Difference between revisions of "2000 AIME I Problems/Problem 6"
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For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>? | For how many [[ordered pair]]s <math>(x,y)</math> of [[integer]]s is it true that <math>0 < x < y < 10^6</math> and that the [[arithmetic mean]] of <math>x</math> and <math>y</math> is exactly <math>2</math> more than the [[geometric mean]] of <math>x</math> and <math>y</math>? | ||
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
<cmath>\begin{eqnarray*} | <cmath>\begin{eqnarray*} | ||
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For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation. | For simplicity, we can count how many valid pairs of <math>(\sqrt{x},\sqrt{y})</math> that satisfy our equation. | ||
− | The maximum that <math>\sqrt{y}</math> can be is <math>10^ | + | The maximum that <math>\sqrt{y}</math> can be is <math>\sqrt{10^6} - 1 = 999</math> because <math>\sqrt{y}</math> must be an integer (this is because <math>\sqrt{y} - \sqrt{x} = 2</math>, an integer). Then <math>\sqrt{x} = 997</math>, and we continue this downward until <math>\sqrt{y} = 3</math>, in which case <math>\sqrt{x} = 1</math>. The number of pairs of <math>(\sqrt{x},\sqrt{y})</math>, and so <math>(x,y)</math> is then <math>\boxed{997}</math>. |
<!-- solution lost in edit conflict - azjps - | <!-- solution lost in edit conflict - azjps - | ||
Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>. | Since <math>y>x</math>, it follows that each ordered pair <math>(x,y) = (n^2, (n+2)^2)</math> satisfies this equation. The minimum value of <math>x</math> is <math>1</math> and the maximum value of <math>y = 999^2</math> which would make <math>x = 997^2</math>. Thus <math>x</math> can be any of the squares between <math>1</math> and <math>997^2</math> inclusive and the answer is <math>\boxed{997}</math>. | ||
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− | === Solution 4 ( | + | === Solution 4 (Similar to Solution 3) === |
Rearranging our conditions to | Rearranging our conditions to | ||
Revision as of 19:26, 15 February 2021
Contents
[hide]Problem
For how many ordered pairs of integers is it true that and that the arithmetic mean of and is exactly more than the geometric mean of and ?
Solutions
Solution 1
Because , we only consider .
For simplicity, we can count how many valid pairs of that satisfy our equation.
The maximum that can be is because must be an integer (this is because , an integer). Then , and we continue this downward until , in which case . The number of pairs of , and so is then .
Solution 2
Let = and =
Then
This makes counting a lot easier since now we just have to find all pairs that differ by 2.
Because , then we can use all positive integers less than 1000 for and .
Without loss of generality, let's say .
We can count even and odd pairs separately to make things easier*:
Odd:
Even:
This makes 499 odd pairs and 498 even pairs, for a total of pairs.
Note: We are counting the pairs for the values of and , which, when squared, translate to the pairs of we are trying to find.
Solution 3
Since the arithmetic mean is 2 more than the geometric mean, . We can multiply by 2 to get . Subtracting 4 and squaring gives
Notice that , so the problem asks for solutions of Since the left hand side is a perfect square, and 16 is a perfect square, must also be a perfect square. Since , must be from to , giving at most 999 options for .
However if , you get , which has solutions and . Both of those solutions are not less than , so cannot be equal to 1. If , you get , which has 2 solutions, , and . 16 is not less than 4, and cannot be 0, so cannot be 4. However, for all other , you get exactly 1 solution for , and that gives a total of pairs.
- asbodke
Solution 4 (Similar to Solution 3)
Rearranging our conditions to
Thus,
Now, let Plugging this back into our expression, we get
There, a unique value of is formed for every value of . However, we must have
and
Therefore, there are only pairs of
Solution by Williamgolly
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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