Great question — this one tests
number theory + fraction manipulation, and the trap is trying random pairs instead of setting up the algebra cleanly.
Step 1: Set up the equation.We're told 1/x + 1/y = 1/3, where x and y are distinct natural numbers.
Rewriting: (x + y) / (xy) = 1/3, which gives us
3(x + y) = xy.
Step 2: Rearrange to isolate one variable.3x + 3y = xy
xy − 3x − 3y = 0
Add 9 to both sides (Simon's Favorite Factoring Trick):
xy − 3x − 3y + 9 = 9
(x − 3)(y − 3) = 9Step 3: Find factor pairs of 9.Since x and y are distinct natural numbers, (x − 3) and (y − 3) must be positive integer factor pairs of 9:
- (1, 9) → x = 4, y = 12 → x + y =
16- (3, 3) → x = 6, y = 6 →
rejected (x and y must be distinct)
- (9, 1) → x = 12, y = 4 → x + y =
16 (same pair, just swapped)
Step 4: Verify.1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3 ✓
The minimum value of x + y =
16.
Common trap: Many students try to guess values like (4, 12) or (6, 6) without a system — this wastes time and risks missing cases. The real trap here is the pair (6, 6), which satisfies the reciprocal equation but violates the "distinct" constraint. Always re-read the stem.
Takeaway: Whenever you see 1/x + 1/y = 1/k with integer constraints, reach for Simon's Favorite Factoring Trick — rearrange, add a constant to both sides, and factor. It turns an open-ended search into a clean factor-pair problem every time.
Answer: D (16)---
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