111,111,111,111 is divisible by

### Concept & Logic Numbers formed by repeating identical digits (repunits) have specific divisibility properties based on their length. A $6$-digit repunit (like $111,111$) is always a multiple of $7$, $11$, and $13$, because it equals $111 \times 1001$. ### Step-by-Step Solution * **Divisibility by 3:** The sum of the digits is $12 \times 1 = 12$, which is divisible by $3$. * **Divisibility by 11:** The difference between the sum of digits in odd places ($6$) and even places ($6$) is $0$, which is divisible by $11$. * **Divisibility by 37 and 111:** Any $3$-digit repunit ($111$) is divisible by $37$ ($37 \times 3 = 111$). Since our $12$-digit number is constructed from blocks of $111$, it is inherently divisible by both $37$ and $111$. * **Divisibility by 1001:** The number $1001$ equals $7 \times 11 \times 13$. A $6$-digit block of $1$s ($111,111$) is mathematically $111 \times 1001$, meaning it is divisible by $1001$. Because $12$ is a multiple of $6$, a $12$-digit block of $1$s is also divisible by $1001$. Since all conditions hold, the number is divisible by all listed factors. ### Exam Strategy & Shortcut Memorize the core factorization rule: $1001 = 7 \times 11 \times 13$. Any number that repeats a $3$-digit block twice (like $abcabc$) is divisible by $1001$. Since $111,111$ repeats a $3$-digit block, it is divisible by $1001$. The $12$-digit version simply scales this up. This immediately confirms option (d). ### Common Pitfall Students often stop testing after finding that the number is divisible by $3$, $11$, and $37$, mistakenly selecting an incomplete option like (b) or (c) because they fail to test for the larger $1001$ block property. ### Final Answer Therefore, the correct answer is **3, 11, 37, 111 and 1001**.