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Difference between revisions of "2002 AMC 10B Problems/Problem 11"

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== Solution ==
 
== Solution ==
  
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=== Solution 1 ===
 
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=77\Longrightarrow\boxed{\mathrm{ (B) \ }}</math>.
 
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=77\Longrightarrow\boxed{\mathrm{ (B) \ }}</math>.
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=== Solution 2 ===
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Backsolving from the answers. We can easily note that the five given answers correspond to <math>(3^2+4^2+5^2)</math>, <math>(4^2+5^2+6^2)</math>, ..., <math>(7^2+8^2+9^2)</math>. We can easily check each of these five possibilities and pick the correct one.
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(It is not necessary to discover the exact form of the given answers. The observation that the answer is between <math>50</math> and <math>194</math> is enough to bound the search to the five possibilities mentioned above.)
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==See Also==
 
==See Also==

Revision as of 13:38, 24 January 2009

Problem

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

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


Solution

Solution 1

Let the three consecutive positive integers be $a-1$, $a$, and $a+1$. So, $a(a-1)(a+1)=a^3-a=24a$. Rearranging and factoring, $a(a+5)(a-5)=0$, so $a=5$. Hence, the sum of the squares is $4^2+5^2+6^2=77\Longrightarrow\boxed{\mathrm{ (B) \ }}$.

Solution 2

Backsolving from the answers. We can easily note that the five given answers correspond to $(3^2+4^2+5^2)$, $(4^2+5^2+6^2)$, ..., $(7^2+8^2+9^2)$. We can easily check each of these five possibilities and pick the correct one.

(It is not necessary to discover the exact form of the given answers. The observation that the answer is between $50$ and $194$ is enough to bound the search to the five possibilities mentioned above.)


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
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