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Chapter 1: Problem 4

Prove that with just 3 -cent and 5 -cent stamps, you can make any amount of postage less than 35 cents (any natural number of cents) except 1 cent, 2 cents, 4 cents, and 7 cents.

Short Answer

Expert verified
We can use mathematical induction to prove we can make all postage values above 7 cents but not 1, 2, 4, or 7 cents.

Step by step solution

01

Understanding the Problem

We need to demonstrate that for any amount of postage less than 35 cents (and greater than 7 cents), we can achieve the exact value using only 3-cent and 5-cent stamps, while explicitly excluding 1, 2, 4, and 7 cents.
02

Identifying Unreachable Values

Let's check small values one by one to identify those that cannot be formed using 3 and 5-cent stamps. It's obvious that we cannot make 1 cent, 2 cents, 4 cents, or 7 cents because each of these values is smaller than the smallest stamp (3 cents) or cannot be composed using the given stamps due to the lack of a combination that sums to these values.
03

Base Cases for Values Above 7

Check numbers from 8 to 12: - 8 = 5 + 3 - 9 = 3 + 3 + 3 - 10 = 5 + 5 - 11 = 5 + 3 + 3 - 12 = 3 + 3 + 3 + 3 Notice that each value between these is achievable using combinations of 3 and 5.
04

Inductive Step Explanation

Let's assume that for some number k (where k < 35), we have shown it is possible to make k, k+1, k+2, k+3, and k+4 using 3 and 5-cent stamps. Now show that k+5 can also be achieved, as adding another 5-cent stamp to the combination for k covers k+5. This inductive step shows that once 8 to 12 are possible, stamping remains possible for all subsequent values up to 34 by continuously adding 5-cent stamps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductive Reasoning
Inductive reasoning is a powerful tool in mathematics that helps prove statements about seemingly infinite possibilities. The process involves two main steps: verifying a base case and then proving an inductive step. The base case is a specific instance of the general situation, and it's the starting point for the reasoning.
  • In our step-by-step example, the base cases were the numbers 8 through 12. We showed that each of these could be created using combinations of 3-cent and 5-cent stamps, such as 8 being 5+3, 9 being 3+3+3, and so forth.

Moving on to the inductive step, if you can prove that if a certain statement holds for a number k, it holds for k+1, k+2, ..., k+4, then adding another logical step (k+5 in this case) means it will continue to hold.
  • This essentially allows us to climb a staircase of reasoning, confirming that each subsequent step can be achieved due to the validity of previous ones.

In our stamp problem, by showing that once we reach 8 to 12 system, using a 5-cent stamp can extend it to 13-17, and so forth up to 34.
Combinatorial Problem Solving
Combinatorial problem solving revolves around finding ways to assemble or combine elements to achieve certain outcomes. In our postage stamp exercise, we deal with two types of stamps (3-cent and 5-cent) and must find all the combinations that add up to every value under 35 except the specified exceptions.
  • The strategy is to identify both direct combinations (like using two 5-cent stamps to make 10) and mixed ones (like using two 3-cent and one 5-cent to make 11).
  • This systematic approach helps us discover the number of possible elements (or stamps, in this case) for each value.

This problem relies heavily on the understanding and manipulation of arithmetic to derive all possible configurations of a particular sum, akin to solving a puzzle with specific rules. By breaking the problem into smaller cases and verifying each one against the constraint (less than 35 cents), we harness combinatorial techniques to solve it efficiently.
Mathematical Proofs
Mathematical proofs are the backbone of establishing the veracity of statements in mathematics. They rely on logic and previously established truths to show that something is always true.
  • In the case of our stamp problem, we prove that all amounts from 8 to 34 can be made with 3-cent and 5-cent stamps.
  • Using a mathematical proof by induction, we extend established smaller truths (our base cases from 8 to 12) to cover larger numbers up to 34 through methodical reasoning.

