Question

3 points
Rita wrote a three digit
number on the white board
and then subtracted the sum
of its digits from the number
itself. Two of the digits of the
resulting number are 3 and 4.
What is the remaining third
digit of resulting number?​

Answer 1

Answer:

Call the 3-digit number xyz, so its value is (100x + 10y + z) and the sum of its

digits is (x + y + x). Subtracting the digit sum from the number itself leaves

(99x + 9y) = 9(11x +y), which is to be equal to the square of the digit sum. Thus

9(11x + y) = (x + y + z)^2, so (11x + y) must be a perfect square. Testing different

values of x starting with x = 1 (x can’t be 0 or xyz would be a 2-digit number), we

have immediate success; 11x + y = 11 + y = 16 (the only 2-digit square with a first

digit of 1), so y = 5 and 9(11x + y) = 9(16) = 144. Therefore, the square of the digit

sum is 144, the digit sum is 12, z must be 6 and the number we were after is 156.

Sure enough, it checks since 156 - 12 = 144. I leave it to you to test other possible

values of x and show that none of them work.