I am currently studying a course in cryptography and sometimes I come across calculations like the following:
$$11013^{10369} \pmod{20413}.$$
The normal technique for such problems is to create a table of powers of $2$ and calculate $11013^{2},11013^{4},\dots \pmod{20413}$ so that you can break down the calculation into smaller numbers. Or maybe even use Euler's theorem, but here $$\varphi(20413)=136 \times 148 = 20128>10369,$$ so it cannot be applied here.
In this example, the table would look like
$$\begin{array}{c|c} n & 11013^n \\ \hline 2 & 12536\\ 4 & 12022 \\ 8 & 4444 \\ 16 & 9765 \\ \vdots & \vdots \\ 8192 & 3225 \end{array}$$
Now,
$$10369 = 8192+2048+128+1.$$
Hence,
$$\begin{align} 11013^{10369}&=11013^{8192}\times 11013^{2048} \times 11013^{128} \times 11013 \\ &\equiv 3225 \times 8965 \times 8510 \times 11013 \pmod{20143}. \end{align}$$
Calculating the last line of the above results in a huge number, and on a standard scientific calculator, it cannot deal with it at all, i.e. rounding errors make it difficult to see what it is congruent to.
How is it possible to deal with something like this?
You do the modular arithmetic along the way. The product of the first two terms is an $8$ digit number. Reduce it modulo $20143$ and you have a $5$ digit number.
This strategy shows you never have to deal with more than the product of two $5$ digit numbers, hence $10$ digit numbers.
You can also ask Wolfram Alpha to do the arithmetic for you.
@didgogns 's comment tells you how to manage with even smaller numbers if you can factor $20143$ - not possible if (you are pretending that) this is part of the public key.