Difference between revisions of "2007 AMC 12A Problems/Problem 3"
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The larger of two consecutive odd integers is three times the smaller. What is their sum? | The larger of two consecutive odd integers is three times the smaller. What is their sum? | ||
| − | + | <math>\mathrm{(A)}\ 4\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 16\qquad \mathrm{(E)}\ 20</math> | |
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| + | ==Solution 1== | ||
| + | Let <math>n</math> be the smaller term. Then <math>n+2=3n \Longrightarrow 2n = 2 \Longrightarrow n=1</math> | ||
| + | *Thus, the answer is <math>1+(1+2)=4 \mathrm{(A)}</math> | ||
| − | Solution | + | |
| − | * By | + | ==Solution 2== |
| + | * By trial and error, 1 and 3 work. 1+3=4. | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2007|ab=A|num-b=2|num-a=4}} | |
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| − | + | [[Category:Introductory Algebra Problems]] | |
| + | {{MAA Notice}} | ||
Latest revision as of 02:09, 16 February 2021
Contents
Problem
The larger of two consecutive odd integers is three times the smaller. What is their sum?
Solution 1
Let 
be the smaller term. Then 
- Thus, the answer is
Solution 2
- By trial and error, 1 and 3 work. 1+3=4.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 2 |
Followed by Problem 4 |
| 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 AMC 12 Problems and Solutions | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
