Problem:
The sum of four two-digit numbers is 221 . None of the eight digits is 0 and no two of them are the same. Which of the following is not included among the eight digits?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
The sum of the digits 1 through 9 is 45 , so the sum of the eight digits is between 36 and 44 , inclusive. The sum of the four units digits is between and , inclusive, and also ends in 1 . Therefore the sum of the units digits is either 11 or 21 . If the sum of the units digits is 11 , then the sum of the tens digits is 21 , so the sum of all eight digits is 32 , an impossibility. If the sum of the units digits is 21 , then the sum of the tens digits is 20 , so the sum of all eight digits is 41 . Thus the missing digit is . Note that the numbers , and 97 sum to 221 .
OR
Each of the two-digit numbers leaves the same remainder when divided by 9 as does the sum of its digits. Therefore the sum of the four two-digit numbers leaves the same remainder when divided by 9 as the sum of all eight digits. Let be the missing digit. Because 221 when divided by 9 leaves a remainder of 5 , and the sum of the digits from 1 through 9 is 45 , the number must leave a remainder of 5 when divided by 9 . Thus .
The problems on this page are the property of the MAA's American Mathematics Competitions