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

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It turns out that there exists such an array satisfying the problem conditions if and only if
 
It turns out that there exists such an array satisfying the problem conditions if and only if
 
<cmath>N\equiv 149 \pmod{744}</cmath>
 
<cmath>N\equiv 149 \pmod{744}</cmath>
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In addition, the first two equation can be written <math>n = 5mod24</math>, and chasing variables in the last three equation gives us <math>89n + 7 = 124e</math>. With these two equations you may skip a lot of rewriting and testing. <math>\boxed{149}</math> still appears as our answer.
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-jackshi2006
  
 
== See also ==
 
== See also ==

Latest revision as of 12:51, 22 July 2020

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 each point $P_i$ be in column $c_i$. The numberings for $P_i$ can now be defined as follows. \begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*}

We can now convert the five given equalities. \begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{align} Equations $(1)$ and $(2)$ combine to form \[N = 24c_2 - 19\] Similarly equations $(3)$, $(4)$, and $(5)$ combine to form \[117N +51 = 124c_3\] Take this equation modulo 31 \[24N+20\equiv 0 \pmod{31}\] And substitute for N \[24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}\] \[18 c_2 \equiv 2 \pmod{31}\]

Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$

The column values can also easily be found by substitution \begin{align*}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{align*} As these are all positive and less than $N$, $\boxed{149}$ is the solution.

Sidenote

If we express all the $c_i$ in terms of $N$, we have \[24c_1=5N+23\] \[24c_2=N+19\] \[124c_3=117N+51\] \[124c_4=73N+35\] \[124c_5=89N+7\]

It turns out that there exists such an array satisfying the problem conditions if and only if \[N\equiv 149 \pmod{744}\]


In addition, the first two equation can be written $n = 5mod24$, and chasing variables in the last three equation gives us $89n + 7 = 124e$. With these two equations you may skip a lot of rewriting and testing. $\boxed{149}$ still appears as our answer.


-jackshi2006

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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