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How do I prove this without using a calculator?

$124235127381 · 34562736458392 \not= 4293905956544262926431352$

What methods should I use?

Or a suggestion in a method to use in order to solve this.

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    What methods have _you_ used? – 2017-02-20
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    I was going to use GCD but the Numbers are to large to not use a calculator – 2017-02-20
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    Hint (I haven't tried). Thinking odd/even is no help. But have you tried modulo 3? Do you know how to compute the remainder when you divide a number by 3? – 2017-02-20

3 Answers 3

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$3$ divides $124235127381$ but not $4293905956544262926431352$

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    How do you know that without a calculator? – 2017-02-20
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    @pillikasnazsaxc Digital roots of the three numbers are 3, 4, 2 respectively. – 2017-02-20
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    by summing digits. Actually you don't need to sum all, just which aren't divisible by $3$ and each time you find a multiple refresh, e.g. 1+2=3=0+= 4+2=6=0+=5+1=6=0+=2+7=9=0+=8+1=9=0. – 2017-02-20
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    To help those who don't know why this works, you could give a quick proof: 1, 10, 100, etc are all a multiple of 9 plus 1. So if the nth digit is $d$, then this represents a multiple of 9 plus $d$. So $\sum d_i 10^i $ equals $\sum d_i$ modulo 9. The sum of the digits must have the same remainder when divided by 9 as the original number. A quick corollary is that this applies to $3$ as well as $9$. – 2017-02-20
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    Thank you @Joel for completing the answer in comments – 2017-02-20
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Use the time-honoured method of casting out nines.

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    A student of mine recalled that her primary school teacher told that “casting out nines works, but nobody knows why”. `;-)` – 2017-02-20
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    No wonder she went on to study mathematics then, to know why! – 2017-02-20
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    @egreg And nowadays nobody knows it works. With a few exceptions, of course. – 2017-02-20
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    There's only one problem. What if $\bmod9$ and $\bmod11$ checks both pass? (I learned both as part of the Trachtenberg system.) – 2017-02-20
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    @ParclyTaxel, the problem could have been construed to defeat the easiest such checks, sure. – 2017-02-20
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    Could you give an example or show how to use it? I have never used casting out nines before – 2017-02-20
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    Oh, sure, although it is well explained in the Wikipedia page. Start adding the digits of the first number, and every time you get $9$, or something greater than $9$, subtract $9$ (that's the *casting out nines*). You get a number $0 \le a < 9$. Do the same for the second number, to get $b$, and for the third number, to get $c$. Now if $c$ does not equal the result of casting out nines in $a * b$, the multiplication is not correct. – 2017-02-20
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"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.

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    I honestly think you're overthinking this a bit. – 2017-02-20
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    @VanDerWarden Indeed the part about actually writing out the product was really splitting hairs about the "use of a calculator," so I've reduced that discussion to a mere mention that it is possible. But I think it's useful to ask whether it would be considered "cheating" to use a calculator in order to find out which non-calculator methods would work. I've tried to clarify the importance of that question. – 2017-02-20