Two different natural numbers are such that their product is less than their sum. One of the numbers must be

## Concept & Logic This is a conceptual number theory question. For natural numbers (positive integers: $1, 2, 3...$), the product of two numbers grows much faster than their sum. We must analyze the boundary conditions to see when the sum can exceed the product. ## Step-by-Step Solution * **Given:** Two distinct natural numbers $x$ and $y$. Condition: $x \times y < x + y$. * **Calculation / Deduction:** * Let's test the smallest possible natural numbers. * Assume neither number is 1. The smallest distinct natural numbers would then be $2$ and $3$. * Test condition: Product $= 2 \times 3 = 6$. Sum $= 2 + 3 = 5$. * Is $6 < 5$? No. * As the numbers get larger (e.g., $3$ and $4$), the product ($12$) outpaces the sum ($7$) even more. * Therefore, the only way the sum can be greater than the product is if we use the absolute smallest natural number, which is $1$. * Let's verify: Let one number be $1$, and the other be any distinct natural number, say $5$. * Product $= 1 \times 5 = 5$. Sum $= 1 + 5 = 6$. * Is $5 < 6$? Yes. The condition holds true. ## Exam Strategy & Shortcut **Logical Substitution:** Instantly test the base cases. Try $2$ and $3$: $2 \times 3 = 6$, $2 + 3 = 5$. $6$ is not less than $5$. The only number smaller than $2$ in the natural number set is $1$. Therefore, one of the numbers absolutely must be $1$ to drag the product down below the sum. ## Common Pitfall A common mistake is confusing "natural numbers" (starts at $1$) with "whole numbers" (starts at $0$) or "integers" (includes negatives). If $0$ or negative numbers were allowed, there would be infinite solutions, but the strict constraint of "natural numbers" makes $1$ the only correct logical anchor. ## Final Answer **Therefore, the correct answer is 1.**