Determine the conditions under which an integer can be uniquely partitioned into two specific multiples using remainder analysis. | Step-by-Step Solution

What We're Solving:

We need to figure out when we can split an integer n into two parts: x (a multiple of i) and y (a multiple of n-i+1), such that x + y = n. Then we'll explore when this split is unique using properties of remainders.

The Approach:

This is a beautiful problem that combines partition theory with modular arithmetic! We're going to:

• 1. Set up the mathematical conditions for such a partition to exist
• 2. Use the Chinese Remainder Theorem concepts to analyze uniqueness
• 3. Explore what happens when we change the parameter i

The key insight is that we're looking for solutions to a system involving two different moduli, which will help us understand when solutions exist and when they're unique.

Step-by-Step Solution:

Step 1: Set up the problem mathematically We want x + y = n where:

• x ≡ 0 (mod i), so x = ki for some integer k
• y ≡ 0 (mod (n-i+1)), so y = j(n-i+1) for some integer j

Substituting: ki + j(n-i+1) = n

Step 2: Analyze existence conditions From ki + j(n-i+1) = n, we can rearrange: ki = n - j(n-i+1) = n(1-j) + j(i-1)

This means n(1-j) + j(i-1) ≡ 0 (mod i) Simplifying: n(1-j) ≡ 0 (mod i)

So we need: n ≡ jn (mod i), which gives us j ≡ 1 (mod i) when gcd(n,i) = i.

Step 3: Examine the uniqueness using GCD properties The uniqueness depends on gcd(i, n-i+1). Here's why:

• If gcd(i, n-i+1) = d, then any solution must satisfy certain divisibility conditions
• When d = 1 (the numbers are coprime), we get unique solutions modulo their product
• When d > 1, we might have multiple solutions or no solutions

Step 4: Apply the Chinese Remainder Theorem framework We're essentially solving:

• n ≡ x (mod i) where x ≡ 0 (mod i)
• n ≡ y (mod (n-i+1)) where y ≡ 0 (mod (n-i+1))

This simplifies to:

• n ≡ 0 (mod i)
• n ≡ 0 (mod (n-i+1))

Step 5: Determine the final conditions For a partition to exist: n must be divisible by both i and (n-i+1) This means n ≡ 0 (mod lcm(i, n-i+1))

For uniqueness: When gcd(i, n-i+1) = 1, the partition is unique modulo i·(n-i+1).

The Answer:

A partition exists if and only if n is divisible by lcm(i, n-i+1).

The partition is unique when:

• gcd(i, n-i+1) = 1, OR
• When considering solutions in the appropriate residue classes

Special cases to note:

• If i and (n-i+1) are coprime, exactly one partition exists when n ≡ 0 (mod i(n-i+1))
• If they share common factors, multiple partitions may exist

Memory Tip:

Think of this like trying to fill two containers of different sizes (i and n-i+1) that together must hold exactly n units. The containers can only be filled in complete "batches" of their respective sizes. The solution exists when n can be perfectly divided between these batch constraints, and it's unique when the batch sizes don't interfere with each other (are coprime)!

This problem beautifully demonstrates how number theory concepts like GCD, LCM, and modular arithmetic work together in partition problems. Great work tackling such an interesting question! 🌟