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Set A contains 3 consecutive prime numbers while set B contains 2 consecutive odd numbers. Sum of largest number of set A and set B is 64 while largest number of set A is 4 less than the 3 times of largest number of set B. Find the smallest number of set A.
- 47
- 41
- 53
- 43
Correct answer: 41
Solution
Let largest of A = a, largest of B = b. Eq1: a + b = 64. Eq2: a = 3b - 4. Substituting: (3b-4) + b = 64 → 4b = 68 → b = 17. So a = 64-17 = 47. Set B = {15, 17} (consecutive odd). Set A = 3 consecutive primes ending at 47: primes are 41, 43, 47. Smallest = 41.
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