Difference between revisions of "2020 AIME II Problems/Problem 12"
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==Solution== | ==Solution== | ||
− | Let us take some cases. Since m and n are odds, and 200 is in the top row and 2000 in the bottom, m has to be 3, 5, 7 or 9. 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>. | + | 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 m | + | 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 m | + | 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 m | + | 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 m | + | Let <math>m = 9</math>. Then <math>n 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$. |
Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath> | Add all of our cases to get <cmath> 165+48+22+13 = \boxed{248} </cmath> | ||
-Solution by thanosaops | -Solution by thanosaops | ||
+ | |||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=11|num-a=13}} | {{AIME box|year=2020|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:58, 8 June 2020
Problem
Let and
be odd integers greater than
An
rectangle is made up of unit squares where the squares in the top row are numbered left to right with the integers
through
, those in the second row are numbered left to right with the integers
through
, and so on. Square
is in the top row, and square
is in the bottom row. Find the number of ordered pairs
of odd integers greater than
with the property that, in the
rectangle, the line through the centers of squares
and
intersects the interior of square
.
Solution
Let us take some cases. Since and
are odds, and
is in the top row and
in the bottom,
has to be
,
,
, or
. Also, taking a look at the diagram, the slope of the line connecting those centers has to have an absolute value of
. Therefore,
.
If ,
can range from
to
. However,
divides
, so looking at mods, we can easily eliminate
and
. Now, counting these odd integers, we get
.
Similarly, let . Then
can range from
to
. However,
, so one can remove
and
. Counting odd integers, we get
.
Take . Then,
can range from
to
. However,
, so one can verify and eliminate
and
. Counting odd integers, we get
.
Let . Then
223
249
225|1800
225
14 - 1 = 13$.
Add all of our cases to get
-Solution by thanosaops
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.