If n is a natural number, then what is the sum of all <strong>DISTINCT</strong> remainders of 4^{n}+6^{n}+9^{n}+11^{n} when divided by 10 for various values of n?

Explanation

Evaluate for different n modulo 10: For n=1, 4+6+9+1 \equiv 20 \equiv 0. For n=2, 6+6+1+1 \equiv 14 \equiv 4. For n=3, 4+6+9+1 \equiv 0. The distinct remainders are 0 and 4. Their sum is 0 + 4 = 4.

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