When the digits of a two-digit number are interchanged, is the resulting number always divisible by 11?

Step 1: Represent the original two-digit number. Let the tens digit be a and the units digit be b. The original number can be written as 10a + b. Here, a is an integer from 1 to 9, and b is an integer from 0 to 9. Step 2: Represent the new number after interchanging digits. When the digits are interchanged, the new tens digit is b and the new units digit is a. The new number can be written as 10b + a. Step 3: Add the original number and the new number. The sum of the original number and the new number is: (10a + b) + (10b + a) Combine like terms: 10a + a + b + 10b 11a + 11b Step 4: Factor out the common term. Factor out 11 from the sum: 11(a + b) Step 5: Determine divisibility. Since the sum is 11(a + b), it is always a multiple of 11. Therefore, the resulting number will always be divisible by 11. Let's check the given options: (a) 9 (b) 3 (c) 11 (d) 5 The sum 11(a+b) is always divisible by 11. It is not necessarily divisible by 9, 3, or 5 for all possible values of a and b. For example, if the original number is 12, then a=1, b=2. Original number = 12 New number = 21 Sum = 12 + 21 = 33. 33 = 11(1+2) = 11(3). 33 is divisible by 11 and 3, but not by 9 or 5. If the original number is 23, then a=2, b=3. Original number = 23 New number = 32 Sum = 23 + 32 = 55. 55 = 11(2+3) = 11(5). 55 is divisible by 11 and 5, but not by 9 or 3. The only number that consistently divides the sum is 11. The correct option is (c). The final answer is 11. That's 2 down. 3 left today — send the next one.