Difference between revisions of "2008 AIME I Problems/Problem 3"
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When readdressing the first equation, we see that if <math>2x+3y</math> will be a multiple of <math>6</math>, <math>4z \equiv 2 \pmod{6} = 5</math>, we get that <math>x=15</math> and <math>y=8</math>, which works because of integer values. Therefore, <math>225+64+25=\boxed{314}</math> | When readdressing the first equation, we see that if <math>2x+3y</math> will be a multiple of <math>6</math>, <math>4z \equiv 2 \pmod{6} = 5</math>, we get that <math>x=15</math> and <math>y=8</math>, which works because of integer values. Therefore, <math>225+64+25=\boxed{314}</math> | ||
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+ | === Solution 4 (Logic) === | ||
+ | Building on top of Solution 3, we can add <math>j+s-2b=17</math> and <math>2b+3j+4s=74</math> (sorry, I used different variables) to get <math>4j+5s=57</math>. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get <math>4j+25=57\implies j=8</math>. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into <math>2b+3j+4s=74</math>, we get <math>b=15</math>. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get <math>5^2+8^2+15^2=\boxed{314}</math>. | ||
== See also == | == See also == |
Revision as of 18:44, 6 August 2022
Problem
Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers kilometers after biking for
hours, jogging for
hours, and swimming for
hours, while Sue covers
kilometers after jogging for
hours, swimming for
hours, and biking for
hours. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.
Contents
Solution
Solution 1
Let the biking rate be , swimming rate be
, jogging rate be
, all in km/h.
We have . Subtracting the second from twice the first gives
. Mod 4, we need
. Thus,
.
and
give non-integral
, but
gives
. Thus, our answer is
.
Solution 2
Let ,
, and
be the biking, jogging, and swimming rates of the two people. Hence,
and
. Subtracting gives us that
. Adding three times this to the first equation gives that
. Adding four times the previous equation to the first given one gives us that
. This gives us that
, and then
and
. Therefore,
.
Solution 3
Creating two systems, we get , and
. Subtracting the two expressions we get
. Note that
is odd, so one of
is odd. We see from our second expression that
must be odd, because
is also odd and
and
are odd. Thus, with this information, we can test cases quickly:
When readdressing the first equation, we see that if will be a multiple of
,
, we get that
and
, which works because of integer values. Therefore,
Solution 4 (Logic)
Building on top of Solution 3, we can add and
(sorry, I used different variables) to get
. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get
. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into
, we get
. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get
.
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.