Difference between revisions of "2000 AIME I Problems/Problem 7"
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Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>. | Thus <math>145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}</math>. So <math>m + n = 1 + 4 = \boxed{5}</math>. | ||
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| + | == Solution 2 == | ||
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| + | Since <math>x+(1/z)=5, 1=z(5-x)=xyz</math>, so <math>5-x=xy</math>. Also, <math>y=29-(1/x)</math> by the second equation. Substitution gives <math>x=1/5</math>, y=24<math>, and </math>z=5/24<math>, so the solution is </math>5$. | ||
== See also == | == See also == | ||
Revision as of 17:51, 6 January 2008
Contents
Problem
Suppose that 
and 
are three positive numbers that satisfy the equations 
and 
Then 
where 
and 
are relatively prime positive integers. Find 
.
Solution
Let 
.
Thus 
. So 
.
Solution 2
Since 
, so 
. Also, 
by the second equation. Substitution gives 
, y=24
z=5/24
5$.
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 | ||
