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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 3"

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Adding these up, we have <math>3x+3y+3z+3w=132</math>, so <math>x+y+z+w=44</math>. Adding the first two equations, we get <math>2x+y+z=27</math>. Subtracting this, we obtain <math>w-x=17</math>. Adding to the third equation, we have <math>2w=38</math>, so <math>w=19</math>. Substituting, we have <math>x=2</math>. <math>y=7</math>, and <math>z=16 \Rightarrow \boxed{(2, 7, 16, 19)}</math> as one of our solutions.
 
Adding these up, we have <math>3x+3y+3z+3w=132</math>, so <math>x+y+z+w=44</math>. Adding the first two equations, we get <math>2x+y+z=27</math>. Subtracting this, we obtain <math>w-x=17</math>. Adding to the third equation, we have <math>2w=38</math>, so <math>w=19</math>. Substituting, we have <math>x=2</math>. <math>y=7</math>, and <math>z=16 \Rightarrow \boxed{(2, 7, 16, 19)}</math> as one of our solutions.
  
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*second solution needed
+
 
 +
*second solution
 +
Let the numbers be <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math>. We have the following 6 equations from the problem statement:
 +
 
 +
<math>x+y=9</math>
 +
 
 +
<math>x+z=18</math>
 +
 
 +
<math>y+z=21</math>
 +
 
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<math>x+w=23</math>
 +
 
 +
<math>y+w=26</math>
 +
 
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<math>z+w=35</math>
 +
 
 +
Adding all of these up:
 +
<math>3x+3y+3z+3w=132</math> gives  <math>x+y+z+w=44</math> : call this equation A
 +
 
 +
Adding first two equations:
 +
<math>2x+y+z=27</math> : call this equation B
 +
 
 +
Subtracting equation B from A gives us <math>w-x=17</math>  : call this equation C
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Adding C to the 4th equation gives <math>2w=40</math>
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<math>w=20</math>
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Substituting w in the original equations gives:
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<math>x=3</math>, <math>y=6</math>, <math>z=15 \Rightarrow \boxed{(3, 6, 15, 20)}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 18:39, 30 June 2019

Problem

A student thinks of four numbers. She adds them in pairs to get the six sums $9,18,21,23,26,35.$ What are the four numbers? There are two different solutions.


Solution

Let the numbers be $x$, $y$, $z$, and $w$. Writing in terms of these, we have:

$x+y=9$

$x+z=18$

$x+w=21$

$y+z=23$

$y+w=26$

$z+w=35$.

Adding these up, we have $3x+3y+3z+3w=132$, so $x+y+z+w=44$. Adding the first two equations, we get $2x+y+z=27$. Subtracting this, we obtain $w-x=17$. Adding to the third equation, we have $2w=38$, so $w=19$. Substituting, we have $x=2$. $y=7$, and $z=16 \Rightarrow \boxed{(2, 7, 16, 19)}$ as one of our solutions.


  • second solution

Let the numbers be $x$, $y$, $z$, and $w$. We have the following 6 equations from the problem statement:

$x+y=9$

$x+z=18$

$y+z=21$

$x+w=23$

$y+w=26$

$z+w=35$

Adding all of these up: $3x+3y+3z+3w=132$ gives $x+y+z+w=44$ : call this equation A

Adding first two equations: $2x+y+z=27$ : call this equation B

Subtracting equation B from A gives us $w-x=17$  : call this equation C

Adding C to the 4th equation gives $2w=40$

$w=20$

Substituting w in the original equations gives: $x=3$, $y=6$, $z=15 \Rightarrow \boxed{(3, 6, 15, 20)}$

See also

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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