0
$\begingroup$

So a mathematician gives me a system like this:

$$ \begin{eqnarray*} B_1 + z_1 M_1 &\equiv& B_2 \pmod {M_2} \\ B_1 + z_1 M_1 + z_2 M_1M_2 &\equiv& B_3 \pmod {M_3} \\ &\cdots& \\ B_1 + z_1M_1 + z_2 M_1M_2 + \cdots + z_{k-1}M_1 M_2 \cdots M_{k-1} &\equiv& B_k \pmod{M_k}. \end{eqnarray*} $$

But I forgot how to solve such systems. So I wonder does $B_1+z_1 M_1 \equiv B_2 \pmod {M_2}$ mean that $z_1 = (B_2 \bmod M_2 - B_1)/M_1$?

$\endgroup$
1
$\begingroup$

$B_1+z_1m_1=B_2(\mod m_2)\Rightarrow$

$\Rightarrow (B_1+z_1m_1)-B_2=am_2\Rightarrow $

$\Rightarrow z_1m_1=B_2-B_1+am_2 \Rightarrow$

$\Rightarrow z_1=\frac{B_2-B_1+am_2 }{m_1}$ , where $a$ is an integer

$\endgroup$

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.