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.