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Difference between revisions of "2001 AIME I Problems/Problem 11"

(solution by chess64)
(→‎Solution)
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We solve the system (the first two equations, and then the latter three) to get
 
We solve the system (the first two equations, and then the latter three) to get
 
<cmath>
 
<cmath>
−
\left(a_{1},a_{2},a_{3},a_{4},a_{5}\right) = \left(\frac {23 + 5N}{24},\frac {19 + N}{24},\frac {51 + 117N}{124},\frac {5 + 73N}{124},\frac {7 + 89N}{124}\right).
+
\left(a_{1},a_{2},a_{3},a_{4},a_{5}\right) = \left(\frac {23 + 5N}{24},\frac {19 + N}{24},\frac {51 + 117N}{124},\frac {35 + 73N}{124},\frac {7 + 89N}{124}\right).
 
</cmath>
 
</cmath>
 
<math>a_{1},a_{2}</math> will be integers iff <math>N\equiv 5\pmod{24}</math> and <math>a_{3},a_{4},a_{5}</math> will be integers iff <math>N\equiv 25\pmod{124}</math>. Solving these [[Congruent (modular arithmetic)|congruence]]s simultaneously by standard methods gives <math>N\equiv \boxed{149}\pmod{744}</math>.
 
<math>a_{1},a_{2}</math> will be integers iff <math>N\equiv 5\pmod{24}</math> and <math>a_{3},a_{4},a_{5}</math> will be integers iff <math>N\equiv 25\pmod{124}</math>. Solving these [[Congruent (modular arithmetic)|congruence]]s simultaneously by standard methods gives <math>N\equiv \boxed{149}\pmod{744}</math>.

Revision as of 16:13, 18 February 2012

Problem

In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth. Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column. Let $y_i$ be the number associated with $P_i$ after the renumbering. It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$

Solution

Let $P_{i} = (i,a_{i})$, where the first coordinate represents the row number and $a_i$ represents the column number. Then $x_{1} = a_{1}$, $x_{2} = N + a_{2}$, $x_{3} = 2N + a_{3}$, etc. and $y_{1} = 5(a_{1} - 1) + 1$, $y_{2} = 5(a_{2} - 1) + 2$, etc. Now we get the system of equations: \par \begin{align}a_{1} & = 5(a_{2} - 1) + 2 \\ N + a_{2} & = 5(a_{1} - 1) + 1 \\ 2N + a_{3} & = 5(a_{4} - 1) + 4 \\ 3N + a_{4} & = 5(a_{5} - 1) + 5 \\ 4N + a_{5} & = 5(a_{3} - 1) + 3 \end{align} We solve the system (the first two equations, and then the latter three) to get \[\left(a_{1},a_{2},a_{3},a_{4},a_{5}\right) = \left(\frac {23 + 5N}{24},\frac {19 + N}{24},\frac {51 + 117N}{124},\frac {35 + 73N}{124},\frac {7 + 89N}{124}\right).\] $a_{1},a_{2}$ will be integers iff $N\equiv 5\pmod{24}$ and $a_{3},a_{4},a_{5}$ will be integers iff $N\equiv 25\pmod{124}$. Solving these congruences simultaneously by standard methods gives $N\equiv \boxed{149}\pmod{744}$.

See also

2001 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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