1993 UNCO Math Contest II Problems/Problem 3

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: $x=3$ , $y=6$ , $z=15 \Rightarrow \boxed{(3, 6, 15, 20)}$

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
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1993 UNCO Math Contest II Problems/Problem 3

Problem

Solution

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