Difference between revisions of "2005 AIME I Problems/Problem 4"
(→Solution 5) |
Phoenixfire (talk | contribs) (→Solution 5) |
||
Line 27: | Line 27: | ||
Let there be <math>m</math> members and <math>n</math> members for the square and <math>c</math> for the number of columns of the other formation. We have <math>n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.</math> | Let there be <math>m</math> members and <math>n</math> members for the square and <math>c</math> for the number of columns of the other formation. We have <math>n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.</math> | ||
− | To maximize this we let <math>2n+ | + | To maximize this we let <math>2n+2c+7 = 68</math> and <math>2n-2c-7 = 1.</math> Solving we find <math>n = 17</math> so the desired number of members is <math>17^2 + 5 = \boxed{294}.</math> |
== See also == | == See also == |
Latest revision as of 00:20, 22 September 2020
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 . Solving gives , so the answer is .
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
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.