A question about divisibility by $7$ 
There is a 6 digit number $abcdef$. Given that it is divisible by $7$, show that $abc-def$ is also divisible by $7$.

Attempt at solution:
$abcdef=10^5a + 10^4b + ... + f=7x$.
Therefore, $7|a, 7|b, ...$.
Hence, $abc=7m, def=7n$ and $7(m-n)$ must be divisible by $7$.
Q.E.D
I still feel this proof is wrong somewhere. Can someone verify?
 A: My first reaction to "If $abcdef$ is divisible by $7$ then $abc-def$ is divisible by $7$" is "really, I had no idea that was true!".  Then I'd think well, gee, if that's true then how could that be true and what would it imply.  How can I get from $abcdef$ to $abc-def$?
$abcdef = 1000abc + def$ is a start.
and $1000abc + def + (abc - def) = 1000abc + abc = 1001(abc)$
So if $abc - def$ and $abcdef$ are both divisible by $7$ then $abcdef + (abc-def) = 1001(abc)$ is divisble by $7$.  Is there any reason why that should be true?  Also $def$ could be anything suppose we had $abc - def$ and $1001abc$ are both divisible by seven but we then change $def$ to $deg$ and $abc-deg$ *wasn't divisible by $7$ then $1001abc$ would still be divisible by $7$.  How could that be possible?
The answer must have something to do with $1001$.  How does that relate to $7$?  Does $7$ divide $1001$ or have an interesting remainder.  So I divide $1001$ by $7$ and get... $1001=7*143$.
So that's that.
Then I'd put it all together.
If $abcdef = 1000abc + def = 1001abc - (abc - def) = 7*143abc - (abc-def)$.
So $7|abcdef\iff 7|(abc-def)$.
A one line proof.
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You commented that you don't know anything about modular arithmetic.
Modular arithmetic is the mathematics of remainders.  I you are only interested in the remainders of things after dividing by $7$ and you wonder what the remainder of $45*73$ will be you don't have to do $45*73 = 3285$ and $3285\div 7 = 469$ with remainder ... um .... $0.2857 ...\times 7 =  2$.  You can think "As far as we are concerned $45$ has remainder $3$ so we might as well say $45$ is $3$ and $73$ is $3$ and so $45*73$ is $3*3$ which equals $9$ which is $2$ as far as we are concerned.  (Algebraically, we can justify this as $45*73 = (6*7 + 3)(7*10 + 3) = (7a + 3)(7b + 3) = 7^2ab + 3*7b + 3*7a + 3*3 = 7M + 3*3 = 7M + 9 = 7M + 7 + 2 = 7N + 2$.  Where $a=6;b=10;M= 7ab+3a+3b;N=M+1$.  But we don't care at all about the actual values of any of those.  We only want the remainders.)
So we write $a \equiv b \mod n$ to mean "$a $ and $b$ have the same remainder when divided by $n$". And it's easy to verify: If $A \equiv a \mod n$ and $B \equiv b\mod n$ then $A+B \equiv a+b \mod n$ and $A*B \equiv a*b \mod n$ and $A^k \equiv a^k \mod n$.
So: $1000 \equiv -1 \mod 7$
So $1000k \equiv -k \mod 7$ for any integer.
So $abcdef = 1000abc + def \equiv -abc + def \equiv -(abc-def)\mod 7$
So $abcdef \equiv 0 \mod 7\iff  abc-def \equiv 0\mod 7$.
And $k \equiv 0 \mod 7$ means "$k$ has remainder $0$ when divided by $7$" which means $7$ divides into $k$.
A: Since $1000\equiv -1\pmod{7}$, we have $$abcdef=1000abc+def\equiv -abc+def\pmod{7}$$ and if the given number is divisible by 7, then by above result $-abc+def$ is divisible by 7.
A: The proof is wrong. Here's a correct one.
We need the values of $10^n\bmod 7$ for small $n$; from $n=0$ they are 1, 3, 2, 6, 4 and 5. Then
$$\overline{abcdef}\equiv0\bmod7$$
$$5a+4b+6c+2d+3e+f\equiv0\bmod7$$
$$-(2a+3b+c)+2d+3e+f\equiv0\bmod7\tag1$$
$$-\overline{abc}+\overline{def}\equiv0\bmod7$$
$$\overline{abc}-\overline{def}\equiv0\bmod7$$
where $(1)$ is because subtracting a multiple of $n$ does not change the residue modulo $n$.
