2005 AMC 12B Problems/Problem 15

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, let's say $\overline{ae}$ , $\overline{bf}$ , $\overline{cg}$ , and $\overline{dh}$ . 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 $e$ , $f$ , $g$ ,and $h$ 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 $45-41=\boxed{\mathrm{(D)}\ 4}$

2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2005 AMC 12B Problems/Problem 15

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