Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 AMC 12B Problems/Problem 7"
(→‎Solution)
 
(One intermediate revision by one other user not shown)
Line 9: Line 9:
 
\qquad\mathrm{(E)}\ 194</math>
 
\qquad\mathrm{(E)}\ 194</math>
 
== Solution ==
 
== Solution ==
−
Let the three consecutive positive integers be <math>a-1</math>, <math>a</math>, and <math>a+1</math>. So, <math>a(a-1)(a+1)=a^3-a=24a</math>. Rearranging and factoring, <math>a(a+5)(a-5)=0</math>, so <math>a=5</math>. Hence, the sum of the squares is <math>4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}</math>.
+
Let the three consecutive positive integers be <math>a-1</math>, <math>a</math>, and <math>a+1</math>. Since the mean is <math>a</math>, the sum of the integers is <math>3a</math>. So <math>8</math> times the sum is just <math>24a</math>. With this, we now know that <math>a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24</math>. <math>24=4\times6</math>, so <math>a=5</math>. Hence, the sum of the squares is <math>4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 22:01, 1 January 2020

The following problem is from both the 2002 AMC 12B #7 and 2002 AMC 10B #11, so both problems redirect to this page.

Problem

The product of three consecutive positive integers is $8$ times their sum. What is the sum of their squares?

$\mathrm{(A)}\ 50 \qquad\mathrm{(B)}\ 77 \qquad\mathrm{(C)}\ 110 \qquad\mathrm{(D)}\ 149 \qquad\mathrm{(E)}\ 194$

Solution

Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. Since the mean is $a$, the sum of the integers is $3a$. So $8$ times the sum is just $24a$. With this, we now know that $a(a-1)(a+1)=24a\Rightarrow(a-1)(a+1)=24$. $24=4\times6$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=\boxed{\mathrm{ (B)}\ 77}$.

See also

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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
2002 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_7&oldid=113995"