Difference between revisions of "2005 AIME I Problems/Problem 4"
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=== 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> | + | 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 20:21, 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 then
and so
. If
is an integer there are no numbers which are 5 more than a perfect square strictly between
and
. Thus, if the number of columns is
, the number of students is
which must be 5 more than a perfect square, so
. In fact, when
we have
, 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 , and the number of rows of the rectangular formation
(so there are
columns). Thus,
. The quadratic formula yields
.
must be an integer, say
. Then
and
. The factors of
are
;
is maximized for the first case. Thus,
, and
. The latter obviously can be discarded, so there are
rows and
columns, making the answer
.
Solution 3
The number of members= for some
and
. Multiply both sides by 4 and complete the square to get
. Thus, we have
. Sine we want to maximize
, set the first factor equal to 69 and the second equal to 1. Solving gives
, so the answer is
.
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 |