What is the remainder when $21^{32}$ is divided by 50?

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Official Explanation

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We can look for a pattern by calculating the remainders of

21^n

divided by 50 for consecutive values of n.

21^1 = 21 \equiv 21 \mod{50}

21^2 = 441 \equiv 41 \mod{50}

21^3 = 21^2 \times 21 = 41 \times 21 = 861 \equiv 11 \mod{50}

21^4 = 21^3 \times 21 = 11 \times 21 = 231 \equiv 31 \mod{50}

21^5 = 21^4 \times 21 = 31 \times 21 = 651 \equiv 1 \mod{50}

Once we reach a remainder of 1, the cycle repeats. The cycle length is 5, meaning the remainders for

21^1, 21^2, 21^3, 21^4, 21^5

are the same as

21^6, 21^7, 21^8, 21^9, 21^{10}

, and so on.…

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What this question tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Forgetting that 1 is not prime and that every integer divides 0.