If you want to be tricky about it, you can do it in one line in Python. Let d be your original integer. Then

    [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))]

will be a list containing the base 10 digits of d as integers.

Technically, you need to do

    from math import log10

to make this work, but you can fit that on the same line with a little hacking around using __import__, such as this quick and very dirty attempt:

    [ [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))] for log10 in [__import__("math", fromlist=["log10"]).log10]][0]

Edit:

Just to carry this a little further, if we interpret "a new integer that's the result of each of the original integer's digits squared" as doing something like this:

    129 -> 1481

(i.e. 9 gets replaced with 81), then we can do it in one line like this:

     int(''.join([str(x**2) for x in [(d%10**(n+1) - d%10**n) / 10**n for n in reversed(xrange(int(log10(d)+1)))]]))

Ooh, nice.

You can detect each digit like this:

For example, take 999 (N). Using only math, how can I determine the value of each digit?

(be sure to take abs(N) and record value of N/abs(N) for later)

res = N - (N - 1). Digit = res + 1; If Digit == 10, Digit = 0.

So, given that function to find each digit, you can build a loop using 10^i to detect each digit (subtracting the sum of all preceding digits and dividing by 10^i to trim the lower order digits you've already resolved).

We'll need a place to store each digit, such as an array.

Of course, this assumes integers in base 10. If they are another base (8, 16), replace the 10^i above with the base.

It's also ambiguous as to if the function should just return a concatenated version of the results or some possibly a sum.

In any case, using a cast to string would be way easier.

Thanks!

Well the mathematical way of doing this involves doing a Mod 10 to pull off a digit and then incrementing a number that multiples the mod 10.

I've also seen incrementing an exponent. And I hear from more math oriented people that there are other ways.