Difference between revisions of "2000 AIME I Problems/Problem 7"
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Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | Suppose that <math>x,</math> <math>y,</math> and <math>z</math> are three positive numbers that satisfy the equations <math>xyz = 1,</math> <math>x + \frac {1}{z} = 5,</math> and <math>y + \frac {1}{x} = 29.</math> Then <math>z + \frac {1}{y} = \frac {m}{n},</math> where <math>m</math> and <math>n</math> are [[relatively prime]] positive integers. Find <math>m + n</math>. | ||
− | + | ==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>. | ||
Line 18: | Line 17: | ||
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>. | ||
− | + | ==Solution 2== | |
− | |||
Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | Let <math>r = \frac{m}{n} = z + \frac {1}{y}</math>. | ||
Line 38: | Line 36: | ||
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) | ||
+ | |||
+ | Because <math>xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz, </math> and <math>\hspace{0.05cm}\frac{1}{z} = xy</math>. Substituting and factoring, we get <math>x(y+1) = 5</math>, <math>\hspace{0.15cm}y(z+1) = 29</math>, and <math>\hspace{0.05cm}z(x+1) = k</math>. Multiplying them all together, we get, <math>xyz(x+1)(y+1)(z+1) = 145k</math>, but <math>xyz</math> is <math>1</math>, and by the Identity property of multiplication, we can take it out. So, in the end, we get <math>(x+1)(y+1)(z+1) = 145k</math>. And, we can expand this to get <math>xyz+xy+yz+xz+x+y+z+1 = 145k</math>, and if we make a substitution for <math>xyz</math>, and rearrange the terms, we get <math>xy+yz+xz+x+y+z = 145k-2</math> This will be important. | ||
+ | |||
+ | |||
+ | Now, lets add the 3 equations <math>x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29 </math>, and <math>\hspace{0.05cm}z(x+1) = k</math>. We use the expand the Left hand sides, then, we add the equations to get <math>xy+yz+xz+x+y+z = k+34</math> 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 <math>145k-2 = k+34</math> We move all constant terms to the right, and all linear terms to the left, to get <math>144k = 36</math>, so <math>k = \frac{1}{4}</math> which gives an answer of <math>1+4 = \boxed{005}</math> | ||
+ | |||
+ | -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> | ||
+ | |||
+ | ==Solution 6== | ||
+ | Rewrite the equations in terms of x. | ||
+ | |||
+ | <math>x+\frac{1}{z}=5</math> becomes <math>z=\frac{1}{x+5}</math>. | ||
+ | |||
+ | <math>y+\frac{1}{x}=29</math> becomes <math>y=29-\frac{1}{x}</math> | ||
+ | |||
+ | Now express <math>xyz=1</math> in terms of x. | ||
+ | |||
+ | <math>\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1</math>. | ||
+ | |||
+ | This evaluates to <math>29x-1=5-x</math>, giving us <math>x=\frac{1}{5}</math>. We can now plug x into the other equations to get <math>y=24</math> and <math>z=\frac{5}{24}</math>. | ||
+ | |||
+ | Therefore, <math>z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}</math>. | ||
+ | |||
+ | <math>1+4=\boxed{5}</math>, and we are done. | ||
+ | ~MC413551 | ||
== See also == | == See also == |
Latest revision as of 17:16, 27 September 2024
Contents
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 .
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
Solution 6
Rewrite the equations in terms of x.
becomes .
becomes
Now express in terms of x.
.
This evaluates to , giving us . We can now plug x into the other equations to get and .
Therefore, .
, and we are done. ~MC413551
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 |
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