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Difference between revisions of "1985 AIME Problems/Problem 7"
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It follows from the givens that <math>a</math> is a [[perfect power | perfect fourth power]], <math>b</math> is a perfect fifth power, <math>c</math> is a [[perfect square]] and <math>d</math> is a [[perfect cube]].  Thus, there exist [[integer]]s <math>s</math> and <math>t</math> such that <math>a = t^4</math>, <math>b = t^5</math>, <math>c = s^2</math> and <math>d = s^3</math>.  So <math>s^2 - t^4 = 19</math>.  We can factor the left-hand side of this [[equation]] as a difference of two squares, <math>(s - t^2)(s + t^2) = 19</math>.  19 is a [[prime number]] and <math>s + t^2 > s - t^2</math> so we must have <math>s + t^2 = 19</math> and <math>s - t^2 = 1</math>.  Then <math>s = 10, t = 9</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d - b = 767</math>.
 
It follows from the givens that <math>a</math> is a [[perfect power | perfect fourth power]], <math>b</math> is a perfect fifth power, <math>c</math> is a [[perfect square]] and <math>d</math> is a [[perfect cube]].  Thus, there exist [[integer]]s <math>s</math> and <math>t</math> such that <math>a = t^4</math>, <math>b = t^5</math>, <math>c = s^2</math> and <math>d = s^3</math>.  So <math>s^2 - t^4 = 19</math>.  We can factor the left-hand side of this [[equation]] as a difference of two squares, <math>(s - t^2)(s + t^2) = 19</math>.  19 is a [[prime number]] and <math>s + t^2 > s - t^2</math> so we must have <math>s + t^2 = 19</math> and <math>s - t^2 = 1</math>.  Then <math>s = 10, t = 9</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d - b = 767</math>.
 
== See also ==
 
== See also ==
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* [[1985 AIME Problems/Problem 6 | Previous problem]]
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{{AIME box|year=1985|num-b=6|num-a=8}}
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* [[1985 AIME Problems/Problem 8 | Next problem]]
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* [[AIME Problems and Solutions]]
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* [[1985 AIME Problems]]
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* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 13:33, 6 May 2007

Problem

Assume that $a$, $b$, $c$, and $d$ are positive integers such that $a^5 = b^4$, $c^3 = d^2$, and $c - a = 19$. Determine $d - b$.

Solution

It follows from the givens that $a$ is a perfect fourth power, $b$ is a perfect fifth power, $c$ is a perfect square and $d$ is a perfect cube. Thus, there exist integers $s$ and $t$ such that $a = t^4$, $b = t^5$, $c = s^2$ and $d = s^3$. So $s^2 - t^4 = 19$. We can factor the left-hand side of this equation as a difference of two squares, $(s - t^2)(s + t^2) = 19$. 19 is a prime number and $s + t^2 > s - t^2$ so we must have $s + t^2 = 19$ and $s - t^2 = 1$. Then $s = 10, t = 9$ and so $d = s^3 = 1000$, $b = t^5 = 243$ and $d - b = 767$.

See also

1985 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AIME Problems and Solutions
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