Difference between revisions of "2000 AIME I Problems/Problem 7"
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note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques | ||
− | + | ==Solution 1== | |
We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>. | We can rewrite <math>xyz=1</math> as <math>\frac{1}{z}=xy</math>. | ||
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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>. | ||
− | + | ==Solution 2== | |
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | ||
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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>. | ||
− | + | == Solution 3 == | |
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>, <math>y=24</math>, and <math>z=5/24</math>, so the answer is 4+1 which is equal to <math>5</math>. | 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>, <math>y=24</math>, and <math>z=5/24</math>, so the answer is 4+1 which is equal to <math>5</math>. | ||
− | + | == Solution 4 == | |
(Hybrid between 1/2) | (Hybrid between 1/2) | ||
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-AlexLikeMath | -AlexLikeMath | ||
− | + | ==Solution 5== | |
Get rid of the denominators in the second and third equations to get <math>xz-5z=-1</math> and <math>xy-29x=-1</math>. Then, since <math>xyz=1</math>, we have <math>\tfrac 1y-5z=-1</math> and <math>\tfrac 1z-29x=-1</math>. Then, since we know that <math>\tfrac 1z+x=5</math>, we can subtract these two equations to get that <math>30x=6\implies x=5</math>. The result follows that <math>z=\tfrac 5{24}</math> and <math>y=24</math>, so <math>z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14</math>, and the requested answer is <math>1+4=\boxed{005}.</math> | Get rid of the denominators in the second and third equations to get <math>xz-5z=-1</math> and <math>xy-29x=-1</math>. Then, since <math>xyz=1</math>, we have <math>\tfrac 1y-5z=-1</math> and <math>\tfrac 1z-29x=-1</math>. Then, since we know that <math>\tfrac 1z+x=5</math>, we can subtract these two equations to get that <math>30x=6\implies x=5</math>. The result follows that <math>z=\tfrac 5{24}</math> and <math>y=24</math>, so <math>z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14</math>, and the requested answer is <math>1+4=\boxed{005}.</math> | ||
Latest revision as of 19:45, 17 April 2021
Problem
Suppose that and are three positive numbers that satisfy the equations and Then where and are relatively prime positive integers. Find .
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques
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 .
Solution 4
(Hybrid between 1/2)
Because and . Substituting and factoring, we get , , and . Multiplying them all together, we get, , but is , and by the Identity property of multiplication, we can take it out. So, in the end, we get . And, we can expand this to get , and if we make a substitution for , and rearrange the terms, we get This will be important.
Now, lets add the 3 equations , and . We use the expand the Left hand sides, then, we add the equations to get Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus We move all constant terms to the right, and all linear terms to the left, to get , so which gives an answer of
-AlexLikeMath
Solution 5
Get rid of the denominators in the second and third equations to get and . Then, since , we have and . Then, since we know that , we can subtract these two equations to get that . The result follows that and , so , and the requested answer is
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.