The question asks for the smallest digit 'x' that makes the 6-digit number 28x219 divisible by both 9 and 11.

Divisibility Rule for 9

A number is divisible by 9 if the sum of its digits is divisible by 9.

• Sum of digits = \(2 + 8 + x + 2 + 1 + 9 = 22 + x\)
• For divisibility by 9, \(22 + x\) must be a multiple of 9.
• Since x is a digit (0 to 9), the possible values for \(22 + x\) range from \(22+0=22\) to \(22+9=31\).
• The only multiple of 9 within the range [22, 31] is 27.
• Therefore, \(22 + x = 27\).
• Solving for x: \(x = 27 - 22 = 5\).

Divisibility Rule for 11

A number is divisible by 11 if the alternating sum of its digits is divisible by 11.

• Alternating sum (from right): \(9 - 1 + 2 - x + 8 - 2 = 16 - x\).
• For divisibility by 11, \(16 - x\) must be a multiple of 11.
• Since x is a digit (0 to 9), the possible values for \(16 - x\) range from \(16-9=7\) to \(16-0=16\).
• The only multiple of 11 within the range [7, 16] is 11.
• Therefore, \(16 - x = 11\).
• Solving for x: \(x = 16 - 11 = 5\).

Conclusion for Digit x

Both divisibility rules require x = 5. Since we are looking for the smallest digit that satisfies both conditions, and we found a single value, x = 5 is the answer.