The positive difference of the cubes of two consecutive positive integers is 111 less than five
times the product of the two consecutive integers. What is the sum of the two consecutive
integers?
(x+1)^3 - x^3 = 5x(x+1) - 111
solve for x and x+1
Why did the two consecutive integers break up? Because they couldn't cube it anymore! Ba dum tss!
Let's assume the two consecutive positive integers are n and n+1.
According to the problem, the positive difference of the cubes of these integers is 111 less than five times their product.
The cube of the first integer (n) is n^3.
The cube of the second integer (n+1) is (n+1)^3.
The positive difference of their cubes can be written as:
|(n^3) - ((n+1)^3)|
The product of the consecutive integers can be written as n * (n+1).
The problem states that this positive difference is 111 less than five times their product, which can be written as:
|(n^3) - ((n+1)^3)| = 5(n * (n+1)) - 111
Next, let's solve this equation step by step.
Expanding the equation:
(n^3) - ((n+1)^3) = 5n^2 + 5n - 111
Since we are looking for the sum of the two consecutive integers, let's express (n+1) as n + 1:
n^3 - (n^3 + 3n^2 + 3n + 1) = 5n^2 + 5n - 111
Simplifying the equation:
-n^2 - 3n - 1 = 5n^2 + 5n - 111
Combining like terms:
6n^2 + 8n - 110 = 0
Let's solve this quadratic equation using factoring.
Dividing the equation by 2 to simplify:
3n^2 + 4n - 55 = 0
Now let's factor this quadratic equation:
(3n - 11)(n + 5) = 0
Therefore, we have two possible solutions for n:
n = 11/3 OR n = -5
Since we are looking for consecutive positive integers, n cannot be -5.
So, n = 11/3 is not a valid solution in the context of this problem.
Therefore, we are left with n = 11.
Now, let's find the consecutive positive integers:
(n, n+1) = (11, 12)
The sum of these two integers is 11 + 12 = 23.
Therefore, the sum of the two consecutive positive integers is 23.
To solve this problem, let's break it down into steps:
Step 1: Define the consecutive integers
Let's represent the two consecutive positive integers as x and x+1.
Step 2: Set up the equation
The positive difference of the cubes of two consecutive positive integers is given by |(x+1)^3 - x^3|, and five times the product of the two consecutive integers is 5(x)(x+1). According to the problem, the positive difference of the cubes is 111 less than five times the product. So we can set up the equation as:
|(x+1)^3 - x^3| = 5(x)(x+1) - 111
Step 3: Simplify the equation
To simplify the equation, let's expand the cubes:
|(x^3 + 3x^2 + 3x + 1) - x^3| = 5(x)(x+1) - 111
|3x^2 + 3x + 1| = 5x^2 + 5x - 111
Step 4: Eliminate the absolute value
To eliminate the absolute value, we can set up two cases: one where the expression inside the absolute value is positive, and another where it is negative.
Case 1: 3x^2 + 3x + 1 = 5x^2 + 5x - 111
Case 2: 3x^2 + 3x + 1 = -(5x^2 + 5x - 111)
Step 5: Solve the equations
Solving each case separately will give us the possible values for x.
Case 1: 3x^2 + 3x + 1 = 5x^2 + 5x - 111
Rearranging the equation:
2x^2 + 2x - 112 = 0
Dividing by 2:
x^2 + x - 56 = 0
Factoring the quadratic equation:
(x + 8)(x - 7) = 0
Solving for x, we have two possible solutions:
x = -8 or x = 7
Case 2: 3x^2 + 3x + 1 = -(5x^2 + 5x - 111)
Rearranging the equation:
8x^2 + 8x - 112 = 0
Dividing by 8:
x^2 + x - 14 = 0
Factoring the quadratic equation:
(x + 4)(x - 3) = 0
Solving for x, we have two more possible solutions:
x = -4 or x = 3
Step 6: Find the sum of the consecutive integers
Out of the four possible solutions we obtained, we only consider the positive integers since we are looking for consecutive positive integers. Therefore, the two positive solutions are x = 7 and x = 3.
So, the sum of the two consecutive integers is 7 + 8 = 15.
Therefore, the sum of the two consecutive positive integers is 15.