The set of residue classes modulo 5 is equipped with the usual operations of multiplication and addition of residue classes. If we denote -a for the additive inverse of element a with respect to addition, and a^-1 for the multiplicative inverse of element a with respect to multiplication, then the following holds: (a) 2^-1 + (-3) = 4 (b) (-4) * 4 = 4 (c) 3 + 3^-1 = 1 (d) (-1) * 1 = 0 (e) (-2) * 4^-1 = 3.

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A: Solution attached below. Summary.... Answer is an = 5(n-1) + 100

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