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Difference between revisions of "1980 AHSME Problems/Problem 29"

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This equation can be rearranged to get <math>67z^2 = 949-13x^2 = 13(73-x^2)</math>.  Since <math>67</math> is not divisible by <math>13</math>, <math>z</math> must be divisible by <math>13</math> and there exists an integer <math>n</math> such that <math>z = 13n</math>.  Substituting <math>z = 13n</math> and dividing both sides of this equation by <math>13</math> yields <math>871n^2 = 73 - x^2</math>.  So <math>871n^2 \leq 73</math>, which is only true if <math>n = 0</math>.  But <math>n = 0</math> would require <math>x^2 = 73</math>, which is impossible since <math>73</math> is not a perfect square.  Therefore, this equation has <math>0</math> integer solutions. <math>\fbox{(A)}</math>
 
This equation can be rearranged to get <math>67z^2 = 949-13x^2 = 13(73-x^2)</math>.  Since <math>67</math> is not divisible by <math>13</math>, <math>z</math> must be divisible by <math>13</math> and there exists an integer <math>n</math> such that <math>z = 13n</math>.  Substituting <math>z = 13n</math> and dividing both sides of this equation by <math>13</math> yields <math>871n^2 = 73 - x^2</math>.  So <math>871n^2 \leq 73</math>, which is only true if <math>n = 0</math>.  But <math>n = 0</math> would require <math>x^2 = 73</math>, which is impossible since <math>73</math> is not a perfect square.  Therefore, this equation has <math>0</math> integer solutions. <math>\fbox{(A)}</math>
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 +
-j314andrews, based on solution by Farenhajt
  
 
== See also ==
 
== See also ==

Revision as of 18:01, 25 June 2025

Problem

How many ordered triples (x,y,z) of integers satisfy the system of equations below?

\[\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array}\]

$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \\ \text{(D)}\ \text{a finite number greater than 2}\qquad\\ \text{(E)}\ \text{infinitely many}$

Solution 1

Sum of three equations,

$x^2-2xy+2y^2+6yz+9z^2 = (x-y)^2+(y+3z)^2 = 175$

(x,y,z) are integers, ie. $175 = a^2 + b^2$,

$a^2$: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169 $b^2$: 174, 171, 166, 159, 150, 139, 126, 111, 94, 75, 54, 31, 6

so there is NO solution

Wwei.yu (talk) 22:09, 28 March 2020 (EDT)Wei

Solution 2 (Eliminate Cross Terms)

First, to eliminate the $xy$ term, multiply both sides of the third equation by $3$ to get $3x^2 + 3xy + 24z^2 = 300$, and then add it to the first equation to get $4x^2 + 2yz + 23z^2 = 331$.

Next, to eliminate the $yz$ term, multiply both sides of the new equation by $-3$ to get $-12x^2 - 6yz - 69z^2 = -993$, and then add it to the second equation to get $-13x^2 - 67z^2 = -949$.

This equation can be rearranged to get $67z^2 = 949-13x^2 = 13(73-x^2)$. Since $67$ is not divisible by $13$, $z$ must be divisible by $13$ and there exists an integer $n$ such that $z = 13n$. Substituting $z = 13n$ and dividing both sides of this equation by $13$ yields $871n^2 = 73 - x^2$. So $871n^2 \leq 73$, which is only true if $n = 0$. But $n = 0$ would require $x^2 = 73$, which is impossible since $73$ is not a perfect square. Therefore, this equation has $0$ integer solutions. $\fbox{(A)}$

-j314andrews, based on solution by Farenhajt

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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