Difference between revisions of "1991 AIME Problems/Problem 1"

Revision as of 23:52, 28 October 2020

Problem

Find $x^2+y^2_{}$ if $x_{}^{}$ and $y_{}^{}$ are positive integers such that

$xy_{}^{}+x+y = 71$

$x^2y+xy^2 = 880^{}_{}.$

Solution

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

@@ Line 36: / Line 36: @@
 <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 ==

1991 AIME (Problems • Answer Key • Resources)
Preceded by First question	Followed by Problem 2
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All AIME Problems and Solutions

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