1
$\begingroup$

So I have that $700=7\cdot2^2\cdot5^2$ and I got that $3^2\equiv1\pmod2$ so then $3^{1442}\equiv1\pmod2$ also $3^2\equiv1\pmod{2^2}$ so $3^{1442}\equiv1\pmod{2^2}$ which covers one of the divisors of $700$. Im not sure if I'm supposed to use $2$ or $2^2$ and I was able to find that $3^2\equiv-1\pmod5$ so $3^{1442}\equiv-1\pmod5$, For mod $7$ I wasn't able to come up with an answer in a way like the other two, and I'm not really sure how to do this to find the least non negative residue

$\endgroup$
4
$\begingroup$

Go $\pmod4$, $\pmod7$ and $\pmod{25}$. We have \begin{align} 3^2 \equiv 1\pmod4\\ 3^6 \equiv 1\pmod7\\ 3^{20} \equiv 1\pmod{25} \end{align} This gives us that \begin{align} 3^{60} \equiv 1\pmod4\\ 3^{60} \equiv 1\pmod7\\ 3^{60} \equiv 1\pmod{25} \end{align} This means $$3^{60} \equiv 1\pmod{700}$$ Note that $3^{1442} = 3^{24\cdot60+2} = \left(3^{60}\right)^{24} \cdot 3^2$. Hence, we obtain $$3^{1442} \equiv 3^2\pmod{700} \equiv9\pmod{700}$$

$\endgroup$
2
$\begingroup$

Since $\phi(700)=240$, therefore from Euler's theorem $$3^{240} \equiv 1 \pmod{700}$$ Now $$1442 =240(6)+2$$ Therefore $$3^{1442} \equiv 3^{240(6)} \cdot 3^{2} \equiv 9 \pmod{700}$$

$\endgroup$
0
$\begingroup$

Carmichael function $\lambda(700)=60$

As $(3,700)=(3,3\cdot233+1)=(3,1)=1,3^{60}\equiv1\pmod{700}$

and $1442=24\cdot60+2\equiv2\pmod{60}\implies3^{1442}\equiv3^2\pmod{700}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.