Find the number of solutions to the equation

1/a+1/b+1/c+1/d=1
where a, b, c, d are positive integers and a≤ b≤ c≤ d.

To find the number of solutions to the equation 1/a + 1/b + 1/c + 1/d = 1, where a, b, c, d are positive integers and a ≤ b ≤ c ≤ d, we can employ a systematic approach.

Let's consider the possible values of a. Since a is a positive integer, the minimum value it can take is 1. Suppose a = 1, then the equation becomes 1/1 + 1/b + 1/c + 1/d = 1, or simply 1 + 1/b + 1/c + 1/d = 1. However, for positive integers b, c, and d, it is impossible for the sum of their reciprocals to yield a value less than 1. Hence, a cannot be 1.

Next, let's consider a = 2. Substituting a = 2 into the equation, we get 1/2 + 1/b + 1/c + 1/d = 1. Rearranging the terms, we have 1/b + 1/c + 1/d = 1/2. Notice that if b, c, and d are all greater than or equal to 3, then the sum of their reciprocals will be less than 1/2. Thus, in order for the equation to hold, either b, c, or d must be 2 or less.

Case 1: b = 2
If b = 2, then the equation becomes 1/2 + 1/c + 1/d = 1/2, which simplifies to 1/c + 1/d = 0. Since c and d are positive integers, this equation has no solutions.

Case 2: c = 2
If c = 2, then the equation becomes 1/b + 1/2 + 1/d = 1/2, which simplifies to 1/b + 1/d = 0. Again, since b and d are positive integers, this equation has no solutions.

Case 3: d = 2
If d = 2, then the equation becomes 1/b + 1/c + 1/2 = 1/2. By simplifying, we get 1/b + 1/c = 0. This implies that b and c must be equal to each other (let's call this value x) and satisfy the equation 2/x = 0, which is not possible for any positive integer x.

From these results, we can conclude that there are no solutions to the equation 1/a + 1/b + 1/c + 1/d = 1, where a, b, c, and d are positive integers satisfying a ≤ b ≤ c ≤ d.