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Difference between revisions of "2010 AIME I Problems/Problem 5"

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Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>.
 
Positive integers <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> satisfy <math>a > b > c > d</math>, <math>a + b + c + d = 2010</math>, and <math>a^2 - b^2 + c^2 - d^2 = 2010</math>. Find the number of possible values of <math>a</math>.
  
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==Solution==
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== Solution 1 ==
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Using the [[difference of squares]], <math>2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010</math>, where equality must hold so <math>b = a - 1</math> and <math>d = c - 1</math>. Then we see <math>a = 1004</math> is maximal and <math>a = 504</math> is minimal, so the answer is <math>\boxed{501}</math>.
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Note: We can also find that <math>b=a-1</math> in another way. We know <cmath>a^{2} - b^{2} + c^{2} - d^{2}=(a+b)+(c+d) \implies (a+b)(a-b)-(a+b)+(c+d)(c-d)-(c+d)=0 \implies (a+b)(a-b-1)+(c+d)(c-d-1)=0</cmath>
  
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=== Solution 1 ===
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Therefore, one of <math>(a+b)(a-b-1)</math> and <math>(c+d)(c-d-1)</math> must be <math>0.</math> Clearly, <math>a+b \neq 0</math> since then one would be positive and negative, or both would be zero. Therefore, <math>a-b-1=0</math> so <math>a=b+1</math>. Similarly, we can deduce that <math>c=d+1.</math>
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Using the [[difference of squares]], <math>2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010</math>, where equality must hold so <math>b = a - 1</math> and <math>d = c - 1</math>. Then we see <math>a = 1004</math> is maximal and <math>a = 504</math> is minimal, so the answer is <math>\boxed{501}</math>.
 
  
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=== Solution 2 ===
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== Solution 2 ==
 
Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so  <math>a</math> has <math>\boxed{501}</math> possible values.
 
Since <math>a+b</math> must be greater than <math>1005</math>, it follows that the only possible value for <math>a-b</math> is <math>1</math> (otherwise the quantity <math>a^2 - b^2</math> would be greater than <math>2010</math>). Therefore the only possible ordered pairs for <math>(a,b)</math> are <math>(504, 503)</math>, <math>(505, 504)</math>, ... , <math>(1004, 1003)</math>, so  <math>a</math> has <math>\boxed{501}</math> possible values.
  
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[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 18:35, 1 March 2020

Problem

Positive integers $a$, $b$, $c$, and $d$ satisfy $a > b > c > d$, $a + b + c + d = 2010$, and $a^2 - b^2 + c^2 - d^2 = 2010$. Find the number of possible values of $a$.

Solution 1

Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d) \ge a + b + c + d = 2010$, where equality must hold so $b = a - 1$ and $d = c - 1$. Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$.


Note: We can also find that $b=a-1$ in another way. We know \[a^{2} - b^{2} + c^{2} - d^{2}=(a+b)+(c+d) \implies (a+b)(a-b)-(a+b)+(c+d)(c-d)-(c+d)=0 \implies (a+b)(a-b-1)+(c+d)(c-d-1)=0\]

Therefore, one of $(a+b)(a-b-1)$ and $(c+d)(c-d-1)$ must be $0.$ Clearly, $a+b \neq 0$ since then one would be positive and negative, or both would be zero. Therefore, $a-b-1=0$ so $a=b+1$. Similarly, we can deduce that $c=d+1.$

Solution 2

Since $a+b$ must be greater than $1005$, it follows that the only possible value for $a-b$ is $1$ (otherwise the quantity $a^2 - b^2$ would be greater than $2010$). Therefore the only possible ordered pairs for $(a,b)$ are $(504, 503)$, $(505, 504)$, ... , $(1004, 1003)$, so $a$ has $\boxed{501}$ possible values.

See Also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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