Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AIME I Problems/Problem 4"

(alternative solution, box)
(→‎Solution)
Line 8: Line 8:
  
 
=== Solution 2 ===
 
=== Solution 2 ===
−
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\displaystyle \sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.
+
Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>.
 +
 
 +
=== Solution 3 ===
 +
The number of members=<math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by 4 and complete the square to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Sine we want to maximize <math>n</math>, set the first factor equal to 69 and the second equal to 1. Solving gives <math>n=14</math>, so the answer is <math>14x21=294</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2005|n=I|num-b=3|num-a=5}}

Revision as of 19:21, 29 December 2007

Problem

The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have.

Solution

Solution 1

If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.

Solution 2

Define the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$.

Solution 3

The number of members=$m^2+5=n(n+7)$ for some $n$ and $m$. Multiply both sides by 4 and complete the square to get $4m^2+69=(2n+7)^2$. Thus, we have $69=((2n+7)+2m)((2n+7)-2m)$. Sine we want to maximize $n$, set the first factor equal to 69 and the second equal to 1. Solving gives $n=14$, so the answer is $14x21=294$.

See also

2005 AIME I (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_4&oldid=21676"