Which of the following numbers is completely divisible by both 9 and 11?

The correct answer is 396.

To determine which number is completely divisible by both 9 and 11, a number must satisfy the divisibility rules for both 9 and 11:

1. Divisibility Rule for 9: A number is divisible by 9 if the sum of its digits is divisible by 9.

2. Divisibility Rule for 11: A number is divisible by 11 if the difference between the sum of the digits at odd positions and the sum of the digits at even positions is either 0 or a multiple of 11.

Alternatively, since 9 and 11 are co-prime numbers, any number divisible by both 9 and 11 must also be divisible by their product:

$9\times 11=99$

Let's check the number 396 using both rules:

Check divisibility by 9:
Sum of the digits of 396:

$3+9+6=18$

Since 18 is divisible by 9 ($18/9=2$), 396 is divisible by 9.

Check divisibility by 11:
Sum of digits at odd positions (1st and 3rd digits):

$3+6=9$

Sum of digit at even position (2nd digit):

$9$

Difference between the sums:

$9-9=0$

Since the difference is 0, 396 is divisible by 11.

Therefore, 396 is completely divisible by both 9 and 11.