A heap of coconuts is divided into groups of 2, 3, and 5, and each time exactly 1 coconut is left over. What is the least possible number of coconuts in the heap?

Introduction / Context:
This question is another example of using LCM with a constant remainder. When a number leaves the same remainder when divided by several divisors, subtracting that remainder from the number gives a value that is divisible by all of those divisors. We then look for the smallest such number.

Given Data / Assumptions:

• N mod 2 = 1
• N mod 3 = 1
• N mod 5 = 1
• N is the least positive integer satisfying these conditions

Concept / Approach:
If N leaves remainder 1 in each case, then N - 1 is divisible by 2, 3, and 5. Therefore:
N - 1 is a multiple of LCM(2, 3, 5)
So we can write:
N = LCM(2, 3, 5) * k + 1
for some positive integer k, and we choose the smallest k that gives a positive N.

Step-by-Step Solution:
Step 1: Compute LCM(2, 3, 5).LCM(2, 3, 5) = 2 * 3 * 5 = 30.Step 2: Express N as N = 30k + 1.Step 3: For the least positive N, take k = 1.Then N = 30 * 1 + 1 = 31.

Verification / Alternative check:
Verify N = 31: 31 mod 2 = 1, 31 mod 3 = 1 (since 30 is divisible by 3), and 31 mod 5 = 1 (since 30 is divisible by 5). Any smaller positive candidate would not satisfy all three congruences simultaneously.

Why Other Options Are Wrong:
27: 27 mod 3 = 0, so it does not leave remainder 1 when divided by 3.16: 16 mod 2 = 0, so it fails for divisor 2.63: 63 mod 3 = 0, so it fails for divisor 3.61: 61 leaves remainder 1 for 2, 3, and 5 but it is not the least, because 31 is smaller.

Common Pitfalls:
Using HCF instead of LCM when considering several divisors at once.Forgetting to subtract the remainder (1) before applying the LCM idea.Not paying attention to the requirement that the number must be the smallest possible.

Final Answer:
31