Difference between revisions of "2000 AIME II Problems/Problem 2"
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<math>(x-y)(x+y)=2000^2=2^8*5^6</math> | <math>(x-y)(x+y)=2000^2=2^8*5^6</math> | ||
| − | We must have <math>(x-y)</math> and <math>(x+y)</math> as both even or else x,y would not be an integer. We first give a factor of two to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6*5^6</math> left. Since there are <math>7*7=49</math> factors of <math>2^6*5^6</math>, | + | We must have <math>(x-y)</math> and <math>(x+y)</math> as both even or else x,y would not be an integer. We first give a factor of two to both <math>(x-y)</math> and <math>(x+y)</math>. We have <math>2^6*5^6</math> left. Since there are <math>7*7=49</math> factors of <math>2^6*5^6</math>, and since both x and y can be negative, this gives us <math>49\cdot2=98</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}} | ||
Revision as of 13:55, 2 March 2008
Problem
A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola 
?
Solution
We must have 
and 
as both even or else x,y would not be an integer. We first give a factor of two to both and . We have 
left. Since there are 
factors of , and since both x and y can be negative, this gives us 
lattice points.
See also
| 2000 AIME II (Problems • Answer Key • Resources) | ||
| 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 | ||
