How to solve $\{ k \in \mathbb Z /266\mathbb Z \;\lvert\; 217k \equiv 210 \bmod 266 \}$ ?
Question: Since $217$ is not invertible in $266$ how can I solve the equation? Usually I would multiply $210$ by $217^{-1}$ mod $266$.
I appreciate every hint.
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1Hint 1: $266=2\cdot 7\cdot 19$. Hint 2: CRT – vadim123 Jan 17 '17 at 01:05
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@vadim123: Thank you. What does CRT mean ? – jublikon Jan 17 '17 at 01:07
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1https://en.wikipedia.org/wiki/Chinese_remainder_theorem – vadim123 Jan 17 '17 at 01:08
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I've edited your post for clarity. Please review to ensure I did not unintentionally change your meaning, and feel free to rollback my edit if you don't like it. – hardmath Jan 17 '17 at 01:20
2 Answers
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$$ \begin{align} &271x\equiv 210\!\!\pmod{\!266}\\[.1em] \iff\ &217x-266y\,=\,210\\[.1em] \iff\ &31x - 38y\,=\,30,\ \ \ {\rm by}\ \ {\rm cancelling}\ \ \ 7 = \gcd(271,210,266)\\[.1em] \iff\ & \color{#c00}{31x\equiv 30}\!\!\pmod{\!38}\end{align}$$
Therefore $\ {\rm mod}\,\ \color{#0a0}{38}\!:\ \ \color{#c00}{x\equiv \dfrac{30}{31}}\equiv\dfrac{-8}{-7}\equiv\dfrac{-40}{-35}\equiv\dfrac{-2}{3}\equiv \dfrac{36}{3}\equiv\color{#0a0}{12}\ $ by Gauss's algorithm.
Hence $\ x = \underbrace{\color{#0a0}{12\!+\!38}\,n = 12\!+\!38(j\!+\!7k)}_{\begin{align}\large {\rm write}\ \ n\ =\ j\ +\ 7k\ \ \text{by division }\\ \large \text{where }\ 0\le j< 7\end{align}} \equiv 12\!+\!38j \pmod{\!266},\,\ j = 0,1,\ldots,6.$
Bill Dubuque
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$217k \equiv 210 \bmod 266$
Divide through by $7$:
$31k\equiv 30 \bmod 38$
Inverse of $31 \bmod 38$ is $-11\equiv 27$
$k\equiv 27\cdot 30 \equiv -26 \equiv 12 \bmod 38$
Therefore $k \in\{12,50,88,126,164,202,240\}$
Joffan
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