Modulo Arithmetic/Examples/Multiplicative Inverse of 41 Modulo 97/41x = 2 Modulo 97
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Example of Modulo Arithmetic
The solution of the integer congruence:
- $41 x \equiv 2 \pmod {97}$
is:
- $x = 45$
Proof
From Multiplicative Inverse of $41 \pmod {97}$ we have:
- $41 \times 71 = 1 \pmod {97}$
Thus:
\(\ds 41 x\) | \(\equiv\) | \(\ds 2\) | \(\ds \pmod {97}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 71 \times 41 x\) | \(\equiv\) | \(\ds 71 \times 2\) | \(\ds \pmod {97}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\equiv\) | \(\ds 142\) | \(\ds \pmod {97}\) | ||||||||||
\(\ds \) | \(\equiv\) | \(\ds 45\) | \(\ds \pmod {97}\) |
Hence the result.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $3$: Equivalence Relations and Equivalence Classes: Exercise $8$