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

(Solution)
 
Line 17: Line 17:
 
= (x-y)^2+(y+3z)^2 = 175</math>
 
= (x-y)^2+(y+3z)^2 = 175</math>
  
(x,y,z) are integers, ie. 175 = a^2 + b^2,  
+
(x,y,z) are integers, ie. <math>175 = a^2 + b^2</math>,  
  
a^2:  1,    4,    9,  16,    25,  36,  49,  64,    81,  100,  121,  144,  169
+
<math>a^2</math>:  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
+
<math>b^2</math>:  174,  171,  166,  159,  150,  139,  126,  111,  94,  75,    54,    31,    6
  
 
so there is NO solution  
 
so there is NO solution  

Latest revision as of 10:14, 15 June 2021

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

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

See also

1980 AHSME (ProblemsAnswer KeyResources)
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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