Difficulty: Medium
Correct Answer: 2
Explanation:
Introduction / Context:
This question connects exponents with the geometry of Pythagorean triples. The numbers 9, 40, and 41 suggest a relation similar to a right triangle with sides forming a triple. The task is to find an exponent a such that 9^a + 40^a equals 41^a. Instead of trying random large exponents blindly, we can look for patterns and test small integer exponents.
Given Data / Assumptions:
Concept / Approach:
Recognise that 9, 40, and 41 resemble a Pythagorean triple because 9^2 + 40^2 = 41^2. This suggests that a = 2 might satisfy the equation exactly. We test a = 2 first. If it works, there is no need to try higher values. In general, for a larger exponent, the largest base dominates, making equality very unlikely unless there is a special relationship like the one for squares.
Step-by-Step Solution:
Test a = 2. Compute 9^2 = 81.Compute 40^2 = 1600.Add them: 81 + 1600 = 1681.Compute 41^2: 41^2 = 1681 as well.Thus 9^2 + 40^2 = 41^2, so a = 2 satisfies the equation exactly.There is no need to test other exponents because this provides a perfect equality.
Verification / Alternative check:
To check that other exponents do not work, note that for a = 1 we have 9^1 + 40^1 = 49 while 41^1 = 41, so the left side is greater. For a = 3, 9^3 + 40^3 = 729 + 64000 = 64729, while 41^3 = 68921, and the right side is greater. As a increases, the term with the largest base 41 will dominate more strongly, so 41^a will grow faster than 9^a + 40^a. Therefore, equality occurs only at a = 2 among positive integer exponents.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
2
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