#math trivia for #August30: #242 has the form abc, a+c=b. Are all such three-digit numbers also divisible by 11? Vice versa?

— 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?

**Answer:** Yes.

**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?

**Answer:** No.

**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.)