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1991 AJHSME Problems/Problem 20

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Problem

In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then C=

unitsize(18);draw((-1,0)--(3,0));draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4));label("$A$",(0.5,2.1),N...

\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 7 \qquad \text{(E)}\ 9

Solution

From this we have 111A+11B+C=300. Clearly, A<3. Since B,C\leq 9, 111A > 201 \Rightarrow A\geq 2. Thus, A=2 and 11B+C=78. From here it becomes clear that B=7 and C=1\rightarrow \boxed{\text{A}}.

See Also

1991 AJHSME (ProblemsResources)
Preceded by
Problem 19
Followed by
Problem 21
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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