Difference between revisions of "2005 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
− | The director of a marching band | + | 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. |
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Revision as of 18:53, 16 December 2019
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
.
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 is for some
and
. Multiply both sides by
and complete the square to get
. Thus, we have
. Since we want to maximize
, set the first factor equal to
and the second equal to
. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.