20. Eva chooses a three-digit positive number and from it she subtracts the sum of its three digits. She finds that the answer is a three-digit number in which all three digits are the same. How many different starting numbers could Eva have chosen?
A 2
B 5
C 10
D 20
E 30
SolUtion
D

Let the three-digit number be ' $a b c$ '. Then, the difference between the number and the sum of its digits is $100 a+10 b+c-(a+b+c)=99 a+9 b$. This difference is divisible by 9 . Therefore the result can only be 333 or 666 or 999 , and it is easy to see that 999 is not possible.

However, 333 corresponds to $11 a+b=37$ so $a=3, b=4$ and we get the numbers $340,341,342, \ldots, 349$. Similarly 666 corresponds to $11 a+b=74$ so that $a=6, b=8$ and we get the numbers $680,681,682, \ldots, 689$. Therefore there are 20 different numbers with this property.