2005 AIME I Problems/Problem 4
Problem
The director of a marching band asks the band members to line up in rows of four, but one is left over. Then she tries to line them up in rows of six, but three are left over. Finally, she tries to line them up in rows of seven, but four are left over. What is the maximum number of members that the 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 |
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