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I wish to find the remainder $823^{823}$ in the division by $11$.

I used the Euler-Fermat theorem that tells that: $a^{\phi(n)} \equiv 1$ $(\mod n)$, whenever $(a,n) =1$ First, Euler- Fermat tells us that $a^{\phi(11)} \equiv 1 (\mod 11)$ whenever $a$ is coprime with to 11. Since 823 is prime, we have, $(823,11) = 1$. Now $823 \equiv 9 (\mod 11)$

But I don't know how to continue.. Because both $823$ and $11$ are prime ; do I have to use Fermat's little theorem?

Any help appreciated

Bill Dubuque
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Joe Goldiamond
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  • It doesn't really matter that $823$ is prime. As you (correctly) point out $823^n\equiv 9^n \pmod {11}$. So now apply Little Fermat to $9^{823}\pmod {11}$. – lulu Feb 13 '19 at 19:38
  • The relevant prime here is 11, so, you can reduce the exponent Mod 10 – Count Iblis Feb 13 '19 at 19:44
  • Fermat's little theorem is a special case of the Euler-Fermat theorem when $n$ is prime – J. W. Tanner Feb 13 '19 at 19:57

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Note that you also have $823\equiv -2\mod11$, which will make calculating powers by hand easier. Now by lil' Fermat, $$823^{823}\equiv (-2)^{823\bmod 10}=-8\equiv 3\mod 11.$$

Bernard
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Note that $$823^5\equiv 1 \mod 11$$ and then you can square the equation.

Dr. Sonnhard Graubner
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