# Difference between revisions of "1991 AJHSME Problems/Problem 20"

## Problem

In the addition problem, each digit has been replaced by a letter.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

$[asy] unitsize(18); draw((-1,0)--(3,0)); draw((-3/4,1/2)--(-1/4,1/2)); draw((-1/2,1/4)--(-1/2,3/4)); label("A",(0.5,2.1),N); label("B",(1.5,2.1),N); label("C",(2.5,2.1),N); label("A",(1.5,1.1),N); label("B",(2.5,1.1),N); label("A",(2.5,0.1),N); label("3",(0.5,-.1),S); label("0",(1.5,-.1),S); label("0",(2.5,-.1),S); [/asy]$

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

## 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 $C=1\rightarrow \boxed{\text{A}}$.

## Solution

Using logic, $a+b+c= 10$, therefore $b+a+1$(from the carry over)$= 10$. So $b+a=9$ $A+1=3$

Thus, $A=2$ and $11B+C=78$. From here it becomes clear that $B=7$ and $C=1\rightarrow \boxed{\text{A}}$.