Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 3"
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== Solution == | == Solution == | ||
| + | Let the numbers be <math>x</math>, <math>y</math>, <math>z</math>, and <math>w</math>. Writing in terms of these, we have: | ||
| + | |||
| + | <math>x+y=9</math> | ||
| + | |||
| + | <math>x+z=18</math> | ||
| + | |||
| + | <math>x+w=21</math> | ||
| + | |||
| + | <math>y+z=23</math> | ||
| + | |||
| + | <math>y+w=26</math> | ||
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| + | <math>z+w=35</math>. | ||
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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. | ||
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| + | *second solution needed | ||
== See also == | == See also == | ||
Revision as of 22:08, 13 February 2018
Problem
A student thinks of four numbers. She adds them in pairs to get the six sums 
What are the four numbers? There are two different solutions.
Solution
Let the numbers be 
, 
, 
, and 
. Writing in terms of these, we have:
.
Adding these up, we have 
, so 
. Adding the first two equations, we get 
. Subtracting this, we obtain 
. Adding to the third equation, we have 
, so 
. Substituting, we have 
. 
, and 
as one of our solutions.
- second solution needed
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 | ||
