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Chapter 4: Problem 4

A wheel of fortune has the integers from 1 to 25 placed on it in a random manner. Show that regardless of how the numbers are positioned on the wheel, there are three adjacent numbers whose sum is at least 39 .

Short Answer

Expert verified
Using the Pigeonhole Principle, regardless of how the numbers are positioned on the wheel, it can be proven that there are always three adjacent numbers whose sum is at least 39.

Step by step solution

01

Understanding the problem

We are given a wheel of fortune with the numbers 1 to 25 placed in a random manner. The problem is to prove that there are always three adjacent numbers whose sum is at least 39, regardless of how the numbers are positioned. We are looking for a sequence of three numbers that, when added together, yield a sum of at least 39.
02

Defining the Sum

To solve this problem, we need to consider the sum of three consecutive minterms. Each minterm, according to the pigeonhole principle, needs to be greater than or equal to the total sum divided by the number of terms (i.e., steps). In this case, we can consider the sum of 1 - 25, using the formula for sum of an arithmetic progression, which is \( \frac{n}{2} * (first \ term + last \ term) = \frac{25}{2} * (1 + 25) = 325. \) The calculated sum needs to be divided by the number of sections the wheel of fortune is divided into. Assuming that each gap between numbers represents a 'pigeonhole', the wheel of fortune is divided into 25 sections. So the average sum of three adjacent numbers has to be \( \frac{325}{25} = 13. \)
03

Evaluating the Sum using the Pigeonhole Principle

However, we are looking for a minimum sum of three adjacent numbers, hence we need to multiply this average by 3, which gives us 39. Hence, the minimum sum of sufficient minterms has to be greater than or equal to 39, thus verifying our conclusion that there are always three such numbers, no matter how they are positioned.

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Most popular questions from this chapter

Determine the prime factorization for each of the following: a) \(8 !\) b) \(10 !\) c) \(12 !\)

Determine those values of \(c \in \mathbf{Z}^{+}, 10

a) Develop a recursive definition for the addition of \(n\) real numbers \(x_{1}, x_{2}, \ldots, x_{n}\), where \(n \geq 2\). b) For any real numbers \(x_{1}, x_{2}\), and \(x_{3}\), the associative law of addition states that \(x_{1}+\) \(\left(x_{2}+x_{3}\right)=\left(x_{1}+x_{2}\right)+x_{3}\). Prove that if \(n, r \in \mathbf{Z}^{+}\), where \(n \geq 3\) and \(1 \leq r

Let \(n \in \mathbf{Z}^{+}\)where \(n \geq 2\). Prove that if \(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n} \in \mathbf{Z}^{+}\)and \(a_{i} \mid b_{i}\) for all \(1 \leq i \leq n\), then \(\left(a_{1} a_{2} \cdots a_{n}\right) \mid\left(b_{1} b_{2} \cdots b_{n}\right) .\)

Give a recursive definition for each of the following integer sequences. a) \(2,4,16,256, \ldots\) (or, \(\left.2,2^{2},\left(2^{2}\right)^{2},\left(\left(2^{2}\right)^{2}\right)^{2}, \ldots\right)\) b) \(2,4,16,65536, \ldots\) (or, \(2,2^{2}, 2^{(2)}, 2^{\left(2\left(2^{2}\right)\right)}, \ldots\) )
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