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Modulo question?
(fyi, the last one was incorrectly typed)
Prove 2^n is congruent to 1 (mod 3) if and only if n is even and nonnegative.
prove this by induction
base
2^0 = 1 congruent to 1 (mod 3)
assumption
assume 2^n is congruent to 1 (mod 3) for non-neg n
inductive steps
then 2^(n+1) = 2 * 2^n which is congruent to 2 (mod 3)
and 2^(n+2) = 4 * 2^n which is congrunt to 4 (mod 3) is congruent to 1 (mod 3)
Nuremberg Toy Fair 2011: CMC Bugatti Type 57