Difference between revisions of "2005 AIME I Problems/Problem 4"
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=== Solution 3 === | === Solution 3 === | ||
− | The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|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>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. | + | The number of members is <math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by <math>4</math> and [[completing the square|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>. Since we want to maximize <math>n</math>, set the first factor equal to <math>69</math> and the second equal to <math>1</math>. Now we have the systems of equations |
+ | <cmath>2n+2m+7 = 69</cmath> | ||
+ | <cmath>2n-2m+7 = 1</cmath> | ||
+ | Adding these two equations gives us <math>4n+14=70</math> which means <math>4n=56</math> so <math>n=14</math>. | ||
+ | Thus, the answer is <math>14\cdot21=294</math>. <math>\square</math> | ||
+ | |||
+ | ~lpieleanu | ||
=== Solution 4 === | === Solution 4 === | ||
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~ CELLSecret | ~ CELLSecret | ||
− | ==Solution 7 (Bash)== | + | ==Solution 7 (Bash-not advisable)== |
Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000). | Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000). | ||
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\end{tabular}</cmath> | \end{tabular}</cmath> | ||
Thus, we can see that our largest valid <math>n(n+7)</math> is <math>\boxed{294}</math>. | Thus, we can see that our largest valid <math>n(n+7)</math> is <math>\boxed{294}</math>. | ||
+ | |||
+ | ~lpieleanu (Minor reformatting and editing) | ||
== See also == | == See also == |
Latest revision as of 11:43, 16 August 2022
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.
Contents
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
. Now we have the systems of equations
Adding these two equations gives us
which means
so
.
Thus, the answer is
.
~lpieleanu
Solution 4
Partially completing the square
Geometrically: Split up the formation of rows and
columns into a square of
rows and
columns and a separate rectangle of the dimensions
rows by
columns. We want to take the rows from the rectangle and add them to the square to get another square and
left over. If we attach exactly
rows on the top and exactly
rows on the side of the
x
square, then we have an
x
square that's missing a
x
corner. For the remaining
to fill this square plus the
extra members,
must be
. If we instead plaster exactly
rows from the
x
formation to two adjacent sides of the
x
square, we have an
x
formation that's missing a
x
corner. For the remaining row of length
to fill this plus five,
. Plugging these in, we find
has a much higher count of members:
Algebraically: We have , where
is the number of members in the band and
is a positive integer. We partially complete the square for
to get
Our goal is to get
because we want
to be
more than a perfect square. From the above,
means
isn't an integer,
means that
, and
means that
. Out of these,
is associated with the highest number of members in the band, so
Solution 5
Let there be members and
members for the square and
for the number of columns of the other formation. We have
To maximize this we let and
Solving we find
so the desired number of members is
Solution 6 (NO ALGEBRA)
Think of the process of moving people from the last column to new rows. Since there are less columns than rows, for each column removed, there are people discarded to the "extra" pile to be placed at the end. To maximize the number of "extra" people to fill in the last few rows. We remove 3 columns and add 4 rows. For the first new row, one more person will be discarded. For the second, 1 extra person are added since there is one more row now, and there are 2 less columns. Thus, there are 3 extra people discarded. Similarly, 5 extra people are discarded for the third column. Now there are people in the extra pile to put as the last row, so there are
people.
~ CELLSecret
Solution 7 (Bash-not advisable)
Note: Only do this if you have a LOT of time (and you've memorized all your perfect squares up to 1000).
We can see that the number of members in the band must be of the form for some positive integer
. When
, this product is
, and since AIME answers are nonnegative integers less than
, we don't have to check any higher
. Also, we know that this product must be 5 more than a perfect square, so we can make a table as shown and bash:
Thus, we can see that our largest valid
is
.
~lpieleanu (Minor reformatting and editing)
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.