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Find a number $a\in[0,50]$ such that $a\equiv (105^{70}+1)^{15}\pmod {51}$

My question is there a simple way to solve this by hand? maybe using Fermat's little theorem or something?, I tried this way:

Since $105\equiv 3\pmod{51}$ I can write $(3^{70}+1)^{15}\pmod{51}$ and then to solve with long way $((3^4)^{10}3^{30}+1)^{15}\pmod{51}...$ with a lot of calculations I got $\color{green}{a=16}$ is a solution

Error 404
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    Not Fermat's little theorem: $51$ isn't a prime, so you must use Euler's theorem. – Arthur Jul 05 '16 at 11:04
  • You can solve this applying the techniques described here to first calculate the remainder $k$ of $3^{70}\pmod {51}$ and then again to find the remainder of $(k+1)^{15}$. A surefire approach is to calculate the remainders modulo $3$ and $17$ separately, and then apply the Chinese Remainder Theorem. There are probably trick-alternatives (consequences of Little Fermat and such). – Jyrki Lahtonen Jul 05 '16 at 11:04
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    You can also use the chinese remainder theorem. In this case, Fermat's little theorem suffices. Note that the remainder modulo $3$ is very easy to calculate. – Peter Jul 05 '16 at 11:05
  • Another hint : $105^{70}\equiv 3^6\equiv -2$ modulo $17$. Take it from here. (I reduced the base modulo $17$ and the exponent modulo $16$). – Peter Jul 05 '16 at 11:09

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By Fermat's theorem, $3^{16} = 1 \mod 17$ and hence $3^{64} = 1\mod 17$. Also $3^5 = 5 \mod 17$. Thus $3^{69} = 5 \mod 17$ and $3^{70} = 15 \mod 51$. What we need therefore is $(15+1)^{15} \mod 51 = 16^{15} \mod 51$ Since $16^2 = 256 = 1 \mod 51$, we have $a = 16$.

Jyrki Lahtonen
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I hope this helps, consider the divisors of 51.

$$105 \equiv 0 \quad mod(3) $$

$$105^{70} \equiv 0 \quad mod(3)$$ $$105^{70} +1 \equiv 1 \quad mod(3)$$ $$[105^{70} +1]^{15} \equiv 1 \quad mod(3)$$ So $a \equiv 1 \quad mod(3)$, so $a$ can be 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49

Now, $$105 \equiv 3 \quad mod(17) $$ $$105^{70} \equiv 3^{70} \equiv [3^{7}]^{10} \equiv 11^{10} \equiv 121^{5}= 2^{5} \equiv 32 \equiv 15 \quad mod(17)$$ $$105^{70}+1 \equiv 16 \equiv -1 \quad mod(17)$$ $$[105^{70}+1]^{15} \equiv [-1]^{15} \equiv -1 \quad mod(17)$$

so $a \equiv -1 \quad mod(17)$ so $a$ can be 16, 33, 50 thus $a$ can only be 16 because it must satisfy both equations.

Burrrrb
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