How many ordered pairs of positive integers (a, b) satisfy 1/a + 1/b = 2/17?

Steps to Solve

1. Eliminate Fractions

To eliminate the fractions in the equation $ \frac{1}{a} + \frac{1}{b} = \frac{2}{17} $, we can find a common denominator for the left side:

$$ \frac{b + a}{ab} = \frac{2}{17} $$

This leads us to:

$$ 17(a + b) = 2ab $$

2. Rearranging the Equation

Rearranging the equation gives us:

$$ 2ab - 17a - 17b = 0 $$

3. Factoring the Equation

Next, we will try factoring the quadratic-like equation. Rewrite it as:

$$ 2ab - 17a - 17b + 289 = 289 $$

Now we can factor the left side as:

$$ (2a - 17)(2b - 17) = 289 $$

4. Finding Factor Pairs of 289

Now, we need to find the pairs of factors of 289. The pairs are:

$$ (1, 289), (17, 17), (289, 1) $$

5. Setting Up Equations From Factor Pairs

Next, we set up equations using each pair. For each factor $(p, q)$, we have:

$$ 2a - 17 = p $$ $$ 2b - 17 = q $$

From this, we solve for $a$ and $b$:

$$ a = \frac{p + 17}{2} $$ $$ b = \frac{q + 17}{2} $$

6. Calculating Values for Each Pair

Now, substitute the factor pairs:

• For $(1, 289)$:

$$ a = \frac{1 + 17}{2} = 9, \quad b = \frac{289 + 17}{2} = 153 $$

• For $(17, 17)$:

$$ a = \frac{17 + 17}{2} = 17, \quad b = \frac{17 + 17}{2} = 17 $$

• For $(289, 1)$:

$$ a = \frac{289 + 17}{2} = 153, \quad b = \frac{1 + 17}{2} = 9 $$

7. Listing Positive Integer Solutions

The positive integer pairs $(a, b)$ that satisfy the equation are:

1. $(9, 153)$
2. $(17, 17)$
3. $(153, 9)$