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1991 AIME Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==
Find <math>x^2+y^2_{}</math> if <math>x_{}^{}</math> and <math>y_{}^{}</math> are positive integers such that
<cmath>\begin{align*}
xy+x+y&=71, \\
x^2y+xy^2&=880.
\end{align*}</cmath>


== Solution ==
== Solution 1 ==
Define <math>a = x + y</math> and <math>b = xy</math>. Then <math>a + b = 71</math> and <math>ab = 880</math>. Solving these two equations yields a [[quadratic equation|quadratic]]: <math>a^2 - 71a + 880 = 0</math>, which [[factor]]s to <math>(a - 16)(a - 55) = 0</math>. Either <math>a = 16</math> and <math>b = 55</math> or <math>a = 55</math> and <math>b = 16</math>. For the first case, it is easy to see that <math>(x,y)</math> can be <math>(5,11)</math> (or vice versa). In the second case, since all factors of <math>16</math> must be <math>\le 16</math>, no two factors of <math>16</math> can sum greater than <math>32</math>, and so there are no integral solutions for <math>(x,y)</math>. The solution is <math>5^2 + 11^2 = \boxed{146}</math>.
 
== Solution 2 ==
Since <math>xy + x + y + 1 = 72</math>, this can be factored to <math>(x + 1)(y + 1) = 72</math>. As <math>x</math> and <math>y</math> are [[integer]]s, the possible sets for <math>(x,y)</math> (ignoring cases where <math>x > y</math> since it is symmetrical) are <math>(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)</math>. The second equation factors to <math>(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11</math>. The only set with a factor of <math>11</math> is <math>(5,11)</math>, and checking shows that it is correct.
 
== Solution 3 ==
 
Let <math>a=x+y</math>, <math>b=xy</math> then we get the equations
<cmath>\begin{align*}
a+b&=71\\
ab&=880
\end{align*}</cmath>
After finding the [[prime factorization]] of <math>880=2^4\cdot5\cdot11</math>, it's easy to obtain the solution <math>(a,b)=(16,55)</math>. Thus
<cmath>x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}</cmath>
Note that if <math>(a,b)=(55,16)</math>, the answer would exceed <math>999</math> which is invalid for an AIME answer.
~ Nafer
 
== Solution 4 ==
From the first equation, we know <math>x+y=71-xy</math>. We factor the second equation as <math>xy(71-xy)=880</math>. Let <math>a=xy</math> and rearranging we get <math>a^2-71a+880=(a-16)(a-55)=0</math>. We have two cases: (1) <math>x+y=16</math> and <math>xy=55</math> OR (2) <math>x+y=55</math> and <math>xy=16</math>. We find the former is true for <math>(x,y) = (5,11)</math>. <math>x^2+y^2=121+25=146</math>.
 
== Solution 5 ==
First, notice that you can factor <math>x^2y + xy^2</math> as <math>xy(x + y)</math>. From this, we notice that <math>xy</math> and <math>x + y</math> is a common occurrence, so that lends itself to a simple solution by substitution. Let <math>xy = b</math> and <math>x + y = a</math>. From this substitution, we get the following system:
<cmath>a + b = 71</cmath>
<cmath>ab = 880</cmath>
Solving that system gives us the following two pairs <math>(a, b)</math>: <math>(16, 55)</math> and <math>(55, 16)</math>. The second one is obviously too big as <math>55^2</math> is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair <math>(16, 55)</math>. This means that <math>x + y = 16</math> and <math>xy = 55</math>
Then, instead of solving the system, we can do a clever manipulation by squaring <math>x + y</math>. Doing so, we get:
<cmath>(x + y)^2 = (x^2 + y^2) + 2xy</cmath>
We see that in this form, we can substitute everything in except for <math>(x^2 + y^2)</math>, which is the desired answer. Substituting, we get:
<cmath>256 = (x^2 + y^2) + 110</cmath>
so <math>x^2 + y^2 = \boxed{146}</math>. (If we were to go with the pair <math>(55, 16)</math>, then the <math>(x + y)^2</math> would be absurdly out of bounds)
 
