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

Revision as of 08:15, 21 July 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 $(x,y) = (5,11)$ . $x^2+y^2=121+25=146$ .

@@ Line 11: / Line 11: @@
 === 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 our solution.
+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 ===

1991 AIME (Problems • Answer Key • Resources)
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