# Squaring is Daring (2)

**Number Theory**Level pending

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 answerâ€¦

If we start with the number 520373885, will the sequence be good or bad?

###### If you stop here, you have finished all my problems regarding this amazing topic. But you won't learn how to independently work and persevere through this unless you continue. If you have made any further progress at all, please feel free to share it with the community. We would love to see your ideas. But remember this: Completely solving the problem and exploring it deeply is not for Brilliant, for points, or for me. It is for you. The choice is yoursâ€¦

**Your answer seems reasonable.**Find out if you're right!

**That seems reasonable.**Find out if you're right!

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