~EricShi1685


== See also ==
== See also ==
* [[1991 AIME Problems]]
{{AIME box|year=1991|before=First question|num-a=2}}
 
[[Category:Intermediate Algebra Problems]]
{{MAA Notice}}

Latest revision as of 21:05, 7 June 2021

Problem

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that \begin{align*} xy+x+y&=71, \\ x^2y+xy^2&=880. \end{align*}

Solution 1

Define $a = x + y$ and $b = xy$. Then $a + b = 71$ and $ab = 880$. Solving these two equations yields a quadratic: $a^2 - 71a + 880 = 0$, which factors to $(a - 16)(a - 55) = 0$. Either $a = 16$ and $b = 55$ or $a = 55$ and $b = 16$. For the first case, it is easy to see that $(x,y)$ can be $(5,11)$ (or vice versa). In the second case, since all factors of $16$ must be $\le 16$, no two factors of $16$ can sum greater than $32$, and so there are no integral solutions for $(x,y)$. The solution is $5^2 + 11^2 = \boxed{146}$.

Solution 2

Since $xy + x + y + 1 = 72$, this can be factored to $(x + 1)(y + 1) = 72$. As $x$ and $y$ are integers, the possible sets for $(x,y)$ (ignoring cases where $x > y$ since it is symmetrical) are $(1, 35),\ (2, 23),\ (3, 17),\ (5, 11),\ (7,8)$. The second equation factors to $(x + y)xy = 880 = 2^4 \cdot 5 \cdot 11$. The only set with a factor of $11$ is $(5,11)$, and checking shows that it is correct.

Solution 3

Let $a=x+y$, $b=xy$ then we get the equations \begin{align*} a+b&=71\\ ab&=880 \end{align*} After finding the prime factorization of $880=2^4\cdot5\cdot11$, it's easy to obtain the solution $(a,b)=(16,55)$. Thus \[x^2+y^2=(x+y)^2-2xy=a^2-2b=16^2-2\cdot55=\boxed{146}\] Note that if $(a,b)=(55,16)$, the answer would exceed $999$ which is invalid for an AIME answer. ~ Nafer

Solution 4

From the first equation, we know $x+y=71-xy$. We factor the second equation as $xy(71-xy)=880$. Let $a=xy$ and rearranging we get $a^2-71a+880=(a-16)(a-55)=0$. We have two cases: (1) $x+y=16$ and $xy=55$ OR (2) $x+y=55$ and $xy=16$. We find the former is true for $(x,y) = (5,11)$. $x^2+y^2=121+25=146$.

Solution 5

First, notice that you can factor $x^2y + xy^2$ as $xy(x + y)$. From this, we notice that $xy$ and $x + y$ is a common occurrence, so that lends itself to a simple solution by substitution. Let $xy = b$ and $x + y = a$. From this substitution, we get the following system: \[a + b = 71\] \[ab = 880\] Solving that system gives us the following two pairs $(a, b)$: $(16, 55)$ and $(55, 16)$. The second one is obviously too big as $55^2$ is clearly out of bounds for an AIME problem (More on that later). Therefore, we proceed with substituting back the pair $(16, 55)$. This means that $x + y = 16$ and $xy = 55$ Then, instead of solving the system, we can do a clever manipulation by squaring $x + y$. Doing so, we get: \[(x + y)^2 = (x^2 + y^2) + 2xy\] We see that in this form, we can substitute everything in except for $(x^2 + y^2)$, which is the desired answer. Substituting, we get: \[256 = (x^2 + y^2) + 110\] so $x^2 + y^2 = \boxed{146}$. (If we were to go with the pair $(55, 16)$, then the $(x + y)^2$ would be absurdly out of bounds)

~EricShi1685

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
First question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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