TITLE: Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers
QUESTION [9 upvotes]: While doing a few math puzzles I noted that you could take any three numbers (integers) and use basic arithmetic operations (Addition, Subtraction, Multiplication and Parentheses only)
to obtain a multiple of 10, using each number only once.
Eg.
Using 29, 73, 36: $73+36-29=80$
Using 2, 4, 7: $2*7-4=10$
Is there a proof for this theory in particular, or is there a more general theorem that applies to this? If not, is there a set of three numbers which disproves this theory?

I found a few clues while solving this:
If one of the numbers is a multiple of 5 (call it $x$) then there definitely exists a solution divisible by 10. This one is easily provable as you merely need an even number to multiply with. If any one of the remaining two numbers is even, you can use that to multiply with $x$ to get number divisible by 10. If both numbers are odd then you can add them to get an even number to multiply with $x$.
It would seem as only the units place digits have any bearing on whether it's a multiple of 10 or not. As you could take any of the examples, add or subtract any multiple of 10 from it and do the same operations to still get a multiple of ten. I feel this is also easily provable but I can't think of a way to do it which isn't long-winded.
EDIT: It can be proven as such, let $a,b$ be two integers s.t $a+b|10$ 
Adding two multiples of 10 ($10m,10n$) to $a$ and $b$ we get $10m+a+10n+b$
Since $a+b|10$ it can be written as $10m+10n+10k$ where $a+b=10k$
Which is divisible by ten
Now let $a,b$ be two integers s.t. $a.b|10$
Adding $10m,10n$ to $a$ and $b$ we get $(10m+a)(10n+b)$
$=100mn+10an+10bm+ab$ or $100mn+10an+10bm+10k$
Which is divisible by ten
Therefore any combination of Addition and Multiplication with the three no.s shouldn't matter.
I realise that could replace 10 with any no. in this proof and it would still work. So we really need to just prove for every single digit no.

I ran a program in python that checked for every combination of single digit no.s, but it didn't find any combination that disproves this.
I'm not sure how this question would be categorised, hence the lack lustre tags.
I'm fairly new to StackExchange. Please forgive me if I've worded this question poorly.

REPLY [2 votes]: Note: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.


$a,b,c\in\mathbb{Z}$


We are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have
\begin{align*}
(a\, \mathrm{mod}\, (10)) + (b\, \mathrm{mod}\, (10)) &\equiv (a+b)\, \mathrm{mod}\, (10) \\
(a\, \mathrm{mod}\, (10)) - (b\, \mathrm{mod}\, (10)) &\equiv (a-b)\, \mathrm{mod}\, (10) \\
(a\, \mathrm{mod}\, (10)) \cdot (b\, \mathrm{mod}\, (10)) &\equiv (a\cdot b)\, \mathrm{mod}\, (10) \\
\end{align*}
it is sufficient to consider $\color{blue}{a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}}$.


$a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$


If two of the three numbers are congruent modulo $10$, let's say $a\equiv b\, \mathrm{mod}\, (10)$ we obtain
\begin{align*}
\color{blue}{(a-b)\cdot c}\equiv 0\cdot c\color{blue}{\equiv 0\, \mathrm{mod}\, (10)}
\end{align*}
and we are done. In the following we may WLOG assume: $a<b<c$


$a,b,c\in\{0,1,2,3,4,5,6,7,8,9\},a<b<c$


If one of them, let's say $a$ is zero we obtain
\begin{align*}
\color{blue}{a\cdot b\cdot c}&\equiv 0\cdot b\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)}
\end{align*}
and we are done.


$a,b,c\in\{1,2,3,4,5,6,7,8,9\}, a<b<c$


If one of them, let's say $a$ is equal to $5$ we consider two cases.
First case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows 
\begin{align*}
a&\equiv 0\, \mathrm{mod}\, (5)\\
b&\equiv 0\, \mathrm{mod}\, (2)\\
ab&\equiv 0\, \mathrm{mod}\, (10)\\
\end{align*}
and we obtain
\begin{align*}
\color{blue}{a\cdot b\cdot c}&\equiv 0\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)}
\end{align*}
Second case: Both, $b$ and $c$ are odd. It follows from
\begin{align*}
b\equiv 1\, \mathrm{mod}\, (2)\\
c\equiv 1\, \mathrm{mod}\, (2)\\
(b+c)\equiv 0\, \mathrm{mod}\, (2)\\
\end{align*}
and we obtain
\begin{align*}
\color{blue}{a\cdot (b+c)} \color{blue}{\equiv 0\, \mathrm{mod}\, (10)}
\end{align*}


$a,b,c\in\{1,2,3,4,6,7,8,9\}, a<b<c$


Since
\begin{align*}
1\equiv (1-10)\equiv (-9)\, \mathrm{mod}\, (10)\\
2\equiv (2-10)\equiv (-8)\, \mathrm{mod}\, (10)\\
3\equiv (3-10)\equiv (-7)\, \mathrm{mod}\, (10)\\
4\equiv (4-10)\equiv (-6)\, \mathrm{mod}\, (10)\\
\end{align*}
we can restrict the attention to $\color{blue}{a,b,c\in\{1,2,3,4\},a<b<c}$.


$a,b,c\in\{1,2,3,4\}, a<b<c$


This case is already nicely considered in the answer from @ErickWong.

Conclusion: Doing arithmetic with three integers $a,b,c\in\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \rightarrow -a$) we can always obtain an integer value which is a multiple of $10$.