Question

Coins are to be put into 7 pockets so that each pocket contains at least one coin. At most 3 of the pockets are to contain the same number of coins, and no two of the remaining pockets are to contain an equal number of coins. What is the least possible number of coins needed for the pockets?

• A. 7
• B. 13
• C. 17
• D. 22
• E. 28

Hint:

For minimization problems with integer constraints, always start by using the smallest possible integers (1, 2, 3, ...). Systematically test the different scenarios allowed by the constraints (e.g., 3 same vs. 2 same vs. all different) to find the absolute minimum.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a minimization problem involving number theory and combinatorics. We need to satisfy a set of conditions while making the total sum as small as possible. The key is to use the smallest possible positive integers for the number of coins in each pocket.
Step 2: Key Formula or Approach:
Let the number of coins in the 7 pockets be \(p_1, p_2, p_3, p_4, p_5, p_6, p_7\).
The conditions are:
1. Each \(p_i \ge 1\).
2. "At most 3 pockets" can have the same number of coins. This means we can have a group of 3 identical pockets, or a group of 2, or no groups (all different).
3. "no two of the remaining pockets" have an equal number of coins.
Our goal is to minimize the sum: \(S = \sum_{i=1}^{7} p_i\).
Step 3: Detailed Explanation:
To minimize the total number of coins, we should use the smallest possible integers (1, 2, 3, ...).
Let's analyze the condition "At most 3 pockets are to contain the same number of coins". This gives us the main constraint to build our set of numbers. To minimize the sum, we want to repeat the smallest number (which is 1) as many times as allowed.
The condition allows up to 3 pockets to be the same. So, let's make 3 pockets have the same number of coins. To minimize the total, we should give them the minimum possible number of coins, which is 1.
Case 1: Three pockets have the same number of coins.
- Three pockets contain 1 coin each: \(\{1, 1, 1\}\).
- The remaining \(7 - 3 = 4\) pockets must all have different numbers of coins, and they cannot be equal to 1.
- To minimize the sum, we must choose the next smallest integers available: 2, 3, 4, 5.
- So, the distribution of coins in the 7 pockets is \(\{1, 1, 1, 2, 3, 4, 5\}\).
- The total number of coins is \(1 + 1 + 1 + 2 + 3 + 4 + 5 = 17\).
Let's check other possibilities to be sure this is the minimum.
Case 2: Two pockets have the same number of coins.
- Two pockets contain 1 coin each: \(\{1, 1\}\).
- The remaining \(7 - 2 = 5\) pockets must all have different numbers of coins, not equal to 1.
- We must choose the next smallest integers: 2, 3, 4, 5, 6.
- The distribution is \(\{1, 1, 2, 3, 4, 5, 6\}\).
- The total is \(1 + 1 + 2 + 3 + 4 + 5 + 6 = 22\). This is greater than 17.
Case 3: No pockets have the same number of coins (all are different).
- The 7 pockets must contain 7 distinct numbers of coins. To minimize, we choose the smallest positive integers.
- The distribution is \(\{1, 2, 3, 4, 5, 6, 7\}\).
- The total is \(1 + 2 + 3 + 4 + 5 + 6 + 7 = 28\). This is also greater than 17.
Step 4: Final Answer:
The least possible number of coins is achieved in the first case, which is 17. This corresponds to option (C).