Consider the following statements for the sequence of numbers given below: 11, 111, 1111, 11111, .... 1. Each number can be expressed in the form $(4m + 3)$, where $m$ is a natural number. 2. Some numbers are squares. Which of the above statements is/are correct?

## Concept & Logic This problem tests two distinct number system concepts: the divisibility rule for 4 and the unit digit/remainder properties of perfect squares. * **Divisibility by 4:** A number is divisible by 4 if its last two digits form a number divisible by 4. * **Properties of Squares:** Any perfect square, when divided by 4, will *always* leave a remainder of exactly 0 (if it is an even square) or 1 (if it is an odd square). A perfect square can **never** leave a remainder of 2 or 3 when divided by 4. ## Step-by-Step Solution * **Evaluate Statement 1:** Can they be expressed as $(4m + 3)$? * The numbers in the sequence all end in "11". * When you divide 11 by 4, the quotient is 2 and the remainder is 3. ($11 = 4 \times 2 + 3$). * For any larger number ending in 11 (like 111, 1111), the hundreds, thousands, etc., places are all multiples of 100. Since 100 is perfectly divisible by 4, the remainder of the entire number depends *only* on the last two digits (11). * Therefore, every number in this sequence leaves a remainder of 3 when divided by 4. * This means they can all be written as $4m + 3$. **Statement 1 is correct.** * **Evaluate Statement 2:** Are some numbers squares? * As proven in Statement 1, every number in this sequence is of the form $4m + 3$. * By the fundamental properties of perfect squares, no perfect square can be of the form $4m + 3$ (they must be $4m$ or $4m+1$). * Therefore, none of these numbers can possibly be perfect squares. **Statement 2 is incorrect.** ## Exam Strategy & Shortcut For Statement 1, don't try to prove it algebraically for the whole sequence. Just test the first term: $11 = 4(2) + 3$. Test the second: $111 = 4(27) + 3$. It holds true. For Statement 2, remember the universal rule: Perfect squares only end in 0, 1, 4, 5, 6, or 9. Since these numbers end in 1, they *could* theoretically be squares based on the unit digit, but the modulo 4 rule ($4m+3$) instantly disqualifies them without any calculation. ## Common Pitfall Students often get confused by the unit digit "1" in Statement 2. Since 121 ($11^2$) ends in 1, and 81 ($9^2$) ends in 1, students guess that somewhere in the infinite sequence, one of these numbers might be a square. Relying on unit digits alone is a trap; always use modulo/remainder rules to verify. ## Final Answer Therefore, the correct answer is 1 only.