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>. | ||
− | + | So <math>b+a=9</math> | |
− | A+1=3 | + | <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>. |
Latest revision as of 16:58, 12 October 2019
Contents
Problem
In the addition problem, each digit has been replaced by a letter. If different letters represent different digits then
Solution
From this we have Clearly, . Since , Thus, and . From here it becomes clear that and .
Solution
Using logic, , therefore (from the carry over). So
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 |
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