Difference between revisions of "1991 AJHSME Problems/Problem 20"
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| − | 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
Contents
Problem
In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then 
![[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]](http://latex.artofproblemsolving.com/8/e/8/8e88799a4ccdbb6c37dec56b6b67f175e419db0b.png)
Solution
From this we have
![\[111A+11B+C=300.\]](http://latex.artofproblemsolving.com/2/a/9/2a9bc8582cda2534a8032588cda0209002437bad.png)
Clearly, 
. Since 
,
![\[111A > 201 \Rightarrow A\geq 2.\]](http://latex.artofproblemsolving.com/7/c/b/7cb3207cd37022d15ebdb849d7e8df2fea60c300.png)
Thus, 
and 
. From here it becomes clear that 
and 
.
Solution
Using logic, 
, therefore 
(from the carry over)
.
so b+a=9
A+1=3
Thus, and . From here it becomes clear that and .
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
