Numerical answer
How many ordered pairs of positive integers (x, y) satisfy xy + x + y = 104?
Show answer and solution
Answer: 6
Add 1 to both sides: xy + x + y + 1 = 105 ⇒ (x + 1)(y + 1) = 105.
Since x, y ≥ 1, both x + 1 and y + 1 are divisors of 105 = 3 × 5 × 7 that are at least 2.
105 has 8 divisors in all: 1, 3, 5, 7, 15, 21, 35, 105. Excluding 1 (which would force the other factor to be 105, making x or y = 0) and excluding 105 itself (same issue the other way), the valid divisors for x + 1 are 3, 5, 7, 15, 21, 35 — 6 values, each giving one ordered pair (x, y).
(e.g. x + 1 = 3 → (x, y) = (2, 34); x + 1 = 35 → (x, y) = (34, 2), etc.)
Tip: Whenever you see xy + x + y (or similar), add 1 to both sides to factor it as (x + 1)(y + 1) — a very common CAT trick.
Since x, y ≥ 1, both x + 1 and y + 1 are divisors of 105 = 3 × 5 × 7 that are at least 2.
105 has 8 divisors in all: 1, 3, 5, 7, 15, 21, 35, 105. Excluding 1 (which would force the other factor to be 105, making x or y = 0) and excluding 105 itself (same issue the other way), the valid divisors for x + 1 are 3, 5, 7, 15, 21, 35 — 6 values, each giving one ordered pair (x, y).
(e.g. x + 1 = 3 → (x, y) = (2, 34); x + 1 = 35 → (x, y) = (34, 2), etc.)
Tip: Whenever you see xy + x + y (or similar), add 1 to both sides to factor it as (x + 1)(y + 1) — a very common CAT trick.
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