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?

Let's assume the three consecutive positive integers as x, x+1, and x+2.

According to the given information, the sum of their reciprocals 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 simplify the equation, we can multiply both sides by x * (x+1) * (x+2):

(x+1)(x+2) + x(x+2) + x(x+1) = 47

Now, let's solve the equation step by step:

Expanding the left side of the equation:
(x^2 + 3x + 2) + (x^2 + 2x) + (x^2 + x) = 47

Combining like terms:
3x^2 + 6x + 2 = 47

Subtracting 47 from both sides:
3x^2 + 6x + 2 - 47 = 0

Simplifying:
3x^2 + 6x - 45 = 0

Dividing both sides by 3 to simplify:
x^2 + 2x - 15 = 0

Now we can factor the quadratic equation:
(x + 5)(x - 3) = 0

Setting each factor equal to zero:
x + 5 = 0 or x - 3 = 0

Solving for x:
If x + 5 = 0, then x = -5
If x - 3 = 0, then x = 3

Since the problem asks for the smallest positive integer, the answer is x = 3.

Therefore, the smallest of the three integers is 3.

To solve this problem, let's first represent the three consecutive positive integers. Let's assume that the smallest of the three integers is "n". Then the next two integers would be "n + 1" and "n + 2".

The sum of the reciprocals of these three consecutive positive integers is given as:

1/n + 1/(n + 1) + 1/(n + 2)

According to the problem, this sum is equal to 47 divided by the product of the integers:

47/(n * (n + 1) * (n + 2))

So, we have the equation:

1/n + 1/(n + 1) + 1/(n + 2) = 47/(n * (n + 1) * (n + 2))

To find the smallest of the three integers, we need to solve this equation.

To simplify the equation, we can cross-multiply to eliminate the fractions:

(n * (n + 1) * (n + 2) * (1 / n)) + (n * (n + 1) * (n + 2) * (1 / (n + 1))) + (n * (n + 1) * (n + 2) * (1 / (n + 2))) = 47

Now, we can simplify further by canceling out common factors:

((n + 1) * (n + 2)) + (n * (n + 2)) + (n * (n + 1)) = 47

Simplifying the left side of the equation gives:

(n^2 + 3n + 2) + (n^2 + 2n) + (n^2 + n) = 47

Combining like terms:

3n^2 + 6n + 2 = 47

Rearranging the equation:

3n^2 + 6n - 45 = 0

To solve this quadratic equation, we can either factor or use the quadratic formula. Factoring this equation, we get:

(3n - 9)(n + 5) = 0

Setting each factor equal to zero:

3n - 9 = 0 or n + 5 = 0

Solving for "n" in each equation:

3n = 9 or n = -5

So, we have two possible values for "n": n = 3 or n = -5.

Since we're looking for the smallest positive integer, n = 3 is the answer.

Therefore, the smallest of the three consecutive positive integers is 3.

1/(x-1) + 1/x + 1/(x+1) = 47/[(x-1)x(x+1)]

put all over a common denominator of (x-1)x(x+1). The numerators then satisfy

x(x+1) + (x-1)(x+1) + (x-1)x = 47
3x^2 - 1 = 47
3x^2 = 48
x = 4

check:
1/3 + 1/4 + 1/5 = 47/60