2005 AMC 12B Problems/Problem 15

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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?

$\mathrm{(A)}\ 1      \qquad \mathrm{(B)}\ 2      \qquad \mathrm{(C)}\ 3      \qquad \mathrm{(D)}\ 4      \qquad \mathrm{(E)}\ 5$

Solution

221 can be written as the sum of eight two-digit numbers, lets say ae, bf, cg, and dh. Then 221= 10(a+b+c+d)+(e+f+g+h). The last digit of 221 is 1, and 10(a+b+c+d) won't affect the units digits, so (e+f+g+h) must end with 1. The smallest value (e+f+g+h) can have is (1+2+3+4)=10, and the greatest value is (6+7+8+9)=30. Therefore, (e+f+g+h) must equal 11 or 21.

Case 1: (e+f+g+h)=11 The only distinct positive integers that can add up to 11 is (1+2+3+5). So, a,b,c, and d must include four of the five numbers (4,6,7,8,9). We have 10(a+b+c+d)=221-11=210, or a+b+c+d=21. We can add all of 4+6+7+8+9=34, and try subtracting one number to get to 21, but to no avail. Therefore, (e+f+g+h) cannot add up to 11.

Case 2: (e+f+g+h)=21 Checking all the values for e,f,g,h each individually may be time-consuming, instead of only having 1 solution like case 1. We can try a different approach by looking at (a+b+c+d) first. If (e+f+g+h)=21, 10(a+b+c+d)=221-21=200, or (a+b+c+d)=20. That means (a+b+c+d)+(e+f+g+h)=21+20=41. We know (1+2+3+4+5+6+7+8+9)=45, so the missing digit is 45-41=4

See also