A heap of coconuts is divided into groups of 2, of 3, and of 5.\nEach time, exactly one coconut is left over.\nWhat is the least possible number of coconuts in the heap?

Introduction:
This is a classic remainder and least common multiple type question, testing your understanding of congruences and modular arithmetic. It asks for the smallest number that leaves the same remainder when divided by several different divisors.

Given Data / Assumptions:

• A heap of coconuts is divided into groups of 2: remainder 1.
• The same heap is divided into groups of 3: remainder 1.
• The heap is also divided into groups of 5: remainder 1.
• We need the least positive number of coconuts satisfying all these conditions.

Concept / Approach:
If a number N leaves remainder 1 when divided by 2, 3, and 5, that means: N ≡ 1 (mod 2) N ≡ 1 (mod 3) N ≡ 1 (mod 5) Equivalently, N - 1 is divisible by 2, 3, and 5. Therefore, N - 1 must be a multiple of the L.C.M. of 2, 3, and 5. The smallest such N is obtained by adding 1 to this L.C.M.

Step-by-Step Solution:
Step 1: Compute LCM(2, 3, 5). Step 2: LCM(2, 3) = 6. Step 3: LCM(6, 5) = 30. Step 4: So N - 1 must be a multiple of 30. Step 5: The smallest positive multiple is 30 itself. Step 6: Therefore, N = 30 + 1 = 31.

Verification / Alternative check:
Check N = 31: 31 ÷ 2 = 15 remainder 1. 31 ÷ 3 = 10 remainder 1. 31 ÷ 5 = 6 remainder 1. All three conditions are satisfied, and since we used the least non-zero multiple of 30, 31 is the least positive solution.

Why Other Options Are Wrong:
63: 63 ÷ 5 leaves remainder 3, not 1. 16: 16 ÷ 3 leaves remainder 1, but 16 ÷ 5 leaves remainder 1, and 16 ÷ 2 leaves 0 remainder, so one condition fails. 27: 27 ÷ 5 leaves remainder 2, not 1. 46: 46 ÷ 3 leaves remainder 1, but 46 ÷ 5 leaves remainder 1 and 46 ÷ 2 leaves 0 remainder, failing the requirement for division by 2.

Common Pitfalls:
Some students test each option directly without recognising the underlying L.C.M. and modular arithmetic shortcut. Others may mistakenly use H.C.F. instead of L.C.M. or forget that N - 1, not N, must be a multiple of 2, 3, and 5.

Final Answer:
The least possible number of coconuts in the heap is 31.