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I'm having quite a bit of confusion over how I should read the congruent sign in a modulus equation. For example, I have seen this example as valid:

$$ 81 \equiv 13 \pmod{17} $$

From what I see, it reads to me as "when $81$ is divided by $17$, the remainder is $13$".

However, in another example, this appears to be valid too:

$$ 9 \equiv 2^6 \pmod{11} $$

Now, if I use the same format as how I read the first example, "when $9$ is divided by $11$, the remainder is $2^6 = 64$", this sounds wrong. The remainder of $9$ divided by $11$ is $9$, not $64$. Instead, it sounds more correct when I read it the opposite way: "when $2^6 = 64$ is divided by $11$, the remainder is $9$".

This is somewhat confusing to me because how should I read the equation? Sometimes it seems like the first term $a$ is the remainder while other times it looks like the second term $b$ is the remainder.

Didier
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xenon
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    Not sure if it helps, but I'm reading those equations as: $81$ and $13$ are congruent with reference to 17. So they are not "really" equal, but something weaker. I think it one could denote it like this $81 \equiv_{17} 13$ – Algebruh Oct 30 '22 at 10:53
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    $a\equiv b\pmod p$ means "$a$ and $b$ have the same remainder when divided by $p$" or (equivalently) "$b-a$ is divisible by $p$". (If one of $a$ or $b$ is negative, you have to still assume the remainder is in the set ${0, 1,2, \ldots, p-1}$.) So neither $a$ nor $b$ needs to be the remainder. –  Oct 30 '22 at 10:53
  • @StinkingBishop Ahhh!! Now I get it! So they aren't exactly the remainders of each other but they have the same remainder. Thank you so much! :D – xenon Oct 30 '22 at 11:09
  • When 12 is divided by 5, there are infinite remainders: 2, 7, 12, -3, etc. However 2 is the least non negative remainder – across Oct 30 '22 at 11:33
  • In Programming languages (& CPU Instruction Sets) , the Modulo is like remainder Eg $9 MOD 6 = 3$ , Particularly , C may be like $9%6 == 3$ where $3%6 == 9$ is not true. Here (A%B) or (A MOD B) will have Some Exact Value. In Math (Number Theory) $A MOD B = C$ (or Equivalently $A = C MOD B$) is more generalized (though it has the historical roots in Integer Division & remainder) & it means $B|(A-C)$ (or Equivalently $(A-C) = DB$) , which is not necessarily the remainder !! – Prem Oct 30 '22 at 11:36
  • Your are confusing the operational vs relational forms of mod. See here in the dupe for further discussion. – Bill Dubuque Oct 30 '22 at 12:45

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