2005 AMC 12B Problem 15

The eight digits are distinct and chosen from $1$ through $9,$ whose total is $45.$ So the eight used digits sum to between $45 - 9 = 36$ and $45 - 1 = 44.$ Let the four units digits sum to $U$ and the four tens digits sum to $T.$ Then $10T + U = 221,$ so $U$ ends in $1.$ Since $1+2+3+4 = 10$ $\le U \le$ $6+7+8+9 = 30,$ we have $U = 11$ or $U = 21.$ If $U = 11,$ then $10T = 210,$ so $T = 21$ and the eight digits sum to $32,$ which is below $36.$ So $U = 21,$ giving $T = 20$ and total $41.$ The missing digit is $45 - 41 = 4.$ For example, $13 + 25 + 86 + 97 = 221.$ Thus, the correct answer is D.