Perhaps this particular problem is not that difficult

for a math-type prodigy, because of its 'regularity'.

I would proceed like this:

365365365365365365^2
= (365 * 1001001001001001)^2

= 365^2 * 1002003004005006005004003002001

Now, 365^2 can be computed 'in the head' very fast

using the well´known trick for squaring numbers ending in 5, namely:

[N]5 ^ 2 = [N^2+N]25 (e.g.: 45^2 = [4^2+4]25 = [16+4]25 = 2025)

so 365^2 is computed like this:

[36^2+36]25 = [1296+36]25 = 133225

at once, requiring only the trivial addition of 36 to the memorized value of 36 squared (any math prodigy worth his/her salt has all squares up to 100^2 perfectly memorized, at the very least).

This being accomplished in a second or so, the prodigy must now compute:

133225 * 1002003004005006005004003002001

but this is much easier to do than it seems, because 133225 is such a small, 6-digit value, and the other factor has only 6 different non-zero digits, arranged in such a regular fashion. The prodigy just needs to form the following 6 partial products:

133225 * 1, 133225 * 2, ... , 133225 * 6

but they are trivially formed by simply starting with 133225, then keep on adding 133225 to the previous result

five times:

133225, 266450, ... , 799350.

Thus, all six of them can be computed mentally in a few seconds at most. Now, it's just simply a matter of arranging them properly for the sum, like this:

133225

266450

.....

.....

266450

133225

-------------------------------------

133491850208566925016658299941583225

and that's it. It's actually much easier (for even a mild prodigy) than it seems at first, and most readers of this forum would be able to accomplish it rather easily with a little training. As you have seen, it has little to do with actual complicated computation but with arranging things properly and doing a few sums of small, 6-digit numbers. Only the final result is really multi-digit.

Best regards from V.