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

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Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.
 
Using logic, <math>a+b+c= 10</math>, therefore <math>b+a+1</math>(from the carry over)<math>= 10</math>.
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so b+a=9
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so <math>b+a=9</math>
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A+1=3
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<math>A+1=3</math>
  
 
Thus, <math>A=2</math> and <math>11B+C=78</math>.  From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.
 
Thus, <math>A=2</math> and <math>11B+C=78</math>.  From here it becomes clear that <math>B=7</math> and <math>C=1\rightarrow \boxed{\text{A}}</math>.

Revision as of 15:58, 12 October 2019

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In the addition problem, each digit has been replaced by a letter.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 (Problems • Answer Key • Resources)
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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