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

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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)</cmath>
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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)</cmath>
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 18:33, 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 \[a^{2} - b^{2} + c^{2} - d^{2}=(a+b)+(c+d) \implies (a+b)(a-b)-(a+b)+(c+d)(c-d)-(c+d)=0 \implies (a+b)(a-b-1)+(c+d)(c-d-1)\]

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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