Difference between revisions of "2020 AIME II Problems/Problem 12"
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− | Let <math>m</math> and <math>n</math> be odd integers greater than <math>1.</math> An <math>m\times n</math> rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers <math>1</math> through <math>n</math>, those in the second row are numbered left to right with the integers <math>n + 1</math> through <math>2n</math>, and so on. Square <math>200</math> is in the top row, and square <math>2000</math> is in the bottom row. Find the number of ordered pairs <math>(m,n)</math> of odd integers greater than <math>1</math> with the property that, in the <math>m\times n</math> rectangle, the line through the centers of squares <math>200</math> and <math>2000</math> intersects the interior of square <math>1099</math> | + | ==Problem== |
+ | Let <math>m</math> and <math>n</math> be odd integers greater than <math>1.</math> An <math>m\times n</math> rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers <math>1</math> through <math>n</math>, those in the second row are numbered left to right with the integers <math>n + 1</math> through <math>2n</math>, and so on. Square <math>200</math> is in the top row, and square <math>2000</math> is in the bottom row. Find the number of ordered pairs <math>(m,n)</math> of odd integers greater than <math>1</math> with the property that, in the <math>m\times n</math> rectangle, the line through the centers of squares <math>200</math> and <math>2000</math> intersects the interior of square <math>1099</math>. | ||
+ | |||
+ | ==Solution== | ||
+ | Let us take some cases. Since <math>m</math> and <math>n</math> are odds, and <math>200</math> is in the top row and <math>2000</math> in the bottom, <math>m</math> has to be <math>3</math>, <math>5</math>, <math>7</math>, or <math>9</math>. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of <math> < 1</math>. Therefore, <math>m < 1800 \mod n < 1800-m</math>. | ||
+ | |||
+ | If <math>m=3</math>, <math>n</math> can range from <math>667</math> to <math>999</math>. However, <math>900</math> divides <math>1800</math>, so looking at mods, we can easily eliminate <math>899</math> and <math>901</math>. Now, counting these odd integers, we get <math>167 - 2 = 165</math>. | ||
+ | |||
+ | Similarly, let <math>m=5</math>. Then <math>n</math> can range from <math>401</math> to <math>499</math>. However, <math>450|1800</math>, so one can remove <math>449</math> and <math>451</math>. Counting odd integers, we get <math>50 - 2 = 48</math>. | ||
+ | |||
+ | Take <math>m=7</math>. Then, <math>n</math> can range from <math>287</math> to <math>333</math>. However, <math>300|1800</math>, so one can verify and eliminate <math>299</math> and <math>301</math>. Counting odd integers, we get <math>24 - 2 = 22</math>. | ||
+ | |||
+ | Let <math>m = 9</math>. Then <math>n</math> can vary from <math>223</math> to <math>249</math>. However, <math>225|1800</math>. Checking that value and the values around it, we can eliminate <math>225</math>. Counting odd integers, we get <math>14 - 1 = 13</math>. | ||
+ | |||
+ | Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath> | ||
+ | |||
+ | -Solution by thanosaops | ||
+ | |||
+ | ==Solution 2 (Official MAA)== | ||
+ | Because square <math>2000</math> is in the bottom row, it follows that <math>\frac{2000}m \le n < \frac{2000}{m-1}</math>. Moreover, because square <math>200</math> is in the top row, and square <math>2000</math> is not in the top row, <math>1 < m \le 10</math>. In particular, because the number of rows in the rectangle must be odd, <math>m</math> must be one of <math>3, 5, 7,</math> or <math>9.</math> | ||
+ | |||
+ | For each possible choice of <math>m</math> and <math>n</math>, let <math>\ell_{m,n}</math> denote the line through the centers of squares <math>200</math> and <math>2000.</math> Note that for odd values of <math>m</math>, the line <math>\ell_{m,n}</math> passes through the center of square <math>1100.</math> Thus <math>\ell_{m,n}</math> intersects the interior of cell <math>1099</math> exactly when its slope is strictly between <math>-1</math> and <math>1</math>. The line <math>\ell_{m,n}</math> is vertical whenever square <math>2000</math> is the <math>200</math>th square in the bottom row of the rectangle. This would happen for <math>m = 3, 5, 7, 9</math> when <math>n = 900, 450, 300, 225</math>, respectively. When <math>n</math> is 1 greater than or 1 less than these numbers, the slope of <math>\ell_{m,n}</math> is <math>1</math> or <math>-1</math>, respectively. In all other cases the slope is strictly between <math>-1</math> and <math>1.</math> The admissible values for <math>n</math> for each possible value of <math>m</math> are given in the following table. | ||
+ | <cmath>\begin{tabular}{|c|c|c|c|c|}\hline | ||
+ | m & minimum n & maximum n & avoided n & number of odd n\\\hline | ||
+ | 3&667&999&899, 900, 901&165\\\hline | ||
+ | 5&400&499&449, 450, 451&48\\\hline | ||
+ | 7&286&333&299, 300, 301&22\\\hline | ||
+ | 9&223&249&224, 225, 226&13\\\hline | ||
+ | \end{tabular}</cmath> | ||
+ | This accounts for <math>165 + 48 + 22 + 13 = 248</math> rectangles. | ||
+ | |||
+ | ==Video Solution 1== | ||
+ | https://www.youtube.com/watch?v=MrtKoO16XLQ ~ MathEx | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/v58SLOoAKTw | ||
+ | |||
+ | ==Video Solution 3== | ||
+ | https://youtu.be/pu_79SSh3mM | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2020|n=II|num-b=11|num-a=13}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:10, 22 September 2022
Contents
Problem
Let and
be odd integers greater than
An
rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers
through
, those in the second row are numbered left to right with the integers
through
, and so on. Square
is in the top row, and square
is in the bottom row. Find the number of ordered pairs
of odd integers greater than
with the property that, in the
rectangle, the line through the centers of squares
and
intersects the interior of square
.
Solution
Let us take some cases. Since and
are odds, and
is in the top row and
in the bottom,
has to be
,
,
, or
. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of
. Therefore,
.
If ,
can range from
to
. However,
divides
, so looking at mods, we can easily eliminate
and
. Now, counting these odd integers, we get
.
Similarly, let . Then
can range from
to
. However,
, so one can remove
and
. Counting odd integers, we get
.
Take . Then,
can range from
to
. However,
, so one can verify and eliminate
and
. Counting odd integers, we get
.
Let . Then
can vary from
to
. However,
. Checking that value and the values around it, we can eliminate
. Counting odd integers, we get
.
Add all of our cases to get
-Solution by thanosaops
Solution 2 (Official MAA)
Because square is in the bottom row, it follows that
. Moreover, because square
is in the top row, and square
is not in the top row,
. In particular, because the number of rows in the rectangle must be odd,
must be one of
or
For each possible choice of and
, let
denote the line through the centers of squares
and
Note that for odd values of
, the line
passes through the center of square
Thus
intersects the interior of cell
exactly when its slope is strictly between
and
. The line
is vertical whenever square
is the
th square in the bottom row of the rectangle. This would happen for
when
, respectively. When
is 1 greater than or 1 less than these numbers, the slope of
is
or
, respectively. In all other cases the slope is strictly between
and
The admissible values for
for each possible value of
are given in the following table.
This accounts for
rectangles.
Video Solution 1
https://www.youtube.com/watch?v=MrtKoO16XLQ ~ MathEx
Video Solution 2
Video Solution 3
~MathProblemSolvingSkills.com
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.