— Burt Kaliski Jr. (@modulomathy) August 31, 2013
Question 1: Are all three-digit numbers having the form abc where a+c=b divisible by 11?
Proof: Let x be a number of the specified form. Because a, b and c are the digits of x, we have
x = 100a+10b+c
Substituting b = a+c, we get
x = 100a+10(a+c)+c
Expanding the expression gives
x = 110a+11c
The result is clearly divisible by 11: x = 11*(10a+c). (In the current example with a = c = 2, we have x = 11*22 = 242.)
Question 2: Do all three-digit numbers that are divisible by 11 have the form abc where a+c=b?
Proof: By contradiction. Consider the number x = 11*19 = 209. It is divisible by 11, but does not have the specified form.
There is a generalization, however. If a three-digit number x is divisible by 11, then it will have the form abc where a+c-b itself is divisible by 11. a+c = b is a special case (where a+c-b = 0, which is divisible by 11). The result also holds for numbers such as 209. Proof left to the reader. (Hint: use a+c-b = 11*d in the equation above.)