
But I have a final question that you should be able to answer…
If we start with the number 520373885, will the sequence be good or bad?