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So I just remembered a question I had a while ago about an explanation of the divisibility by $7$ trick where you double the last digit and subtract it from the rest of the number. I think the explanation was something about $-2$ being the multiplicative of $10$ $\pmod{7}$, and I'm not quite sure what that means. Could someone explain?

Adola
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Bob Wang
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    The divisibility rule by $7$ is impractical. It's often always faster to do long division (without subtracting) mentally instead. – Adola Jun 17 '20 at 08:12
  • Related. Not voting to close as a duplicate of that because I answered that version. – Jyrki Lahtonen Jun 20 '20 at 19:09
  • Anyway, I also agree with @Adola. For pencil & paper work it is faster, for example to use the fact that $7\mid 1001$. Therefore $$141592653589793$$ is divisible by $7$ if and only if $$141-592+653-589+793$$ is. Conveniently, the exact same procedure works for $11$ and $13$ as $1001=7\cdot11\cdot13$ :-) – Jyrki Lahtonen Jun 20 '20 at 19:12

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$10a+b \mod 7=0 \iff -10a-b\mod7=0 \iff -20a-2b\mod 7=0 \iff a-2b\mod 7=0$, where the last "iff" comes from the fact that $21a \mod 7=0$.

Adola
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mathemagician99
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  • I understand that, but what does -2 being the multiplicative of 10 (mod 7) have to do with that, what does multiplicative even mean here? – Bob Wang Jun 17 '20 at 07:52
  • $-2$ is the multiplicative inverse of $10$ (mod $7$). – paw88789 Jun 17 '20 at 07:59
  • @Bernard maybe this is just wierd Russian notation? We write $a\vdots b$ instead of $b|a$ – Alexey Burdin Jun 17 '20 at 08:01
  • @paw88789 why is that so? Am I missing something? Also what does that have to do with divisibility by 7? – Bob Wang Jun 17 '20 at 08:07
  • @AlexeyBurdin: I doubt that – $a:b$ is perfectly understandable as an ellipsis for ‘the quotient $a/b$ is an integer’, whereas here, the numbers are swapped. I'd rather think of a lapsus calami. – Bernard Jun 17 '20 at 08:08
  • @Bernard You are completely right. I wanted to use $\mod$ anyway but got it mixed up. – mathemagician99 Jun 17 '20 at 08:10
  • @BobWang Well, I think in your explanation this might mean something like $-3\cdot 10 \mod 7=-30 \mod 7=-2 \mod 7$. So $-2$ is a mulitplicative of $10 \mod 7$. You could have taken a different route and used this fact but I didn't. – mathemagician99 Jun 17 '20 at 08:13
  • @mathemagician99 Yeah, I'm that's why I was confused with why the multiplicative stuff even made sense. Why does it even make sense or have anything to do with the divisibility rule in simple terms – Bob Wang Jun 17 '20 at 08:44
  • @BobWang I'm afraid, I don't fully understand your question. Do you ask, why one has to deal with multiplicatives when dealing with divisibility? – mathemagician99 Jun 17 '20 at 08:50
  • @mathemagician99 I'm not quite sure what multiplicative means in this case anymore, I thought it was just a number you multiply by in mod n to get 1? but -2*10 mod 7 = -6. Like for example 9 would be the multiplicative inverse of 3 mod 26 right? – Bob Wang Jun 17 '20 at 08:53
  • @BobWang You are right in the group $(\mathbb{Z}/ 26\mathbb{Z},\cdot)$ $9$ is the inverse of $3$. However I used the word multiplicative as the multiple of some number so to speak. But in your example above $-2$ is even the inverse of $10$ in $(\mathbb{Z}/ 7\mathbb{Z},\cdot)$, since $-2\cdot 10\mod 7=-20\mod 7=1\mod 7$. – mathemagician99 Jun 17 '20 at 10:35
  • @mathemagician99 sorry, I must've been screwed up in the head while typing that. I understand the modulo part of this now, but how does it relate to the divisibility rule? – Bob Wang Jun 17 '20 at 10:55
  • You just use this fact in order to establish the calculation above. We have $-2\cdot 10 \mod 7$ and therefore this is the same as $1\mod 7$. "Multiplicative inverse" is a little bit overkill in terminology. You would have gotten just as easy along without using this term. – mathemagician99 Jun 17 '20 at 11:32