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?

  • 2
    $\begingroup$ Yeah, that's wrong. You can't infer that $7|a$, $7|b$, etc from $7|10^5a+10^4b+...$ A simpler example: $7|42$, but $7\not | 4$ and $7 \not | 2$ $\endgroup$ – Bram28 May 7 '17 at 18:25
  • $\begingroup$ In general if $x+y$ is divisible by $7$, it does not mean that $x$ and $y$ are each divisible by $7$. $\endgroup$ – angryavian May 7 '17 at 18:25
  • $\begingroup$ Then what is the correct proof? $\endgroup$ – Mainak Roy May 7 '17 at 18:27
  • $\begingroup$ @MainakRoy I would say that you should go back to the drawing board for a bit :) Don't give up so quickly! But as a hint, think about this: why the 'break' right at the 1000 mark? How do 7 and 1000 relate? $\endgroup$ – Bram28 May 7 '17 at 18:28
  • $\begingroup$ Well, I really have no idea about modular arithmetic (and I'm in grade 10...). But I'll have another go at it, using algebraic methods. $\endgroup$ – Mainak Roy May 7 '17 at 18:49

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.


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 $k$.

  • $\begingroup$ Surely "100abc + abc = 1001(abc)" was meant to be "1000abc + abc = 1001(abc)". (I suggested that edit, but apparently multiple moderators rejected it for "deviating from the original intent" and "does not make the post even a little bit ... more accurate".) $\endgroup$ – jamesdlin May 8 '17 at 5:25
  • $\begingroup$ Surely you are correct. $\endgroup$ – fleablood May 8 '17 at 5:34
  • $\begingroup$ How did I not notice that $abcdef = 1000abc + def$?.... Thank you so much! $\endgroup$ – Mainak Roy May 8 '17 at 9:41

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.


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 $n$.


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