Problem 23
Question
Three gentlemen - Mr. Blue, Mr. Gray, and Mr. White-have shirts and ties that are blue, gray, and white, but not necessarily in that order. No person's clothing has the same color as his last name. Mr. Blue's tie has the same color as Mr. Gray's shirt. What color is Mr. White's shirt? (Mathematics Teacher, 1986)
Step-by-Step Solution
Verified Answer
Mr. White's shirt is blue.
1Step 1: List the given information.
We are given the following information:
1. No person's clothing has the same color as his last name.
2. Mr. Blue's tie has the same color as Mr. Gray's shirt.
2Step 2: Translate information into possible combinations.
Since each gentleman has two clothing items (shirt and tie) with three possible colors (blue, gray, and white), we can label the possible combinations as follows:
1. Mr. Blue's shirt (B_s) and tie (B_t)
2. Mr. Gray's shirt (G_s) and tie (G_t)
3. Mr. White's shirt (W_s) and tie (W_t)
3Step 3: Use the given information to eliminate possibilities.
Given the fact that no person's clothing has the same color as his last name, we get:
1. B_s ≠ Blue; B_s can only be White or Gray.
2. G_s ≠ Gray; G_s can only be Blue or White.
3. W_s ≠ White; W_s can only be Blue or Gray.
Now, from the second given piece of information:
1. Mr. Blue's tie (B_t) has the same color as Mr. Gray's shirt (G_s) i.e., B_t = G_s.
4Step 4: Apply logic to find Mr. White's shirt color.
Since Mr. Blue's tie (B_t) has the same color as Mr. Gray's shirt (G_s), and Mr. Blue is not wearing a blue tie, the only possibilities are:
1. B_t = G_s = White OR
2. B_t = G_s = Gray
If B_t = G_s = White, then Mr. Gray's shirt (G_s) is white, which is not possible since G_s can only be Blue or Gray.
Therefore, B_t = G_s = Gray. This means Mr. Gray's shirt (G_s) is gray.
Now, W_s ≠ White and W_s ≠ Gray (eliminating this from the previous possibilities). So, W_s can only be Blue.
5Step 5: Summarize the findings.
Mr. White's shirt (W_s) is Blue.
Key Concepts
Discrete MathematicsProblem-Solving StrategiesCombinatorics
Discrete Mathematics
Discrete mathematics is a branch of mathematics dealing with objects that can assume only distinct, separated values. It encompasses a wide range of topics, including logic puzzles, graph theory, set theory, combinatorics, and more. In our problem with Mr. Blue, Mr. Gray, and Mr. White, discrete mathematics is at play, particularly in the use of logic to solve the puzzle.
The scenario presents a finite number of possibilities and requires an understanding of how to manage these discrete options logically. By analyzing the information given and methodically eliminating impossibilities, we arrive at one single outcome. This bears the hallmark of a mathematical proof, in which discrete mathematics often finds its utility, turning this fun puzzle into an exercise in logical deduction.
A solid grasp of discrete mathematics empowers students to tackle complex problems by breaking them down into smaller, more manageable parts. It can benefit those working in computer science, cryptography, and information theory, where such skills are essential.
The scenario presents a finite number of possibilities and requires an understanding of how to manage these discrete options logically. By analyzing the information given and methodically eliminating impossibilities, we arrive at one single outcome. This bears the hallmark of a mathematical proof, in which discrete mathematics often finds its utility, turning this fun puzzle into an exercise in logical deduction.
A solid grasp of discrete mathematics empowers students to tackle complex problems by breaking them down into smaller, more manageable parts. It can benefit those working in computer science, cryptography, and information theory, where such skills are essential.
Problem-Solving Strategies
Problem-solving strategies in mathematics are systematic approaches to unraveling puzzles, equations, or any problems that require a solution.
In our logic puzzle, the first step is to comprehend the premise and constraints. Here, we consider the unique stipulation that no one's clothing matches their last name, along with the connection between Mr. Blue's tie and Mr. Gray's shirt.
Next, we organize the puzzle's details. This means representing the possible shirt and tie color combinations for each gentleman, as was done in the step-by-step solution.
Utilizing the constraints, we employ the process of elimination. As shown in the solution, this involves removing impossible color combinations based on the provided facts.
Finally, we use logical reasoning to conclude what must be true given the remaining possibilities, ultimately determining the color of Mr. White's shirt. These strategies not only lead to the solution of puzzles but also strengthen the problem-solving capabilities that are crucial in various scientific and engineering disciplines.
Understanding the Problem
In our logic puzzle, the first step is to comprehend the premise and constraints. Here, we consider the unique stipulation that no one's clothing matches their last name, along with the connection between Mr. Blue's tie and Mr. Gray's shirt.
Organizing Information
Next, we organize the puzzle's details. This means representing the possible shirt and tie color combinations for each gentleman, as was done in the step-by-step solution.
Process of Elimination
Utilizing the constraints, we employ the process of elimination. As shown in the solution, this involves removing impossible color combinations based on the provided facts.
Logical Deduction
Finally, we use logical reasoning to conclude what must be true given the remaining possibilities, ultimately determining the color of Mr. White's shirt. These strategies not only lead to the solution of puzzles but also strengthen the problem-solving capabilities that are crucial in various scientific and engineering disciplines.
Combinatorics
Combinatorics, one of the key arms of discrete mathematics, explores counting, arrangement, and combination of sets of elements. The field is fundamental in probability and is used in logic puzzles like the one posed by the color of the gentlemen's shirts and ties.
We are working with a finite set of elements: colors and clothing items. The exercise has us calculate the number of possible ways these elements can be arranged, adhering to certain rules. Our problem particularly deals with permutations – the different ways in which a set of items can be ordered or arranged.
In the inital scenario, we could think of 3! (3 factorial) possible ways to assign shirt and tie colors, which equals 6 permutations. But the given constraints drastically reduce the possibilities, showcasing the power of combinatorial rules in simplifying complex problems. By applying combinations selectively and ruling out invalid ones, combinatorics is instrumental in reaching the correct solution with precision and efficiency.
We are working with a finite set of elements: colors and clothing items. The exercise has us calculate the number of possible ways these elements can be arranged, adhering to certain rules. Our problem particularly deals with permutations – the different ways in which a set of items can be ordered or arranged.
In the inital scenario, we could think of 3! (3 factorial) possible ways to assign shirt and tie colors, which equals 6 permutations. But the given constraints drastically reduce the possibilities, showcasing the power of combinatorial rules in simplifying complex problems. By applying combinations selectively and ruling out invalid ones, combinatorics is instrumental in reaching the correct solution with precision and efficiency.
Other exercises in this chapter
Problem 22
Rewrite each sentence symbolically, where the UD consists of real numbers. There are real numbers \(x\) and \(y\) such that \(x=2 y\)
View solution Problem 22
Prove by contradiction, where \(p\) is a prime number. \(\log _{10} 2\) is an irrational number.
View solution Problem 23
Prove by cases, where \(n\) is an arbitrary integer and \(|x|\) denotes the absolute value of \(x\). \(n^{2}+n\) is an even integer.
View solution Problem 23
Construct a truth table for each proposition. $$\sim p \vee \sim q$$
View solution