A prime number $N$, in the range 10 to 50, remains unchanged when its digits are reversed. The square of such a number is

### Concept & Logic This problem combines number properties (palindromes) with prime identification. If a two-digit number remains unchanged when reversed, its tens digit must be equal to its units digit (e.g., $11, 22$). ### Step-by-Step Solution The condition states the number $N$ is between $10$ and $50$. It also states that reversing the digits yields the exact same number. Let the two-digit number be $xy$. If reversed, it becomes $yx$. For $xy = yx$, it must be true that $x = y$. Let's list all two-digit numbers with identical digits between $10$ and $50$: * $11$ * $22$ * $33$ * $44$ The problem specifies that $N$ is a **prime number**. Let's evaluate our list: * $22$ is divisible by $2$ and $11$. * $33$ is divisible by $3$ and $11$. * $44$ is divisible by $2, 4$, and $11$. The only prime number in this list is $11$. Therefore, $N = 11$. The question asks for the square of such a number: $$N^2 = 11^2 = 121$$ ### Exam Strategy & Shortcut Any two-digit number with repeating digits (like $22, 33, 44...$) is inherently a multiple of $11$. Therefore, none of them can be prime, *except* for $11$ itself. You can bypass listing the numbers and immediately realize $N$ must be $11$. Since $11^2 = 121$, the answer is found instantly. ### Common Pitfall A student might read the question too quickly, find $N = 11$, and look for $11$ in the options. Realizing it's missing, they panic. Always double-check what the question is asking for—in this case, it asks for the *square* of the number, not the number itself. ### Final Answer Therefore, the correct answer is 121.