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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.)

1 Answer

4 votes

Answer: 231

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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)}

answered
User IsidroGH
by
8.1k points
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