Difference between revisions of "2000 AIME II Problems/Problem 2"

 
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== Problem ==
 
== Problem ==
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola <math>x^2 - y^2 = 2000^2</math>?
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A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola <math>x^2 - y^2 = 2000^2</math>?
  
 
== Solution ==
 
== Solution ==
<math>(x-y)(x+y)=2000^2=2^8*5^6</math>
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<cmath>(x-y)(x+y)=2000^2=2^8 \cdot 5^6</cmath>
  
Since there are <math>7*9=63</math> factors of 2000^2, we have 63 lattice points.
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Note that <math>(x-y)</math> and <math>(x+y)</math> have the same [[parity|parities]], so both must be even. We first give a factor of <math>2</math> to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6 \cdot 5^6</math> left. Since there are <math>7 \cdot 7=49</math> factors of <math>2^6 \cdot 5^6</math>, and since both <math>x</math> and <math>y</math> can be negative, this gives us <math>49\cdot2=\boxed{098}</math> lattice points.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2000|n=II|num-b=1|num-a=3}}
 
{{AIME box|year=2000|n=II|num-b=1|num-a=3}}
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[[Category:Intermediate Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 19:30, 4 July 2013

Problem

A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola $x^2 - y^2 = 2000^2$?

Solution

\[(x-y)(x+y)=2000^2=2^8 \cdot 5^6\]

Note that $(x-y)$ and $(x+y)$ have the same parities, so both must be even. We first give a factor of $2$ to both $(x-y)$ and $(x+y)$. We have $2^6 \cdot 5^6$ left. Since there are $7 \cdot 7=49$ factors of $2^6 \cdot 5^6$, and since both $x$ and $y$ can be negative, this gives us $49\cdot2=\boxed{098}$ lattice points.

See also

2000 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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