Proofs can follow different styles, such as direct proof, contradiction, or induction, like in our problem. The beauty of a mathematical proof lies in its rigor and its ability to establish a concept's truth indefinitely once the proof is correctly applied and the foundational assumptions are valid.
Stamp Problem
The stamp problem is a classic exercise in discrete mathematics, often used to illustrate the concepts of inductions, combinatorics, and problem-solving.
  • In our scenario, the challenge is to use only two types of stamps to produce all possible postage combinations under 35 cents, while understanding and stating why certain small numbers can't be formed.
  • By demonstrating the combinations that work and proving through induction that you can always get the next value by adding more 5-cent stamps, we solve the problem systematically.

This problem enhances the understanding of constructing combinations and finding patterns in seemingly abstract tasks. Real-world applications of such problems are numerous, ranging from algorithm design to resource allocation, where similar logic and reasoning are essential to find efficient solutions.

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Most popular questions from this chapter

Recall that in the definition of a boolean algebra, we did not require that \(T, \perp,\) and each \(\neg x\) be specified; we merely said they must exist. So, it is natural to ask whether there might be several clements that could equally well be chosen as \(T\) or \(\perp\) or, for some clement \(x\) of the boolean algebra, several different possible choices for \(\neg x\), Show that in a complemented laftice: (a) There is only one possible choice of elements \(T\) and \(\perp\) satisfying the definition of a complemented lattice, (Hint: Suppose there were two possible choices for \(T\). say, \(T_{1}\) and \(T_{2}\). Evaluate \(T_{1} \wedge T_{2}\) in two different ways. ) (b) For each element \(x\) of a complemented, distributive lattice, there is only one possible choice for \(\neg x\) that satisfies the definition of \(\neg x\). (Hint: Suppose there were two choices, say, \(\neg x_{1}\) and \(\neg x_{2},\) for \(\neg x\). Find two ways to evaluate \(\neg x_{1} \wedge x \vee \neg x_{2}\).)

Challenge: Exactly where is the mistake in the following proof that all personal computers are the same brand? Let \(\mathcal{T}=\\{n \in \mathbb{N}: n \geq 1\) and in every set of \(n\) personal computers, all the personal computers are the same brand \(\\} .\) Prove by induction that for every natural number \(n\) such that \(n \geq 1\) is in \(T\). (Base step) \(1 \in T\), since, trivially, if a set of personal computers contains only one computer, then every (one) computer in the set has the same brand. (Inductive step) Suppose \(n \in T\). We need to show \(n+1 \in T\). So, let \(P\) be any set of \(n+1\) personal computers. Pick any computer \(c \in P ;\) we need to show that every computer in \(P\) is the same brand as \(c\). So, let \(d\) be any computer in \(P\). If \(d=c\), then, trivially, \(d\) and \(c\) are the same brand. Otherwise, \(c \in P-(d\\} .\) The set \(P-(d)\) contains \(n\) computers, so by inductive hypothesis, all the computers in \(P-(d]\) are the same brand. Furthermore, \(d \in P-\mid c\\},\) and. also by inductive hypothesis, all the computers in \(P-\\{c\\}\) are the same brand. Now, let \(e\) be a computer in both \(P-\mid c\\}\) and \(P-[d\\}\). Then, \(d\) is the same brand as \(e,\) and \(c\) is the same brand as \(e\). Therefore, \(d\) is the same brand as \(c\).

Find the number of integers between 1 and 1000 , including 1 and 1000 , that are not divisible by any of \(4,5,\) or 6.

For natural number exponents and nonzero bases, most of the familiar laws of exponents can be proved by induction on the exponents using the facts that \(b^{0}=1\) (for \(b \neq 0\) ) and \(b^{n+1}=b \cdot b^{n}\), Assuming that \(m\) and \(n\) are natural numbers and both \(r\) and \(s\) are nonzero real numbers, prove the following: (a) \(r^{m+n}=r^{m} \cdot r^{n}\) (b) \(r^{m n}=\left(r^{m}\right)^{n}\). (c) If \(r>1,\) then \(r^{m}>r^{n}\) if and only if \(m>n\). (d) If \(n, r, s>0,\) then \(r^{n}>s^{n}\) if and only if \(r>s\).

The terms of a sequence are given recursively as \(a_{0}=2, a_{1}=6,\) and \(a_{n}=2 a_{n-1}+\) \(3 a_{n-2}\) for \(n \geq 2\). Find the first eight terms of this sequence.
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