HELP ASAP
[tex]Given that $33^{-1} \equiv 77 \pmod{508}$, find $11^{-1} \pmod{508}$ as a residue modulo 508. (Give an answer between 0 and 507, inclusive.)[/tex]

Respuesta :

Answer: 231

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Work Shown:

[tex]33^{-1} \equiv 77 \text{ (mod 508)}\\\\(3*11)^{-1} \equiv 77 \text{ (mod 508)}\\\\3^{-1}*11^{-1} \equiv 77 \text{ (mod 508)}\\\\3*3^{-1}*11^{-1} \equiv 3*77 \text{ (mod 508)}\\\\11^{-1} \equiv 231 \text{ (mod 508)}\\\\[/tex]

Notice how 33*77 = 2541 and 11*231 = 2541

[tex]2541 \equiv 1 \text{ (mod 508)}[/tex] since 2541/508 has a remainder of 1.

So effectively [tex]33*77 \equiv 1 \text{ (mod 508)}[/tex] and [tex]11*231 \equiv 1 \text{ (mod 508)}[/tex]