The problem asks us to find which of the given numbers (11,116, 16,462, 21,240, 28,814) is divisible by both 2 and 6.

1. Problem Description

The problem asks us to find which of the given numbers (11,116, 16,462, 21,240, 28,814) is divisible by both 2 and

6.

2. Solution Steps

A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8).

A number is divisible by 6 if it is divisible by both 2 and

3. A number is divisible by 3 if the sum of its digits is divisible by

3.

First, let's check if the given numbers are divisible by

2. All the given numbers are even, so they are all divisible by

2.

Now, let's check the divisibility by 3 for each number.

- 11,116: Sum of digits is $1+1+1+1+6 = 10$. 10 is not divisible by

3. Thus, 11,116 is not divisible by 3, and therefore not divisible by

6. - 16,462: Sum of digits is $1+6+4+6+2 = 19$. 19 is not divisible by

3. Thus, 16,462 is not divisible by 3, and therefore not divisible by

6. - 21,240: Sum of digits is $2+1+2+4+0 = 9$. 9 is divisible by

3. Thus, 21,240 is divisible by

3. Since it is also divisible by 2, it is divisible by

6. - 28,814: Sum of digits is $2+8+8+1+4 = 23$. 23 is not divisible by

3. Thus, 28,814 is not divisible by 3, and therefore not divisible by

6.

Therefore, only 21,240 is divisible by both 2 and

6.

3. Final Answer

21,240