Difference between revisions of "2012 AMC 10A Problems/Problem 16"
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Plugging in <math>500</math> for <math>0.4t</math> we get <math>t=1250</math>, but this does not work for <math>0.6t</math> (<math>750</math> isn't a multiple of <math>500</math>). Plugging in <math>0.4t=1000</math>, we get <math>t=2500</math>, and this does work for <math>0.6t</math>. | Plugging in <math>500</math> for <math>0.4t</math> we get <math>t=1250</math>, but this does not work for <math>0.6t</math> (<math>750</math> isn't a multiple of <math>500</math>). Plugging in <math>0.4t=1000</math>, we get <math>t=2500</math>, and this does work for <math>0.6t</math>. | ||
− | Therefore, <math>t=2500</math> and the answer is <math>\textbf{(C) } 2500</math>. | + | Therefore, <math>t=2500</math> and the answer is <math>\boxed{\textbf{(C) } 2500}</math>. |
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+ | *Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer | ||
+ | |||
+ | ==Solution 5== | ||
+ | Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition... | ||
+ | After <math>t</math> seconds, respectively the runners would've ran <math>4.4t, 4.8t,</math> and <math>5t</math> meters. These three values are congruent <math>\pmod{500}</math>, so | ||
+ | <cmath>4.4t \equiv 4.8t \equiv 5t \pmod{500}</cmath>. Subtract <math>4.4t</math> from all three sides to get <math>0, 0.4t,</math> and <math>0.6t</math> are congruent. Now all we need to find is a value of <math>t</math> for which <math>0.4t</math> and <math>0.6t</math> are congruent <math>\pmod{500}</math>. Subtract <math>0.4t</math> from both sides to get <math>0.2t</math> and <math>0</math> are congruent mod <math>500</math>, or that <math>0.2t=\dfrac{t}{5}</math> is a multiple of <math>500</math>. Let <math>t=500k</math>, so we want <math>100k</math> to be a multiple of <math>500</math>, or <math>k</math> to be a multiple of <math>5</math>. Therefore, the smallest value of <math>t</math> is when <math>k=5</math>, and when <math>t=500k=500(5)=2500 \space \boxed{(\text{C})}</math> | ||
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+ | - Solution by mathchampion1 | ||
== See Also == | == See Also == |
Latest revision as of 00:21, 14 June 2024
Problem
Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
Solution 1
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let be the time these runners run in seconds.
Because is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of and results in the 5.0 m/s runner to end up on the start line. Every
seconds, that fastest runner goes
meters. In
seconds, he goes
meters. Therefore the runners run
seconds.
Solution 2
Working backwards from the answers starting with the smallest answer, if they had run seconds, they would have run
meters, respectively. The first two runners have a difference of
meters, which is not a multiple of
(one lap), so they are not in the same place.
If they had run seconds, the runners would have run
meters, respectively. The last two runners have a difference of
meters, which is not a multiple of
.
If they had run seconds, the runners would have run
meters, respectively. The distance separating each pair of runners is a multiple of
, so the answer is
seconds.
Solution 3
Let be the time run in seconds, then the difference in meters run between the three runners is
. For them to be at the same location all of them need to be multiples of 500. It is now easy to see that
, so
.
Solution 4
After seconds, respectively the runners would've ran
and
meters. Their current positions on the track are these values
. We're trying to find the value of
such that
Subtracting
on all sides, we get
Now, we must find a value for
such that both
and
are simultaneously multiples of
.
Plugging in for
we get
, but this does not work for
(
isn't a multiple of
). Plugging in
, we get
, and this does work for
.
Therefore, and the answer is
.
- Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer
Solution 5
Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition...
After seconds, respectively the runners would've ran
and
meters. These three values are congruent
, so
. Subtract
from all three sides to get
and
are congruent. Now all we need to find is a value of
for which
and
are congruent
. Subtract
from both sides to get
and
are congruent mod
, or that
is a multiple of
. Let
, so we want
to be a multiple of
, or
to be a multiple of
. Therefore, the smallest value of
is when
, and when
- Solution by mathchampion1
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.