Difference between revisions of "2007 AMC 10A Problems/Problem 4"

m (See also: gah insert key was on)
(Problem)
 
(4 intermediate revisions by 4 users not shown)
Line 5: Line 5:
  
 
== 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{(B)}</math>.
+
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 ==
Line 11: Line 11:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 00:58, 28 April 2021

Problem

The larger of two consecutive odd integers is three times the smaller. What is their sum?

$\text{(A)}\ 4 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 20$

Solution

Let the two consecutive odd integers be $a$, $a+2$. Then $a+2 = 3a$, so $a = 1$ and their sum is $4\ \mathrm{(A)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png