If a number is divisible by both 11 and 13, then it must be necessarily

## Concept & Logic If a number is completely divisible by two **co-prime** integers, it must also be perfectly divisible by their product. ## Step-by-Step Solution **Calculation / Deduction:** * **Step 1: Identify the properties of the given divisors.** Both 11 and 13 are prime numbers. * **Step 2: Determine their relationship.** Because they are both prime, they share no common factors other than 1. This classifies them as co-prime to each other. * **Step 3: Apply number theory rules.** If a number $N$ is divisible by $x$ and $y$, and the greatest common divisor $gcd(x, y) = 1$, then $N$ is strictly divisible by $(x \times y)$. * **Step 4: Conclusion.** Therefore, any number divisible by both 11 and 13 must necessarily be divisible by their product, $(11 \times 13)$. ## Exam Strategy & Shortcut This is a foundational theoretical rule of number systems. You should not need to perform any calculations. Recognize the prompt mentions "both primes" and instantly select the option containing their "product". ## Common Pitfall A common trap is selecting option (a) `429` because $429 = 11 \times 13 \times 3$. While 429 *is* divisible by both, a number divisible by 11 and 13 doesn't *have* to be 429 (for example, it could be 143 or 286). The question asks what condition must *necessarily* be true for all such numbers. ## Final Answer **Therefore, the correct answer is divisible by (11 × 13).**