Difference between revisions of "1989 AIME Problems/Problem 4"

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== Problem ==
 
== Problem ==
If <math>a<b<c<d<e^{}_{}</math> are consecutive positive integers such that <math>b+c+d^{}_{}</math> is a perfect square and <math>a+b+c+d+e^{}_{}</math> is a perfect cube, what is the smallest possible value of <math>c^{}_{}</math>?
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If <math>a<b<c<d<e^{}_{}</math> are [[consecutive]] [[positive]] [[integer]]s such that <math>b+c+d^{}_{}</math> is a [[perfect square]] and <math>a+b+c+d+e^{}_{}</math> is a [[perfect cube]], what is the smallest possible value of <math>c^{}_{}</math>?
  
 
== Solution ==
 
== Solution ==
{{solution}}
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Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know <math>b + c + d = 3c</math> and <math>a + b + c + d + e = 5c</math>. Thus, <math>c</math> must be in the form of <math>3 \cdot x^2</math> based upon the first part and in the form of <math>5^2 \cdot y^3</math> based upon the second part, with <math>x</math> and <math>y</math> denoting an [[integer]]s. <math>c</math> is minimized if it’s [[prime factorization]] contains only <math>3,5</math>, and since there is a cubed term in <math>5^2 \cdot y^3</math>, <math>3^3</math> must be a factor of <math>c</math>. <math>3^35^2 = 675</math>, which works as the solution.
  
 
== See also ==
 
== See also ==
* [[1989 AIME Problems/Problem 5|Next Problem]]
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{{AIME box|year=1989|num-b=3|num-a=5}}
* [[1989 AIME Problems/Problem 3|Previous Problem]]
 
* [[1989 AIME Problems]]
 

Revision as of 20:57, 26 February 2007

Problem

If $a<b<c<d<e^{}_{}$ are consecutive positive integers such that $b+c+d^{}_{}$ is a perfect square and $a+b+c+d+e^{}_{}$ is a perfect cube, what is the smallest possible value of $c^{}_{}$?

Solution

Since the middle term of an arithmetic progression with an odd number of terms is the average of the series, we know $b + c + d = 3c$ and $a + b + c + d + e = 5c$. Thus, $c$ must be in the form of $3 \cdot x^2$ based upon the first part and in the form of $5^2 \cdot y^3$ based upon the second part, with $x$ and $y$ denoting an integers. $c$ is minimized if it’s prime factorization contains only $3,5$, and since there is a cubed term in $5^2 \cdot y^3$, $3^3$ must be a factor of $c$. $3^35^2 = 675$, which works as the solution.

See also

1989 AIME (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
All AIME Problems and Solutions
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