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What is the inverse of 17 modulo 19 and how is it computed quickly?

The inverse of 17 modulo 19 is 9. It is computed quickly by rewriting 17 as -2 modulo 19. Since 20≡1(mod19)20 \equiv 1 \pmod{19} and 20=(−2)(−10)20 = (-2)(-10), the inverse of -2 is -10. Converting -10 to a positive residue by adding 19 gives 9.

Conditions

  • Modulus is 19.
  • Residue a=17a = 17.

Reasoning, step by step

  1. Rewrite 17 as -2 modulo 19.
  2. Find the inverse of -2 modulo 19. Note that (−2)(−10)=20≡1(mod19)(-2)(-10) = 20 \equiv 1 \pmod{19}, so the inverse is -10.
  3. Convert the negative inverse -10 to a positive representative by adding 19: −10+19=9-10 + 19 = 9.
  4. Conclude that 17−1≡9(mod19)17^{-1} \equiv 9 \pmod{19}.

Example

The video shows the board writing 17 ≡ -2 (mod 19), (-2)^{-1} ≡ -10 (mod 19), and 17−117^{-1} ≡ 9 (mod 19).

Common misconceptions

  • Trying to use the slow brute-force method of adding 19 repeatedly until a multiple of 17 is found.
  • Forgetting to convert the negative answer -10 back to the positive residue 9.

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