the sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of the integers. what is the smallest of the three integers?
To solve this problem, let's assume that the three consecutive positive integers are x, x+1, and x+2.
The reciprocal of a number is obtained by dividing 1 by the number.
According to the problem, the sum of the reciprocals of these three consecutive positive integers is equal to 47 divided by the product of the integers:
1/x + 1/(x+1) + 1/(x+2) = 47/(x * (x+1) * (x+2))
To make the equation easier to solve, we can multiply both sides of the equation by x * (x+1) * (x+2):
(x+1)(x+2) + x(x+2) + x(x+1) = 47
Expanding and simplifying the equation:
(x^2 + 3x + 2) + (x^2 + 2x) + (x^2 + x) = 47
Combining like terms:
3x^2 + 6x + 2 = 47
Rearranging the equation:
3x^2 + 6x - 45 = 0
Dividing both sides of the equation by 3:
x^2 + 2x - 15 = 0
Factoring the quadratic equation:
(x + 5)(x - 3) = 0
Setting each factor equal to 0 and solving for x:
x + 5 = 0 OR x - 3 = 0
x = -5 OR x = 3
Since we are looking for positive integers, we can disregard the solution x = -5.
Therefore, the smallest of the three consecutive positive integers is x = 3.