Understanding the Problem: Finding Natural Number Solutions

The question asks for the number of distinct possible values for the sum \((p + q + r)\), where p, q, and r are natural numbers that satisfy the equation \(1/p + 1/q + 1/r = 1\). Natural numbers are positive integers (1, 2, 3, ...).

Solving the Equation 1/p + 1/q + 1/r = 1 for Natural Numbers

We are looking for positive integer solutions \((p, q, r)\) to the equation \( \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1 \). These kinds of equations are sometimes referred to as Diophantine equations involving unit fractions.

Since p, q, and r are natural numbers, they must be greater than or equal to 1. The terms \(1/p\), \(1/q\), and \(1/r\) are all positive.

Let's assume without loss of generality that \( p \le q \le r \). This ordering helps us systematically find all unique sets of values for \((p, q, r)\). Once we find these sets, we can find their sums \((p+q+r)\) and count the number of distinct sum values.

Establishing Bounds for p, q, and r

Since \( p \le q \le r \), we have \( \frac{1}{p} \ge \frac{1}{q} \ge \frac{1}{r} \). Using this in the equation:

\( 1 = \frac{1}{p} + \frac{1}{q} + \frac{1}{r} \le \frac{1}{p} + \frac{1}{p} + \frac{1}{p} = \frac{3}{p} \)

This inequality \( 1 \le \frac{3}{p} \) implies \( p \le 3 \). So, p can only be 1, 2, or 3.

However, if \( p = 1 \), the equation becomes \( \frac{1}{1} + \frac{1}{q} + \frac{1}{r} = 1 \), which simplifies to \( 1 + \frac{1}{q} + \frac{1}{r} = 1 \). This gives \( \frac{1}{q} + \frac{1}{r} = 0 \). Since q and r are natural numbers, \(1/q\) and \(1/r\) are positive, so their sum cannot be 0. Therefore, p cannot be 1.

Thus, the only possible values for p (under the assumption \( p \le q \le r \)) are 2 and 3.

Case 1: p = 2

If \( p = 2 \), the equation becomes \( \frac{1}{2} + \frac{1}{q} + \frac{1}{r} = 1 \). Subtracting \(1/2\) from both sides, we get \( \frac{1}{q} + \frac{1}{r} = 1 - \frac{1}{2} = \frac{1}{2} \).

Now, considering \( p \le q \le r \), we have \( 2 \le q \le r \).

Using \( \frac{1}{q} + \frac{1}{r} = \frac{1}{2} \) and \( q \le r \), we have \( \frac{1}{2} = \frac{1}{q} + \frac{1}{r} \le \frac{1}{q} + \frac{1}{q} = \frac{2}{q} \).

This inequality \( \frac{1}{2} \le \frac{2}{q} \) implies \( q \le 4 \). So, the possible values for q when \( p = 2 \) and \( q \ge p \) are 2, 3, or 4.

• If \( q = 2 \): \( \frac{1}{2} + \frac{1}{r} = \frac{1}{2} \implies \frac{1}{r} = 0 \). This is not possible for a natural number r. So q cannot be 2.
• If \( q = 3 \): \( \frac{1}{3} + \frac{1}{r} = \frac{1}{2} \implies \frac{1}{r} = \frac{1}{2} - \frac{1}{3} = \frac{3 - 2}{6} = \frac{1}{6} \). So \( r = 6 \). This gives the solution \((p, q, r) = (2, 3, 6)\). This satisfies \( 2 \le 3 \le 6 \).
• If \( q = 4 \): \( \frac{1}{4} + \frac{1}{r} = \frac{1}{2} \implies \frac{1}{r} = \frac{1}{2} - \frac{1}{4} = \frac{2 - 1}{4} = \frac{1}{4} \). So \( r = 4 \). This gives the solution \((p, q, r) = (2, 4, 4)\). This satisfies \( 2 \le 4 \le 4 \).

So, for \( p=2 \), the unique ordered solutions with \( p \le q \le r \) are \((2, 3, 6)\) and \((2, 4, 4)\).

Case 2: p = 3

If \( p = 3 \), the equation becomes \( \frac{1}{3} + \frac{1}{q} + \frac{1}{r} = 1 \). Subtracting \(1/3\) from both sides, we get \( \frac{1}{q} + \frac{1}{r} = 1 - \frac{1}{3} = \frac{2}{3} \).

Now, considering \( p \le q \le r \), we have \( 3 \le q \le r \).

Using \( \frac{1}{q} + \frac{1}{r} = \frac{2}{3} \) and \( q \le r \), we have \( \frac{2}{3} = \frac{1}{q} + \frac{1}{r} \le \frac{1}{q} + \frac{1}{q} = \frac{2}{q} \).

This inequality \( \frac{2}{3} \le \frac{2}{q} \) implies \( q \le 3 \). So, the only possible value for q when \( p = 3 \) and \( q \ge p \) is 3.

• If \( q = 3 \): \( \frac{1}{3} + \frac{1}{r} = \frac{2}{3} \implies \frac{1}{r} = \frac{2}{3} - \frac{1}{3} = \frac{1}{3} \). So \( r = 3 \). This gives the solution \((p, q, r) = (3, 3, 3)\). This satisfies \( 3 \le 3 \le 3 \).

So, for \( p=3 \), the unique ordered solution with \( p \le q \le r \) is \((3, 3, 3)\).

Case 3: p > 3

As shown earlier, the inequality \( p \le 3 \) must hold for natural number solutions. Therefore, there are no solutions where \( p > 3 \).

Listing the Unique Solutions (p, q, r)

The unique sets of natural numbers \((p, q, r)\) satisfying \( \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1 \) (when ordered \( p \le q \le r \)) are:

• (2, 3, 6)
• (2, 4, 4)
• (3, 3, 3)

Any permutation of these sets will also satisfy the original equation, but they will result in the same sum \(p+q+r\).

Calculating Possible Values of (p + q + r)

Now we calculate the sum \( p + q + r \) for each of the unique sets:

• For \((2, 3, 6)\): \( p + q + r = 2 + 3 + 6 = 11 \)
• For \((2, 4, 4)\): \( p + q + r = 2 + 4 + 4 = 10 \)
• For \((3, 3, 3)\): \( p + q + r = 3 + 3 + 3 = 9 \)

The possible values for \((p + q + r)\) are 9, 10, and 11.

There are 3 distinct possible values for \((p + q + r)\).

The set of possible sum values is \{9, 10, 11\}. The number of possible values is the number of elements in this set.

Number of possible values = 3.

Revision Table: Key Steps to Finding Solutions

Additional Information: Unit Fractions and the Erdos-Straus Conjecture

The equation \( \frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1 \) is a specific example of an equation involving unit fractions (fractions with a numerator of 1). This type of equation appears in number theory.

Finding integer solutions to equations involving sums of unit fractions is related to covering systems and Egyptian fractions.

A famous unsolved problem related to unit fractions is the Erdos-Straus Conjecture, which states that for all integers \(n \ge 2\), the equation \( \frac{4}{n} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) has positive integer solutions for x, y, and z. Our problem is a simpler case where the sum equals 1.

Our systematic approach using the ordered assumption \(p \le q \le r\) is a standard method for finding all integer solutions to such equations where the number of terms is fixed and the sum is positive.