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I am having trouble, I tried using the fact that the $gcd(30, 22) = 2$ but I have been stuck here for a bit now.

$22^{201} \equiv x \mod (30)$

$22^{201} \equiv 22*22^{200} mod (30)$

How can I proceed?

JOX
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  • Chinese Remainder Theorem is your friend. Have you heard of it? If not, then you can start calculating the remainders of $22^n$ modulo $30$ for small values of $n$ up to the point where you notice a repeating pattern, and go from there. – Jyrki Lahtonen Feb 13 '15 at 05:43
  • I am doing the second thing, unfortunately after a small set of values (3) it gets tough to compute that. – JOX Feb 13 '15 at 05:45
  • @JOX, Try http://math.stackexchange.com/questions/1143012/modular-exponentiation-854/1144678#1144678 – lab bhattacharjee Feb 13 '15 at 05:46
  • Using the CRT amounts to the following. Determine what is the remainder of $22^{201}$ A) modulo $2$, B) modulo $3$, C) modulo $5$. Probably you can do all those by the usual methods. But, because $2,3,5$ are all coprime to each other, there is only one residue class modulo $30$ that matches those three residue classes modulo $2,3,5$. – Jyrki Lahtonen Feb 13 '15 at 05:49
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    If you don't want to use CRT, the best way to start the powers will be$$22^{201}\equiv(-8)^{201}\equiv-2^{603}\ .$$If you calculate a few powers of $2$ you will see the pattern very soon. – David Feb 13 '15 at 05:50
  • And I don't see why it would become excessively difficult to calculate those remainders: $22^2=484\equiv 4\pmod {30}$ so $22^3= 22^2\cdot22\equiv 4\cdot22=88\equiv 28\pmod {30}$, so $22^4=22^2\cdot 22^2\equiv 4\cdot4=16\pmod{30}$, so $22^5\equiv 16\cdot22=352\equiv 22\pmod{30}$,... David's idea keeps the numbers even smaller :-) – Jyrki Lahtonen Feb 13 '15 at 05:52
  • Yes, I started to notice a pattern with that @David, however, how could I use the CRT? It sounds interesting but although I just read about it, I can't seem to find how to use it for this? Out of interest could someone post the solution using that theorem? I haven't learned about it but it would be good to know how to use it. – JOX Feb 13 '15 at 05:53
  • Mod $2$ the result is $0$. Modulo $15$, use Euler's Theorem, $\varphi(15)=8$ so $22^{200}\equiv 1\pmod{15}$, so the the power $201$ it is congruent to $22$ mod $15$. This happens to be even, so the result modulo $30$ is $22$. – André Nicolas Feb 13 '15 at 06:36
  • @AndréNicolas why does $\phi(15) = 8$ imply the other thing? Can you expand in an answer? This seems pretty helpful since I didn't realize I could apply the totients function. – JOX Feb 13 '15 at 06:43
  • We have $22^{200}=(22^8)^{25}$. By Euler's Theorem, we have $22^8\equiv 1\pmod{15}$. And $1^{25}\equiv 1\pmod{15}$. It follows that $22^{201}=22^{200}\cdot 22\equiv 22\pmod{15}$. – André Nicolas Feb 13 '15 at 06:47
  • Ultimately you can take one or preferably more than one approach that has been suggested and write up an answer. – André Nicolas Feb 13 '15 at 06:49

2 Answers2

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$[22^{201}]_{30}$

$ = [484^{100}]_{30} \cdot [22]_{30}$ and $[484] = 16 \cdot 30 + [4]$

$ = [64^{33}]_{30} \cdot [22 \cdot 4]_{30}$ and $[64] = 2 \cdot 30 + [4]$ and $[88] = 2 \cdot 30 + [28]$

$ = [4^{33}]_{30} \cdot [28]_{30}$

$ = [64^{11}]_{30} \cdot [28]_{30}$

$ = [4^{11}]_{30} \cdot [28]_{30}$

$ = [64^{3}]_{30} \cdot [28 \cdot 4 \cdot 4]_{30}$ and $[112] = 3 \cdot 30 + [22]$

$ = [4^{3}]_{30} \cdot [22 \cdot 4]_{30}$

$ = [64]_{30} \cdot [28]_{30}$

$ = [4 \cdot 28]_{30}$

$ = [22]_{30}$

Vinícius Ferraz
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We have $22^2=484\equiv 4\pmod{30}$ Then $4^3\equiv 4\pmod{30}$ and this means $22^6\equiv 4\pmod{30}$ and $201=33\times 6+3$ so $22^{201}=(22^6)^{33}+22^3$. This gives $$22^{201}\equiv 4^{33}\times 22^3\pmod{30}$$ Using the above $4^{33}\equiv 4^{11}\equiv (4^3)^{3}\times 4^2\equiv 4$ and $22^3\equiv4\times22\equiv 28\pmod{30}$ and we can conclude that $$22^{201}\equiv 4\times 28\equiv -8\equiv 22 \pmod{30}$$

marwalix
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