#math trivia #242 solution

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.)

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