Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of—

Let the three-digit number be represented as 'abc', which algebraically equals 100a + 10b + c [t2][t6]. The problem states that the middle digit (b) is the sum of the other two digits (a and c), so b = a + c. Substituting this into the algebraic form, we get 100a + 10(a + c) + c. Simplifying this expression results in 100a + 10a + 10c + c, which equals 110a + 11c. Factoring out 11, the expression becomes 11(10a + c) [t2]. Since the number can be expressed as a product of 11 and an integer, it is a multiple of 11. This aligns with the divisibility rule for 11, which states that a number is divisible by 11 if the alternating sum of its digits (a - b + c) is 0 or a multiple of 11 [t1][t3][t6]. Here, a - (a + c) + c = 0, confirming divisibility [t4].