Difference between revisions of "2001 AIME I Problems/Problem 11"
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== Solution == | == Solution == | ||
+ | |||
+ | Let each point <math>P_i</math> be in column <math>c_i</math>. The numberings for <math>P_i</math> can now be defined as follows. | ||
+ | <cmath>\begin{align*}x_i &= (i - 1)N + c_i\\ | ||
+ | y_i &= (c_i - 1)5 + i | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | We can now convert the five given equalities. | ||
+ | <cmath>\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}</cmath> | ||
+ | Equations <math>(1)</math> and <math>(2)</math> combine to form | ||
+ | <cmath>N = 24c_2 - 19</cmath> | ||
+ | Similarly equations <math>(3)</math>, <math>(4)</math>, and <math>(5)</math> combine to form | ||
+ | <cmath>117N +51 = 124c_3</cmath> | ||
+ | Take this equation modulo 31 | ||
+ | <cmath>24N+20\equiv 0 \pmod{31}</cmath> | ||
+ | And substitute for N | ||
+ | <cmath>24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}</cmath> | ||
+ | <cmath>18 c_2 \equiv 2 \pmod{31}</cmath> | ||
+ | |||
+ | Thus the smallest <math>c_2</math> might be is <math>7</math> and by substitution <math>N = 24 \cdot 7 - 19 = 149</math> | ||
+ | |||
+ | The column values can also easily be found by substitution | ||
+ | <cmath>\begin{align*}c_1&=32\\ | ||
+ | c_2&=7\\ | ||
+ | c_3&=141\\ | ||
+ | c_4&=88\\ | ||
+ | c_5&=107 | ||
+ | \end{align*}</cmath> | ||
+ | As these are all positive and less than <math>N</math>, <math>\boxed{149}</math> is the solution. | ||
== See also == | == See also == |
Revision as of 14:25, 8 May 2014
Problem
In a rectangular array of points, with 5 rows and columns, the points are numbered consecutively from left to right beginning with the top row. Thus the top row is numbered 1 through
the second row is numbered
through
and so forth. Five points,
and
are selected so that each
is in row
Let
be the number associated with
Now renumber the array consecutively from top to bottom, beginning with the first column. Let
be the number associated with
after the renumbering. It is found that
and
Find the smallest possible value of
Solution
Let each point be in column
. The numberings for
can now be defined as follows.
We can now convert the five given equalities.
Equations
and
combine to form
Similarly equations
,
, and
combine to form
Take this equation modulo 31
And substitute for N
Thus the smallest might be is
and by substitution
The column values can also easily be found by substitution
As these are all positive and less than
,
is the solution.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.