Find the two-digit numbers A and B such that the number 65AB0 is divisible by…

To determine the two-digit numbers A and B such that the number 65AB0 is divisible by 66, we must use the divisibility rules for 6 and 11, because 66 = 6 * 11.

Step-by-Step Calculation

1. Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3.

• Divisibility by 2: The number 65AB0 ends in 0, so it is already divisible by 2.

• Divisibility by 3: The sum of all digits must be divisible by 3. Sum = 6 + 5 + A + B + 0 = 11 + A + B. For this sum to be divisible by 3, (11 + A + B) must be 12, 15, 18, 21, 24, 27, or 30.

2. Divisibility by 11: A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Alternating sum = 6 - 5 + A - B + 0 = 1 + A - B. For this to be divisible by 11, (1 + A - B) must be 0, 11, or -11.

• Case: 1 + A - B = 0 => B = A + 1. If we test this in our divisibility by 3 sum (11 + A + B), we get 11 + A + (A + 1) = 12 + 2A. If A=17, B=18 (from Option D): 12 + 2(17) = 46 (not divisible by 3). If A=17, B=17 (from Option C): 12 + 2(17) = 46 (not divisible by 3).

• Case: 1 + A - B = -11 => B = A + 12. If A=17, B=29 (not two-digit numbers as required by standard digit placement, but let's re-examine the alternating sum provided in Option C).

3. Evaluating Option C (A=17, B=17): The number becomes 6517170.

• Alternating sum: 6 - 5 + 1 - 7 + 1 - 7 + 0 = -11. Since -11 is divisible by 11, the divisibility rule for 11 is satisfied.

• Sum of digits: 6 + 5 + 1 + 7 + 1 + 7 + 0 = 27. Since 27 is divisible by 3, the divisibility rule for 3 is satisfied.

Since the number satisfies both rules for 6 and 11, it is divisible by 66.