Understanding the Chocolate Exchange Problem

This problem involves three individuals, A, B, and C, exchanging chocolates in a specific sequence. The key to solving such problems is understanding how the amount of chocolates changes for each person in a round and recognizing that the total number of chocolates remains constant throughout the entire process.

Let's denote the number of chocolates A, B, and C have at any given point as \(A\), \(B\), and \(C\), respectively. The problem describes a unique exchange rule: when a person (say, X) gives chocolates, they give to each of the other two (Y and Z) half the number of chocolates Y and Z already possess.

Let's verify if the total number of chocolates remains constant with this rule. Suppose before a player X gives, the amounts are \(A_{old}\), \(B_{old}\), \(C_{old}\). When X gives:

• X gives \(Y_{old}/2\) chocolates to Y. Y's chocolates become \(Y_{old} + Y_{old}/2 = 3Y_{old}/2\).
• X gives \(Z_{old}/2\) chocolates to Z. Z's chocolates become \(Z_{old} + Z_{old}/2 = 3Z_{old}/2\).
• X's chocolates decrease by the total amount given: \(X_{old} - (Y_{old}/2 + Z_{old}/2)\).

The total chocolates after the exchange would be:

\(X_{new} + Y_{new} + Z_{new} = (X_{old} - Y_{old}/2 - Z_{old}/2) + (3Y_{old}/2) + (3Z_{old}/2)\)

\(= X_{old} - Y_{old}/2 - Z_{old}/2 + 3Y_{old}/2 + 3Z_{old}/2\)

\(= X_{old} + (3Y_{old}/2 - Y_{old}/2) + (3Z_{old}/2 - Z_{old}/2)\)

\(= X_{old} + Y_{old} + Z_{old}\)

Indeed, the total number of chocolates remains constant. This is crucial for solving the problem by working backwards.

Chocolates Distribution: Working Backwards

The problem states that at the end, each person has the same number of chocolates. Let this final amount be \(F\) for each person. So, after all turns, A, B, and C each have \(F\) chocolates. The total number of chocolates will be \(3F\).

Let's denote the amounts before C's turn as \(A_2, B_2, C_2\), before B's turn as \(A_1, B_1, C_1\), and the initial amounts as \(A_0, B_0, C_0\).

Step-by-Step Calculation (Backward)

Round 3: After C's Turn (Final State)

• \(A_3 = F\)
• \(B_3 = F\)
• \(C_3 = F\)

Total chocolates = \(A_3 + B_3 + C_3 = F + F + F = 3F\).

Round 2: Just Before C's Turn

C gave chocolates to A and B. A received \(A_2/2\), and B received \(B_2/2\). So:

• \(A_3 = A_2 + A_2/2 = 3A_2/2 \implies A_2 = 2A_3/3 = 2F/3\)
• \(B_3 = B_2 + B_2/2 = 3B_2/2 \implies B_2 = 2B_3/3 = 2F/3\)

Since the total chocolates remain \(3F\), \(C_2\) can be found:

• \(C_2 = 3F - A_2 - B_2 = 3F - 2F/3 - 2F/3 = 3F - 4F/3 = (9F - 4F)/3 = 5F/3\)

So, just before C's turn, the amounts were: \(A_2 = 2F/3, B_2 = 2F/3, C_2 = 5F/3\).

Round 1: Just Before B's Turn

B gave chocolates to A and C. A received \(A_1/2\), and C received \(C_1/2\). So:

• \(A_2 = A_1 + A_1/2 = 3A_1/2 \implies A_1 = 2A_2/3 = 2(2F/3)/3 = 4F/9\)
• \(C_2 = C_1 + C_1/2 = 3C_1/2 \implies C_1 = 2C_2/3 = 2(5F/3)/3 = 10F/9\)

Since the total chocolates remain \(3F\), \(B_1\) can be found:

• \(B_1 = 3F - A_1 - C_1 = 3F - 4F/9 - 10F/9 = 3F - 14F/9 = (27F - 14F)/9 = 13F/9\)

So, just before B's turn, the amounts were: \(A_1 = 4F/9, B_1 = 13F/9, C_1 = 10F/9\).

Initial State: Just Before A's Turn

A gave chocolates to B and C. B received \(B_0/2\), and C received \(C_0/2\). So:

• \(B_1 = B_0 + B_0/2 = 3B_0/2 \implies B_0 = 2B_1/3 = 2(13F/9)/3 = 26F/27\)
• \(C_1 = C_0 + C_0/2 = 3C_0/2 \implies C_0 = 2C_1/3 = 2(10F/9)/3 = 20F/27\)

Since the total chocolates remain \(3F\), \(A_0\) can be found:

• \(A_0 = 3F - B_0 - C_0 = 3F - 26F/27 - 20F/27 = 3F - 46F/27 = (81F - 46F)/27 = 35F/27\)

So, the initial amounts were: \(A_0 = 35F/27, B_0 = 26F/27, C_0 = 20F/27\).

Determining the Minimum Total Chocolates

For the number of chocolates to be whole numbers (integers) at the start and throughout the process, the value of \(F\) must be a multiple of 27 (the largest denominator in the initial fractions).

To find the minimum number of chocolates they have among themselves initially, we need to find the smallest positive integer value for \(F\) that makes \(A_0, B_0, C_0\) integers.

The smallest such value for \(F\) is 27.

If \(F = 27\):

• \(A_0 = (35 \times 27)/27 = 35\) chocolates
• \(B_0 = (26 \times 27)/27 = 26\) chocolates
• \(C_0 = (20 \times 27)/27 = 20\) chocolates

The minimum total number of chocolates they have among themselves initially is \(A_0 + B_0 + C_0 = 35 + 26 + 20 = 81\).

Verification of Chocolate Exchange

Let's verify the process with the initial amounts \(A_0=35, B_0=26, C_0=20\). The total is 81. The final amount for each should be \(81/3 = 27\).

The final amounts are 27 for each person, and all intermediate amounts are integers. This confirms our calculated minimum initial total of 81 chocolates.