Directions: For a 5-digit number, without repetition of digits, the following information is available. (i) The first digit is more than 5 times the last digit. (ii) The two-digit number formed by the last two digits is the product of two prime numbers. (iii) The first three digits are all odd. (iv) The number does not contain the digits 3 or 0 and the first digit is also the largest. The largest digit in the number is

### Concept & Logic To find the largest digit, we must combine the constraint that the first three digits are odd with the constraint that the first digit itself is the largest in the entire number. ### Step-by-Step Solution Let the 5-digit number be $A B C D E$. From statement (iv), $A$ (the first digit) is the largest digit in the entire 5-digit number. From statement (iii), $A, B$, and $C$ are all odd digits. From statement (iv), the digits $3$ and $0$ are not allowed. This means the pool of available odd digits is $1, 5, 7$, and $9$. We know from statement (i) that $A > 5E$. Since $E$ cannot be $0$, $E$ must be at least $1$. If $E = 1$, then $A > 5$. The odd digits greater than $5$ are $7$ and $9$. Since the first three digits must all be distinct odd digits (no repetition), we must choose three digits from the set $\{1, 5, 7, 9\}$. Because $E=1$ (as derived from the inequality $A > 5E$), the digit $1$ is already used for the last position. Therefore, the first three digits $A, B$, and $C$ must exactly be $5, 7$, and $9$ in some order. Since statement (iv) explicitly states that the first digit ($A$) is the largest, and our available digits for $A, B, C$ are $\{5, 7, 9\}$, $A$ must be $9$. ### Exam Strategy & Shortcut You don't need to solve for the whole number. The first three digits are odd and unique. The highest odd digit is $9$. If the first digit is the largest digit overall, it is highly probable to be $9$. Since it must be larger than $5 \times E$ (where $E \ge 1$), it has to be $>5$. The only available odd digits are $7$ and $9$. If $A=7$, you can't fit three unique odd digits in the front (you'd need to use $5, 7$, and what else? $1$ is taken by $E$). Thus, $A$ must be $9$. ### Common Pitfall Ignoring the "no repetition" rule. If repetition were allowed, you could have multiple $9$s, but since all digits must be unique, the limited pool of odd digits forces the largest one to take the front position. ### Final Answer Therefore, the correct answer is 9.