Answer :
To solve this problem, we need to find the largest possible remainder when the larger of the two numbers is divided by the smaller one.
Let's consider a 2-digit number, where the digits are represented as x and y, where x is the tens digit and y is the ones digit.
The original 2-digit number is xy, and the reversed number is yx.
The larger of the two numbers is the larger of xy and yx, and the smaller of the two numbers is the smaller of xy and yx.
The possible combinations of x and y are:
- x = 1, 2, 3, 4, 5, 6, 7, 8, 9
- y = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
Now, let's find the largest possible remainder when the larger of the two numbers is divided by the smaller one.
The largest possible remainder occurs when the larger number is 99, and the smaller number is 10, which gives a remainder of 9.
Therefore, the largest possible remainder is 9, and the correct answer is 1.
The largest possible remainder when a 2-digit number is reversed and divided by the smaller one is 9, based on maximizing the remainder while ensuring that the tens digit is only one greater than the ones digit. The correct option is a.
The question asks us to find the largest possible remainder when a 2-digit number is reversed and the larger of the two numbers is divided by the smaller one. To find the largest possible remainder, we have to consider when the 2-digit number has a tens digit that is exactly one greater than the ones digit. For example, the number 21 reversed is 12. Dividing 21 by 12 gives us a remainder of 9, which cannot be greater because if the tens digit was any greater, the resulting number could be divisible by its reversed counterpart, reducing the remainder.
Therefore, the largest possible remainder when one 2-digit number is divided by its reversal is 9. This is based on the principle that if the tens digit is one greater than the ones digit, the remainder will be maximized when dividing the two permutations of the digits. The correct option is a.
