Question

HELP ASAP


`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.)`

Answer 1

Answer: 231

===================================================

Work Shown:

`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)}\\\\`

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

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

So effectively
`33*77 \equiv 1 \text{ (mod 508)}` and
`11*231 \equiv 1 \text{ (mod 508)}`