Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

\begin{align*} 5x & \equiv 1 \pmod{1303} \\ 5x & \equiv 10 \pmod{30} \end{align*} Can you post a step by step solution? I am reviewing for my finals.

share|improve this question
    
I would have thought there would be lots of tutorials on simultaneous linear congruences already.... –  Simon Hayward Nov 29 '12 at 20:40
    
Found one math.stackexchange.com/questions/79282/… –  Simon Hayward Nov 29 '12 at 20:41
add comment

2 Answers

$$5x \equiv 10 \pmod{30} \implies x \equiv 2 \pmod6$$ $$5x \equiv 1 \pmod{1303} \implies 5x = 1303k + 3910 \implies x = 1303 \ell + 782$$ Since $6 \vert x -2$, we have that $6 \vert 1303 \ell \implies 6 \vert l$. Hence, $$x = 7818 M + 782$$

share|improve this answer
add comment

More generally, replace $1303$ by any integer $\rm\:c = \color{#C00}{13} + 30\,n.\:$ Then

$\ $ Easy CRT $\rm\,\Rightarrow\, 5x \equiv 1\! +\! c\,\left[\dfrac{9}{c} mod\ 30\right]\! \equiv 1\! +\! c\,\left[\dfrac{39}{\color{#C00}{13}} mod\ 30\right]\! \equiv 1\!+\!3c\pmod{ 30c}$

So we infer that $\rm\ x \equiv \dfrac{1+3c}5\equiv \dfrac{40+90n}{5}\,\equiv\, 9+18n\pmod{6c}$

So $\rm\,c\! =\! 1303\:\Rightarrow\:x \equiv \dfrac{1+3\!\cdot\!1303}{5}\equiv\dfrac{3910}5\equiv 782 \pmod{6\cdot 1303}$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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