Solving the Reciprocal Sum of Three Consecutive Integers: A Step-By-Step Guide

The problem at hand is to find the sum of three consecutive integers whose reciprocals add up to 37/60. This article explores various methods to solve this problem through algebraic manipulation, demonstrating the application of brute force calculation and trial and error techniques.

Understanding the Problem

The given equation is:

[frac{1}{x-1} frac{1}{x} frac{1}{x 1} frac{37}{60}]

Method 1: Brute Force Calculation

We denote the three consecutive integers as ((n-1)), (n), and ((n 1)). We start by expressing the left side of the equation as a single fraction:

[frac{1}{n-1} frac{1}{n} frac{1}{n 1} frac{n(n 1) n(n-1) (n-1)(n 1)}{n(n-1)(n 1)} frac{n^2 n n^2 - n n^2 - 1}{n(n^2 - 1)} frac{3n^2 - 1}{n^3 - n}]

Setting this equal to 37/60, we get:

[frac{3n^2 - 1}{n^3 - n} frac{37}{60}]

Multiplying both sides by (60(n^3 - n)) yields:

[60(3n^2 - 1) 37(n^3 - n)]

Simplifying, we have:

[180n^2 - 60 37n^3 - 37n]

Bringing all terms to one side, we get:

[37n^3 - 180n^2 - 37n 60 0]

We then factor this equation:

[(n - 5)(37n^2 5n - 12) 0]

The quadratic equation (37n^2 5n - 12 0) does not yield integer solutions. Therefore, the only integer solution is (n 5).

The three consecutive integers are ((5-1)), (5), and ((5 1)), or 4, 5, and 6. Their sum is (4 5 6 15).

Method 2: Trial and Error

To use trial and error, we start by dividing the sum by 3 to find an approximate middle value:

[frac{37}{60} div 3 frac{37}{180} approx frac{1}{5}]

The reciprocal of 5 is (frac{1}{5}). Testing if 5 is the middle integer:

[frac{1}{4} frac{1}{5} frac{1}{6} frac{15}{60} frac{12}{60} frac{10}{60} frac{37}{60}]

The values 4, 5, and 6 satisfy the condition, confirming that the sum of these integers is 15.

Algebraic Verification

We can also solve the equation algebraically using the rational root theorem. Let the integers be (x-1), (x), and (x 1). We have:

[frac{1}{x-1} frac{1}{x} frac{1}{x 1} frac{37}{60}]

Multiplying through by (6(x-1)(x 1)), we get:

[6(x 1) 60(x-1)(x 1) 6(x-1) 37x(x-1)(x 1)]

Simplifying, we have:

[6^2 6 6^2 - 60 6^2 - 6 37(x^3 - x)]

Combining like terms, we obtain:

[18^2 - 60 37x^3 - 37x]

Bringing all terms to one side, we get:

[37x^3 - 18^2 - 37x 60 0]

Testing (x 4), we find that it is a root:

[(4 - 5)(37(4)^2 5(4) - 12) 0]

The values (x 4), (x - 1 3), and (x 1 5) give the integers 3, 4, and 5, which sum to 12. However, the initial method correctly identifies the integers 4, 5, and 6, which sum to 15.

Conclusion

The three consecutive integers whose reciprocals sum to 37/60 are 4, 5, and 6. Therefore, the sum of these integers is:

[4 5 6 15]