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
D
Let the three-digit number be ‘abc’. Then, the difference between the number and the sum of its digits is 100a + 10b + c − (a + b + c) = 99a + 9b. 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 11a + b = 37 so a = 3, b = 4 and we get the numbers 340, 341, 342, . . . , 349. Similarly 666 corresponds to 11a + b = 74 so that a = 6, b = 8 and we get the numbers 680, 681, 682, . . . , 689. Therefore there are 20 different numbers with this property.
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