A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will always be divisible by :

A two-digit number can be algebraically represented as 10a + b, where 'a' is the tens digit and 'b' is the units digit [t2]. When the digits are interchanged, the new number becomes 10b + a [t2][t6]. According to the problem, adding the original number to the reversed number results in the expression (10a + b) + (10b + a). Simplifying this expression yields 11a + 11b, which can be factored as 11(a + b) [t2][t3][t6]. This mathematical proof demonstrates that the resulting sum is always a multiple of 11, regardless of the specific digits chosen for 'a' and 'b' [t2][t3]. For example, if the number is 13, its reverse is 31; their sum is 44, which is divisible by 11 [t2]. Similarly, for 57, the sum 57 + 75 = 132 is also divisible by 11. Thus, the resulting number is always divisible by 11.