Difference between revisions of "2007 AMC 10A Problems/Problem 4"
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== Solution == | == Solution == | ||
| − | Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{( | + | Let the two consecutive odd integers be <math>a</math>, <math>a+2</math>. Then <math>a+2 = 3a</math>, so <math>a = 1</math> and their sum is <math>4\ \mathrm{(A)}</math>. |
== See also == | == See also == | ||
| − | {{AMC10 box|year=2007|ab=A|num-b=3|num-a=5} | + | {{AMC10 box|year=2007|ab=A|num-b=3|num-a=5}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:58, 27 April 2021
Problem
The larger of two consecutive odd integers is three times the smaller. What is their sum?
Solution
Let the two consecutive odd integers be 
, 
. Then 
, so 
and their sum is 
.
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
