Difference between revisions of "2000 AIME I Problems/Problem 7"
Flyhawkeye (talk | contribs) m (→Solution 3) |
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We can then substitute once again to get | We can then substitute once again to get | ||
<cmath>x=\frac15</cmath> | <cmath>x=\frac15</cmath> | ||
− | <cmath> | + | <cmath>z=\frac{5}{24}.</cmath> |
Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>. | Thus, <math>z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}</math>, so <math>m+n=\boxed{005}</math>. | ||
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===Solution 2=== | ===Solution 2=== |
Revision as of 05:22, 10 January 2019
Problem
Suppose that
and
are three positive numbers that satisfy the equations
and
Then
where
and
are relatively prime positive integers. Find
.
Solution 1
We can rewrite as
.
Substituting into one of the given equations, we have
We can substitute back into to obtain
We can then substitute once again to get
Thus,
, so
.
Solution 2
Let .
Thus . So
.
Solution 3
Since , so
. Also,
by the second equation. Substitution gives
,
, and
, so the answer is 4+1 which is equal to
.
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.