Difference between revisions of "1985 AIME Problems/Problem 7"
(→Solution) |
|||
Line 2: | Line 2: | ||
Assume that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are [[positive integer]]s such that <math>a^5 = b^4</math>, <math>c^3 = d^2</math>, and <math>c - a = 19</math>. Determine <math>d - b</math>. | Assume that <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are [[positive integer]]s such that <math>a^5 = b^4</math>, <math>c^3 = d^2</math>, and <math>c - a = 19</math>. Determine <math>d - b</math>. | ||
== Solution == | == Solution == | ||
− | 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 = | + | 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 = 3</math> and so <math>d = s^3 = 1000</math>, <math>b = t^5 = 243</math> and <math>d - b = 757</math>. |
== See also == | == See also == |
Revision as of 17:36, 21 October 2007
Problem
Assume that , , , and are positive integers such that , , and . Determine .
Solution
It follows from the givens that is a perfect fourth power, is a perfect fifth power, is a perfect square and is a perfect cube. Thus, there exist integers and such that , , and . So . We can factor the left-hand side of this equation as a difference of two squares, . 19 is a prime number and so we must have and . Then and so , and .
See also
1985 AIME (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 | ||
All AIME Problems and Solutions |