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

(Solution)
(Solution)
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==Solution==
 
==Solution==
  
Using logic, a+b+c= 10, therefore b+a+1(from the carry over)= 10.
+
Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.
 
so b+a=9
 
so b+a=9
 
A+1=3
 
A+1=3

Revision as of 23:02, 17 April 2017

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In the addition problem, each digit has been replaced by a letter.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 $C=1\rightarrow \boxed{\text{A}}$.

See Also

1991 AJHSME (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions

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