Let a number with three digits has for its middle digit the sum of the other two digits. Then this number is a multiple of

Let the three-digit number be represented as (100a + 10b + c), where a, b, and c are the digits, and the middle digit b is given as the sum of the other two digits.

Thus, we can write:

( b = a + c )

Now the number becomes:

⇒ (100a + 10b + c = 100a + 10(a + c) + c)

Which simplifies to:

⇒ (100a + 10a + 10c + c = 110a + 11c)

This shows that the number can be written as:

⇒ {11(10a + c)}

Therefore, the number is always a multiple of 11.

The correct answer is "Option 1".