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I would like to know if there is a shorter, more elegant and compact way to prove the statement with respect to the one I used to develop my proof, which is the following.

Assume $10 \vert abc$, meaning $2 \vert abc$ and $5 \vert abc$. Since $2$ and $5$ are both primes, it follows, without loss of generality, that $2 \vert ab$ or $2 \vert c$ and $5 \vert ab$ or $5 \vert c$. This means we need to consider four possible non-trivial combinations: $2 \vert ab$ and $5 \vert ab$, $2 \vert ab$ and $5 \vert c$, $2 \vert c$ and $5 \vert ab$, $2 \vert c$ and $5 \vert c$. A combination is assumed to be "trivial" if it assumes one of the four listed to be true. Then, I proceeded to prove that the statement holds in all the listed cases, applying the definition of divisibility and the already exploited property of prime numbers.

I feel this proof to be a little bit heavy and cumbersome. Let me know if you have a cleaner approach to this, and thanks in advance for any feedback.

  • $2\mid abc \overset{\rm wlog}\Rightarrow 2\mid a \Rightarrow 2\mid\color{#c00}{ab},ac,\ $ so $\ 5\mid 10\mid (ab)(ac)\overset{\rm wlog}\Rightarrow 5\mid\color{#c00}{ab},,$ so $,2,5\mid\color{#c00}{ab}\Rightarrow 2\cdot 5\mid\color{c00}{ab}\ \ $ – Bill Dubuque Nov 17 '24 at 21:03

2 Answers2

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As you know, $10$ can be factorized into the primes $2$ and $5$. That means that $2$ (and $5$ respectively) divides one of the numbers $a$, $b$ or $c$. Then the product (which was $10$) must divide the product of those two numbers (if they divide the same, you can select the other number randomly).

Henrik supports the community
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If 10 already divides $a$ (or $b$ or $c$), then you would be done directly. Thus assume for the sequel that this would be not true.

If 2 divides $a$ (say), then 5 now needs to divide $b$ (or $c$). But then clearly 10 divides $ab$ (or $ac$). And this is what was asked to be proved.

There aren't further possibilities, because 10 itself has no other factorisation.

--- rk

Dr. Richard Klitzing
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