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Chapter 3: Problem 3

Show that the sum of any three consecutive integers is divisible by 3 .

Short Answer

Expert verified
The sum of any three consecutive integers is divisible by 3 because it can be expressed as 3(n + 1).

Step by step solution

01

Identify the Three Consecutive Integers

Let the three consecutive integers be denoted as , n+1, and n+2 where n is an integer.
02

Write the Sum of the Three Integers

The sum of these three integers can be written as: + (n+1) + (n+2).
03

Simplify the Expression

Combine like terms in the expression: + (n + 1) + (n + 2) = 3n + 3.
04

Factor the Sum

Factor out the common term: 3n + 3 = 3(n + 1).
05

Conclude Divisibility by 3

Since the sum 3(n + 1) is clearly a multiple of 3, it follows that the sum of any three consecutive integers is divisible by 3.

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Key Concepts

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

Consecutive Integers
Consecutive integers are numbers that follow each other in order. For example, in the sequence 3, 4, and 5, each number increases by one. In mathematical exercises, consecutive integers are often represented using a variable and its increments. If we let the first integer be denoted as , then the next two consecutive integers can be written as +1 and +2. This representation is simple and effective in solving problems that involve relationships between these numbers.
Divisibility
Divisibility refers to the ability of one number to be divided by another without leaving a remainder. For example, 6 is divisible by 3 because 6 梅 3 = 2 with no remainder. In the context of our problem, we need to show that the sum of any three consecutive integers is divisible by 3. To understand this concept better, consider the sum we derived in the exercise: + (n+1) + (n+2) = 3n + 3. By factoring out the 3, we get 3(n + 1). The presence of 3 as a factor ensures that the expression is divisible by 3, since any number multiplied by 3 is naturally divisible by 3. This is a fundamental property of numbers and is key to solving many integer-related problems.
Factoring
Factoring is the process of breaking down an expression into a product of its simpler factors. This process simplifies complex expressions and is crucial in proving properties such as divisibility. In our exercise, after writing the sum of three consecutive integers as + (n+1) + (n+2), we simplified it to 3n + 3. The next step involved factoring out the common term, which in this case is 3. Factoring out 3 from 3n + 3, we get: 3(n + 1). This is a simpler expression that clearly shows the original sum is a multiple of 3. Such techniques make it easier to demonstrate and understand various mathematical properties.

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Most popular questions from this chapter

Prove that the sum of two rational numbers is a rational number

Prove that the sum of an odd number and an even number is odd.

Recall that a quadratic equation \(a x^{2}+b x+c=0\) has two real solutions if and only if the discriminant \(b^{2}-4 a c\) is positive. Prove that if \(a\) and \(c\) have different signs then the quadratic equation has two real solutions.

The trichotomy property of the real numbers simply states that every real number is either positive or negative or zero. Trichotomy can be used to prove many statements by looking at the three cases that it guarantees. Develop a proof (by cases) that the square of any real number is non-negative.

Prove that every prime number other than 2 and 3 has the form \(6 q+1\) or \(6 q+5\) for some integer \(q\). (Hint: this problem involves thinking about cases as well as contrapositives.)
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