Small question about proof
$3$ divides $124235127381$ but not $4293905956544262926431352$
Use the time-honoured method of casting out nines.
"How do I prove this without using a calculator?"
What do we mean by "using"?
The answer you're looking for appears to be something along the following lines: guess a way in which the inequality might be proved without computing all the digits (or even very many of the digits) of the product on the left-hand side; then check to see if this method of proof works. If it does not work, guess another method.
Multiplying just the last $n$ digits on the left-hand side will tell you the remainder of each side on division by $10^n.$ But for any value of $n$ you're likely to want to try, this turns out not to work. There are quick tests for division by other numbers you can perform, however; casting out $3$s works, as it turns out, as does casting out $9$s.
Note that you could be guessing and checking a lot of methods, depending on how cleverly the person posing the problem has protected it against different methods and on your luck at guessing one of the methods that works. The proof is a lot harder to guess in this way if we change the right-hand side to $4293905966547263926431352,$ for example.
Another interpretation of the problem is that the proof must not mention any calculations performed by a calculator. This leaves open the possibility that the discovery of the proof might be assisted by a calculator. We find that $$124235127381 \cdot 34562736458392 = 42939059\underline{6}6544262926431352,$$ which agrees with the given right-hand side except for the single underlined digit, which is $6$ in the correct product and $5$ in the given number. A difference of one in one digit implies that most of the quick tests for divisibility will work, especially if the test also gives you a unique result for each remainder modulo the divisor. For example, the test for divisibility by $3$ will definitely show remainders on the left-hand side whose product is inconsistent with the remainder on the right-hand side (which is $r+1,$ where $r$ is the "correct" remainder). So we immediately know this test will work.
On the other hand, if the right-hand side in the question had been $4293905966547263926431352,$ we would immediately have been able to rule out remainder-modulo-$3$ and several other methods of attempted proof, and would have a good clue as to what methods might work instead. (I might use the fact that $1000\equiv1 \pmod{37},$ or that $10000\equiv1 \pmod{101}.$)
By the way, the given numbers are tedious to multiply out by hand, but not impossible. You probably wouldn't do this for question $5$ out of $40$ on a three-hour examination, but it's not going to take hours. Possibly the less said about that idea the better, however.