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I've seen this type of problem circulated around on this high school math competition called Number Sense, where the aim is to solve as many problems as fast as possible and more importantly, all mentally. The type of problem looks something like the following: $$20^6\div11\mathrm{\space has\space a \space remainder\space of\space ?}$$ My first approach was to this problem was saying that we can utilize the fact that $20^{10}\div 11\equiv 1\pmod{11}$ and essentially say that now if we take the square root of this, we get $20^5\div 11\equiv 1\pmod{11}$, and therefore $20^6\pmod{11}=20\pmod{11}=9$, giving us 9 as the answer. However, I have noticed that this fails to work in some instances. Take the following problem: $$8^7\div 13\mathrm{\space has\space a\space remainder\space of\space ?}$$ In this instance, if we take the exact same approach, utilizing $8^{12}\div 13\equiv 1\pmod{13}$, when we take the square root of $8^{12}$ we get $8^6$ so theoretically, $8^6\div 13\equiv 1\pmod{13}$ which can then lead us to $8^6\times 8\div 13$ having a remainder of 8. I noticed that whenever you plug something like this into Wolfram Alpha though, that the answer seems to be 5, and the reason being for this is that $8^6\div 13\equiv -1\pmod{13}$ rather than positive 1 which would make the remainder the 5 that we see in Wolfram Alpha rather than the 8 that we expect. Can anybody think of a method to approach this problem in terms of how we are able to solve for if the modulus will be a negative 1 or a positive 1? I have been trying to figure this out for a day or two now but being a high-school student that doesn't know a whole lot about things like this, I haven't been able to come to some solid conclusions (though I have had ideas). Other posts, such as the one referenced with the $2^11\div 23$ question have had some responses but have only clarified how you can get to the fact that the mod is -1 or +1, rather than how you can actually determine which one it is to get the correct answer to the problem. I'm not sure if determining whether it is -1 or +1 lies in Fermat's Little Theorem, but all ideas are welcome. I've also looked into Euler's Toitent function but that has proved to be little help in solving the problem as well. Thanks to all!

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    The square root of $1\pmod p$ is $\pm1 \pmod p$. – lulu May 03 '24 at 00:08
  • Right but is it possible to determine which one it is? Because of this fact, that leaves two possible answers to the problem based off of the method that I utilized and I am trying to understand if there is a way to determine whether you need to take the negative or positive mod. It might not even involve the method that I am using right now, but I'm trying to gain some direction for this problem. – Steven Livingston May 03 '24 at 00:10
  • Extracting roots is hard. Here , $20\equiv -2\pmod {11}$ so straight exponentiation is easy. – lulu May 03 '24 at 00:10
  • In general, look up Iterated Squaring for a good, fast way to exponentiate. – lulu May 03 '24 at 00:12
  • Right, exponentation would be the way that I would traditionally do it and it wouldn't be too bad here. However, there are instances where you can't have something as simple as -2 and it will instead be something like 13 (I can find those kinds of problems if you want). In instances like these, exponentation would not be as easy, and considering that the average time per question in this competition is around 7 seconds, it would be very tedious and time-consuming to do such exponentation. If there is truly not workaround that's find again I just want to explore my options. – Steven Livingston May 03 '24 at 00:18
  • I think the easiest way to do this one quickly is to say $20\equiv -2 \pmod {11}\cdots 20^3 \equiv -8 \equiv 3\pmod {11}\cdots 20^6 \equiv 3^2 \equiv 9 \pmod {11}$ – user317176 May 03 '24 at 00:45
  • As for you second line of reasoning $8^{12} \equiv 1 \pmod {13} \implies 8^6\equiv \pm 1 \pmod {13}$ – user317176 May 03 '24 at 00:48

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