# 1993 UNCO Math Contest II Problems/Problem 3

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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 byProblem 2 Followed byProblem 4 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 All UNCO Math Contest Problems and Solutions