A heap of coconuts is divided into groups of 2, 3 and 5 and each time one coconut is left over. The least number of Coconuts in the heap is_______?
Correct answer: A. 31
- A. 31
- B. 41
- C. 51
- D. 61
Explanation
The number must leave remainder 1 when divided by 2, 3 and 5, so it is 1 more than a multiple of LCM(2, 3, 5) = 30. The least heap-sized number greater than these divisors is 30 + 1 = 31 coconuts. The likely mistake is option b, 41, which is not 1 more than a multiple of 30.
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