Difference between revisions of "2000 AIME I Problems/Problem 7"
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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>, <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>. | ||
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+ | === 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. | ||
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+ | 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> | ||
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+ | -AlexLikeMath | ||
== See also == | == See also == |
Revision as of 16:27, 26 June 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
.
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
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.