<p>If <i>x</i> and <i>y</i> are positive integers and 2<i>xy</i> = 2009 − 3<i>y</i>, then the number of ordered pairs (<i>x</i>, <i>y</i>) is</p>
Step-by-Step Solution
Key Concept: Rearrange to isolate one variable, then use divisibility: the denominator must divide the constant term. Find all valid divisors and check which yield integer solutions.
<p><strong>Step 1:</strong> Rearrange: 2<i>xy</i> + 3<i>y</i> = 2009, so <i>y</i>(2<i>x</i> + 3) = 2009.</p><p><strong>Step 2:</strong> Thus, <i>y</i> = 2009/(2<i>x</i> + 3). For <i>y</i> to be a positive integer, 2<i>x</i> + 3 must divide 2009.</p><p><strong>Step 3:</strong> Since 2009 = 7² × 41, the divisors are: 1, 7, 49, 41, 287, 2009.</p><p><strong>Step 4:</strong> For each divisor <i>d</i> of 2009: 2<i>x</i> + 3 = <i>d</i>, so <i>x</i> = (<i>d</i> − 3)/2. We need <i>x</i> to be a positive integer, so <i>d</i> must be odd and <i>d</i> > 3.</p><p>• 2<i>x</i> + 3 = 7 ⟹ <i>x</i> = 2, <i>y</i> = 287 ✓</p><p>• 2<i>x</i> + 3 = 49 ⟹ <i>x</i> = 23, <i>y</i> = 41 ✓</p><p>• 2<i>x</i> + 3 = 41 ⟹ <i>x</i> = 19, <i>y</i> = 49 ✓</p><p>• 2<i>x</i> + 3 = 287 ⟹ <i>x</i> = 142, <i>y</i> = 7 ✓</p><p>• 2<i>x</i> + 3 = 2009 ⟹ <i>x</i> = 1003, <i>y</i> = 1 ✓</p><p><strong>Step 5:</strong> There are 5 ordered pairs: (2, 287), (23, 41), (19, 49), (142, 7), (1003, 1).</p><p>∴ Answer is (B).</p>
Correct Answer: B