Difficulty: Easy
Correct Answer: 56
Explanation:
Introduction / Context:
A classic numeric classification is to check whether items are all powers of a base (most commonly 2 or 3). Here, three numbers are powers of 2; one is not. This yields an immediate 3-to-1 split.
Given Data / Assumptions:
Concept / Approach:
List nearby powers of 2 and compare: 2^1 = 2, 2^4 = 16, 2^7 = 128. Check 56 against this set. If one is not of the form 2^k, it is the outlier.
Step-by-Step Solution:
2 is 2^1 → power of two.16 is 2^4 → power of two.128 is 2^7 → power of two.56 = 7 * 8 = 7 * 2^3, not a pure power of two.
Verification / Alternative check:
Prime factorization confirms: 56 = 2^3 * 7 (extra prime 7), while genuine powers of two have only the prime factor 2. Therefore, 56 uniquely fails the criterion.
Why Other Options Are Wrong:
Common Pitfalls:
Equating multiples of powers of two with exact powers of two. Multiples introduce additional prime factors and break the definition.
Final Answer:
56
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