How many ordered quadruples of distinct positive integers (a,b,c,d) are there such that 1/a+1/b+1/c+1/d=1?
multiply all by abcd to get
bcd+acd+abd+abc = abcd.
To find the number of ordered quadruples (a, b, c, d) such that 1/a + 1/b + 1/c + 1/d = 1, we can use a combinatorial approach.
First, let's understand the problem:
We are looking for distinct positive integers: a, b, c, and d.
The sum of their reciprocals should equal 1: 1/a + 1/b + 1/c + 1/d = 1.
The quadruples are ordered, which means (a, b, c, d) and (b, a, c, d) are distinct.
Now, let's break down the problem into simpler steps:
Step 1: Rearrange the equation
1/a + 1/b + 1/c + 1/d = 1
Multiply both sides of the equation by (abcd) to eliminate the denominators:
(bcd) + (acd) + (abd) + (abc) = abcd
Step 2: Count the solutions using combinatorics
We need to find the number of solutions to the equation (bcd) + (acd) + (abd) + (abc) = abcd, where a, b, c, and d are distinct positive integers.
To count the ordered quadruples, we will fix the value of a and then determine the values of b, c, and d.
For each fixed value of a, we can choose b, c, and d from the remaining (a-1) numbers.
Since b, c, and d are distinct positive integers, there are (a-1) choices for b, (a-2) choices for c, and (a-3) choices for d.
The number of ordered quadruples for each value of a is:
(a-1) * (a-2) * (a-3)
Step 3: Sum up the possibilities for different values of a
To find the total number of ordered quadruples, we need to sum up the possibilities for different values of a.
Since a can take on any value from 4 to infinity, we sum up the values of (a-1)*(a-2)*(a-3) as a goes from 4 to infinity.
The total number of ordered quadruples is:
∑[(a-1)*(a-2)*(a-3)], where a goes from 4 to infinity.
Note: Calculating this infinite sum is not practical. In practice, we can determine the number of quadruples by limiting a to a certain large value, which would give us an approximation of the total count.