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 1

$221$ can be written as the sum of four 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 none of them work. 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 $45-41=\boxed{\mathrm{(D)}\ 4}$

Solution 2

Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d$, where $d$ is some digit. Clearly, $\boxed{d = 4}$.

See also

2005 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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