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I've calculated the following equation and I've got this: My Solution

Does there exist an easier solution?

Martin Sleziak
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MathCracky
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    You might have a look at How do I compute $a^b\bmod c$ by hand?. – Martin Sleziak Mar 17 '16 at 17:41
  • $2^{48} \equiv x mod 20 \implies 4|x$. Let $y = x/4$. $2^{46} \equiv y \mod 5$. $2^2 \equiv -1 \mod 5$ so $2^4 \equiv -1 \mod 5$ so $2^2 = 4 \equiv y \mod 5$ so $2^48 \equiv 16 \mod 20$.... – fleablood Mar 17 '16 at 18:08
  • Actually, I'm kind of perplexed by why you were given this problem. Have you learned that if $\gcd(m,a) = 1$ that $a^{\phi(m)} \equiv 1 \mod m$ yet? I'm assuming not as this seems to be in introduction to the idea but... – fleablood Mar 17 '16 at 18:11
  • $2^{\phi(140)} = (2^{\phi(35)})^{\phi(4)} \equiv 1 \mod 35$ So $2^{\phi(140)} \equiv 1 + 35k \mod 140$. As $4|2^{\phi(140)}$ and $4|140$ then $4|1 + 35k$. So $1 + 35k = 36$. – fleablood Mar 17 '16 at 18:17

3 Answers3

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$2^{48} = (2^8)^6 = 256^6 \equiv(-24)^6 = 576^3 \equiv 16^3 = 4096 \equiv 36 \pmod {140}$

SiXUlm
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It seems that you have obtained the exponent as $48=\varphi(140)$.

We could use Euler's theorem in combination with Chinese remainder theorem.

Namely we have $$2^{\varphi(35)}\equiv 1 \pmod{35}$$ by Fermat's theorem. Consequently, we also have $2^{\varphi(140)}\equiv 1 \pmod{35}$. It is clear that $4\mid 2^{48}$, hence $2^{\varphi{140}}\equiv 0 \pmod4$.

The system of congruences \begin{align*} 2^{\varphi(140)}&\equiv 1 \pmod{35}\\ 2^{\varphi(140)}&\equiv 0 \pmod{4} \end{align*} has a unique solution modulo $4\cdot 35=140$. It is not difficult to find that it is $$2^{\varphi(140)}\equiv 36 \pmod{140},$$ since we only have four possibilities $1$, $36$, $71$ and $106$ fulfilling the first congruence.

Martin Sleziak
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$2^{48} \equiv 0 \bmod 4$

$2^{48} = (2^2)^{24} \equiv (-1)^{24} \equiv 1 \bmod 5$

$2^{48} = (2^3)^{16} \equiv (1)^{24} \equiv 1 \bmod 7$

So $2^{48} \bmod 140$ must be a multiple of $4$ of the form $1+35k$. We get $36$ for $k=1$.

lhf
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