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Difference between revisions of "2002 AMC 12B Problems/Problem 7"

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{{duplicate|[[2002 AMC 12B Problems|2002 AMC 12B #7]] and [[2002 AMC 10B Problems|2002 AMC 10B #11]]}}
 
== Problem ==
 
== Problem ==
 
The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of their [[perfect square|squares]]?
 
The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of their [[perfect square|squares]]?
Line 8: Line 9:
 
\qquad\mathrm{(E)}\ 194</math>
 
\qquad\mathrm{(E)}\ 194</math>
 
== Solution ==
 
== Solution ==
−
Let the three consecutive integers be <math>x-1, x, x+1</math>; then
+
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>.
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<cmath>(x-1)(x)(x+1) = x(x^2 - 1) = 8(x-1 + x + x+1) = 24x</cmath>
 
−
 
 
−
Since <math>x \neq 0</math>, we have <math>x^2 = 25</math>, with the positive solution being <math>x = 5</math>. Then <math>4^2 + 5^2 + 6^2 = 77\ \mathrm{(B)}</math>.
 
  
 
== See also ==
 
== See also ==
 +
{{AMC10 box|year=2002|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2002|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2002|ab=B|num-b=6|num-a=8}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

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

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