Problem 60 - Combinatorics Through Guided Discovery

Question

American coins are all marked with the year in which they were made. How many coins do you need to have in your hand to guarantee that on two (at least) of them, the date has the same last digit? (When we say "to guarantee that on two (at least) of them,..." we mean that you can find two with the same last digit. You might be able to find three with that last digit, or you might be able to find one pair with the last digit 1 and one pair with the last digit \(9,\) or any combination of equal last digits, as long as there is at least one pair with the same last digit.)

Step-by-Step Solution

Answer

At least 11 coins are needed.

Step 1: Understand the problem

Determine how many coins are needed to guarantee that at least two coins have dates ending with the same last digit.

Step 2: Identify the range of last digits

American coins can have dates ending in any of the digits from 0 to 9. Therefore, there are 10 possible last digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Step 3: Use the pigeonhole principle

According to the pigeonhole principle, if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Here, the 'pigeons' are the coins, and the 'pigeonholes' are the possible last digits.

Step 4: Calculate the minimum number of coins needed

To ensure at least two coins share the same last digit, we must have more coins than there are unique last digits. Since there are 10 unique last digits, we need at least 11 coins. With 11 coins, we are guaranteed that at least two coins will share the same last digit.

Key Concepts

Combinatorics

Combinatorics is the branch of mathematics dealing with counting, arrangement, and combination of objects. In this exercise, combinatorics helps us understand how to distribute coins based on their last digits. The pigeonhole principle, a fundamental concept in combinatorics, states that if you have more items than containers, at least one container must hold more than one item. For our problem, the 'items' are the coins and the 'containers' are the possible last digits (0 through 9). With 10 unique last digits, having at least 11 coins means one digit will surely repeat.

Number Theory

Number theory explores the properties and relationships of numbers, especially integers. A key number theory concept in this problem is the 'last digit' of a number, which provides a simplified way to classify and compare numbers. By focusing on the last digit of the coin's year, we reduce the problem's complexity. Each coin's year can end in one of 10 digits (0-9), forming a finite and manageable set. Thus, understanding modular arithmetic, or congruence, helps in dealing with repetitive cycles like last digits.

Problem Solving

Problem solving in mathematics involves understanding the problem, devising a plan, carrying out the plan, and reviewing/reflecting on the solution. Let's break down the steps for this exercise:

1. **Understand the Problem:** We need to guarantee at least two coins have the same last digit.

2. **Devise a Plan:** Use the pigeonhole principle to determine the minimum number of coins needed.

3. **Carry Out the Plan:** Identify that there are 10 possible last digits (0-9). Calculate using the pigeonhole principle: with 11 coins, at least two must share the same last digit.

4. **Review the Solution:** Verify by ensuring all digits 0-9 can be unique with up to 10 coins. Adding the 11th coin forces a repeat of one digit.