A positive number, which when added to $1000$, gives a sum which is greater than when it is multiplied by $1000$. This positive integer is

### Concept & Logic This problem sets up a basic linear inequality comparing the addition and multiplication of an unknown variable with a constant. ### Step-by-Step Solution * **Given:** * Let the positive integer be $x$. * Condition: The sum of the number and $1000$ is strictly greater than their product. * **Calculation / Deduction:** 1. Translate the given condition into an algebraic inequality: $$x + 1000 > 1000x$$ 2. Isolate the terms with $x$ on one side by subtracting $x$ from both sides: $$1000 > 999x$$ 3. Solve for $x$ by dividing both sides by $999$: $$x < \frac{1000}{999}$$ 4. Evaluate the resulting fraction: $\frac{1000}{999}$ is slightly greater than $1$ (approximately $1.001$). 5. The problem explicitly states $x$ is a "positive integer". The only positive integer strictly less than $1.001$ is $1$. ### Exam Strategy & Shortcut Option verification is incredibly fast here and avoids algebra entirely. Test the smallest option first. Let's try $x = 1$. The sum is $1 + 1000 = 1001$. The product is $1 \times 1000 = 1000$. Since $1001 > 1000$, option (a) satisfies the condition immediately. You can quickly see that for any integer $2$ or greater, the multiplication side will grow much faster ($2000 > 1002$). ### Common Pitfall Students might overcomplicate the phrasing and set up an equality ($=$) instead of an inequality ($>$). Additionally, ignoring the constraint that the number must be a "positive integer" could cause confusion if one tries to solve the inequality abstractly without restricting the domain of $x$. ### Final Answer Therefore, the correct answer is 1.