Find the last two digits of the number 2151^415.

Introduction / Context:
This question involves modular arithmetic and cyclic patterns in the last digits of powers. Instead of computing the full value of 2151^415, which is enormous, we only need the last two digits. This is a typical number theory problem where understanding remainders and repeating patterns makes the work manageable.

Given Data / Assumptions:

• We need the last two digits of 2151^415.
• The last two digits of a number are determined by its remainder when divided by 100.
• 2151 and 100 are relatively large, but modular arithmetic allows simplification.

Concept / Approach:
The last two digits of a number depend only on the last two digits of its base. Therefore, 2151^415 and 51^415 have the same last two digits. We then analyze the pattern of powers of 51 modulo 100. If this pattern repeats with a short cycle, we can reduce the exponent 415 modulo the cycle length and find the result with a few small computations.

Step-by-Step Solution:
Reduce the base modulo 100: 2151 ≡ 51 (mod 100). So we only need the last two digits of 51^415.Compute 51^1 mod 100: 51.Compute 51^2 = 51 × 51 = 2601, so 51^2 ≡ 2601 mod 100 ≡ 1 (mod 100).Thus, 51^2 ≡ 1 (mod 100). This means every even power of 51 is congruent to 1 modulo 100 and every odd power is congruent to 51 modulo 100.Since 415 is odd, we can write 415 = 2k + 1. Therefore, 51^415 = 51^(2k + 1) = (51^2)^k × 51 ≡ 1^k × 51 ≡ 51 (mod 100).Hence, the last two digits of 51^415, and therefore of 2151^415, are 51.

Verification / Alternative check:
We can verify the pattern further by checking 51^3 = 51 × 51^2. Since 51^2 ≡ 1 (mod 100), we have 51^3 ≡ 51 × 1 = 51 (mod 100), consistent with the odd-power rule.Similarly, 51^4 = (51^2)^2 ≡ 1^2 = 1 (mod 100), confirming the cycle of length 2 in the last two digits.

Why Other Options Are Wrong:
Any alternative last two digits like 81, 61 or 91 would require a different remainder pattern modulo 100, which contradicts the strong cycle 51, 01, 51, 01 already observed for consecutive powers of 51.Since we have a clear algebraic argument based on 51^2 ≡ 1 (mod 100), the options 81, 61 and 91 cannot be correct.

Common Pitfalls:
A common mistake is to look only at the last digit and assume a pattern based on modulo 10 instead of modulo 100. That fails when the question asks specifically for two last digits.Another pitfall is attempting to compute extremely large powers directly or with approximate calculators, which is unnecessary and error prone.

Final Answer:
The last two digits of 2151^415 are 51.