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2005 AMC 12B Problem 15

Problem 15 of 25IntermediateNumber Theory

The sum of four two-digit numbers is 221.221. None of the eight digits is 00 and no two of them are the same. Which of the following is not included among the eight digits?

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Solution

The eight digits are distinct and chosen from 11 through 9,9, whose total is 45.45. So the eight used digits sum to between 459=3645 - 9 = 36 and 451=44.45 - 1 = 44. Let the four units digits sum to UU and the four tens digits sum to T.T. Then 10T+U=221,10T + U = 221, so UU ends in 1.1. Since 1+2+3+4=101+2+3+4 = 10 U\le U \le 6+7+8+9=30,6+7+8+9 = 30, we have U=11U = 11 or U=21.U = 21. If U=11,U = 11, then 10T=210,10T = 210, so T=21T = 21 and the eight digits sum to 32,32, which is below 36.36. So U=21,U = 21, giving T=20T = 20 and total 41.41. The missing digit is 4541=4.45 - 41 = 4. For example, 13+25+86+97=221.13 + 25 + 86 + 97 = 221. Thus, the correct answer is D.

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Concepts: digits · place value · casework

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