Which of the following numbers are completely divisible by 7? I. $195195$ II. $181181$ III. $120120$ IV. $891891$

## Concept & Formula Any six-digit number formed by repeating a three-digit block (format $abcabc$) is inherently a multiple of $1001$. Since $1001 = 7 \times 11 \times 13$, any such number is always divisible by 7. ## Step-by-Step Solution **Given:** Four six-digit numbers: $195195$, $181181$, $120120$, and $891891$. **Calculation / Deduction:** * **Step 1: Observe the structural pattern.** All four numbers follow the exact format of repeating triplets: $abcabc$. * **Step 2: Apply the expansion rule.** Recall that $abcabc = (abc \times 1000) + abc = abc \times 1001$. * **Step 3: Evaluate the prime factors.** Factor $1001$ into its primes: $7 \times 11 \times 13$. * **Step 4: Conclude.** Because $7$ is a prime factor of the base multiplier $1001$, every single number formatted as $abcabc$ is mathematically guaranteed to be completely divisible by $7$. Thus, I, II, III, and IV all qualify. ## Exam Strategy & Shortcut Never perform manual division for 7 on these types of numbers. The divisibility rule for 7 is notoriously tedious to execute under time pressure. The moment you spot the repeating 3-digit triplet, know instantly that it passes the divisibility tests for 7, 11, and 13. You can answer this question visually in 2 seconds. ## Common Pitfall The biggest mistake is wasting precious exam time manually applying the standard divisibility rule for 7 to all four numbers sequentially instead of recognizing the overarching property of the number format. ## Final Answer **Therefore, the correct answer is All are divisible.**