Modulo 97

The modulo of any number rounded to a power of 10 (be it 1000, 100000, or, as here, 10,000,000) is relatively easy to calculate. Thus, any number added to this rounding must incorporate this modulo, and any multiple of this number will have a modulo multiplied in the same way.

For example:
1000 % 97 = 30
(1000 / 97 = 10.309278 then 0.309278 * 97 = 30)
Therefore, 10,000 % 97 must be 10x 30 (which still needs to be taken % 97, etc.).
And thus 10,001 % 97 must be (10x 30 +1) % 97.
So, by limiting the input to 4 or 5 digits, any number "aaa.bbb" modulo 97 equals (aaa x 30 + bbb) modulo 97

Since a standard calculator only has 8 digits, we first do 10,000,000 / 97 = 103092.78 then 0.78 x 97 = 76.
So any number "a.aaa.aab.bbb.bbb" modulo 97 will yield the same result as ("aaaaaaa" x 76 + "bbbbbbb") modulo 97.

And for more than 13 digits, we can continue:
a.aaa.aab.bbb.bbc.ccc.ccc % 97 => ???
First, 10,000,000,000,000 % 97 = (1,000,000 x 76 + 0) % 97 = 15
So this gives (aaaaaa x 15 + bbbbbbccccccc) % 97.
The calculation aaaaaa x 15 + bbbbbbccccccc will necessarily yield a result with 13 digits (or 14 if bbbbbb = 999999, but the rest remains manageable for an 8-digit calculator), and then we redo it with 76 as above.
So yes, we would need to break the calculation into two parts to do aaaaaa x 15 + bbbbbbccccccc (or do the addition on paper...), but it remains relatively simple to carry out. And we can continue this indefinitely - as long as we have the time to do it - by adding the same steps one after another.

Or, another way to look at the same question:

100 % 97 = 3
therefore 1,000 % 97 = (3 * 10) % 97 = 30
therefore 10,000 % 97 = (3 * 10^2) % 97 = 9
therefore 100,000 % 97 = (3 * 10^3) % 97 = 90
therefore 1,000,000 % 97 = (3 * 10^4) % 97 = 27
therefore 10,000,000 % 97 = (3 * 10^5) % 97 = 76

from which it is clear that:
if (x % y) = z
then (10x % y) = (10z % y)
or more generally: (nx % y) = (n(x % y) % y) = (nz % y)

Since ((x+p) % y) = (((x % y) + p) % y) (for the modulo "y" of two added digits, one or both can be replaced by their modulo "y" without changing the result)
therefore ((nx + p) % y) = ((n(x % y) + p) % y)
and again, if (x % y) = z,
((nx+p) % y) = (nz + p) % y

Thus for any very large multiple n of x (in this case, the 6 largest digits of a 13-digit number, which are thus a multiple "n" of "x" where x = 10,000,000), if we know the result "z" of x % y (in this case, 76 if y = 97), the equation for a digit "a.aaa.aab.bbb.bbb" % 97 simply becomes (n * 76 + p) % 97
= ("aaaaaa" * 76 + bbbbbbb) % 97


And finally, for a 31-digit number "a.aaa.aab.bbb.bbc.ccc.ccd.ddd.dde.eee.eee" % y ...
= ("aaaaaa" * (10^25 % y) + bbbbbbccccccddddddeeeeeee) % y
= ("aaaaaa" * (10^25 % y) + ("bbbbbb" * (10^19 % y) + ccccccddddddeeeeeee) % y) % y
and so on, which ultimately leads to:
= ("aaaaaa" * (10^25 % y) + ("bbbbbb" * (10^19 % y) + ("cccccc" * (10^13 % y) + ("dddddd" * (10^7 % y) + "eeeeeee")))) % y

For modulo 97, we know 10^7 % 97 = 76, thus "a.aaa.aab.bbb.bbc.ccc.ccd.ddd.dde.eee.eee" % 97
= ("aaaaaa" * (10^18 * 76 % 97) + ("bbbbbb" * (10^12 * 76 % 97) + ("cccccc" * (10^6 * 76 % 97) + ("dddddd" * 76 + "eeeeeee")))) % 97
It is easy to calculate 10^6 * 76 % 97 = 15
= ("aaaaaa" * (10^12 * 15 % 97) + "bbbbbb" * (10^6 * 15 % 97) + "cccccc" * 15 + "dddddd" * 76 + "eeeeeee") % 97
Again, 10^6 * 15 % 97 = 17
= ("aaaaaa" * (10^6 * 17 % 97) + "bbbbbb" * 17 + "cccccc" * 15 + "dddddd" * 76 + "eeeeeee") % 97
And finally, 10^6 * 17 % 97 = 71
= ("aaaaaa" * 71 + "bbbbbb" * 17 + "cccccc" * 15 + "dddddd" * 76 + "eeeeeee") % 97
And since the result ("ffgggggg") % 97 is likely to exceed the limits of a basic calculator, we redo (ff * 76 + ggggggg) % 97 to fit everything within 8 digits.