Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | {{ | + | The problem doesn't tell us that <math>a, b, c,</math> and <math>d</math> are distinct integers, but we can prove it easily. If we assume that these four positive integers are all distinct, then we will obtain six distinct pair-wise sums, since the number of ways to choose two objects out of four total where order does not matter is <math>(4 * 3)/2 = 6</math>. If any of the elements <math>a, b, c,</math> or <math>d</math> are equal to another, at least one of these distinct pair-wise sums would disappear, because at least two of the sums would be identical. However, we can see that the pair-wise sums we are given are all distinct, and that there are 6 of them, meaning that elements <math>a, b, c,</math> and <math>d</math> are all distinct. |
| + | |||
| + | Before we move on to considering cases, it would help to consider what information might help save time. If we assume that <math>a < b < c < d</math>, it is obvious that <math>a + b = 5</math> and <math>c + d = 19</math>, as those are the minimum and maximum pair-wise sums. We can do better though, because we know that <math>a + c = 10</math> and <math>b + d = 14</math>, since those are the next smallest and next largest pair-wise sums. | ||
| + | |||
| + | With this knowledge, we can start considering cases. If <math>a = 1</math> and <math>b = 4</math>, we can use the equations we found to determine that <math>c = 9</math> and <math>d = 10</math>. Similarly, if <math>a = 2</math> and <math>b = 3</math>, <math>c = 8</math> and <math>d = 11</math>. Both of these solutions are valid, and are also the only solutions, because the only two possibilities for <math>a</math> and <math>b</math> were exhausted. | ||
| + | |||
| + | Thus, the answer is | ||
| + | <math>\{1,4,9,10\} </math> or <math>\{2,3,8,11\}</math> | ||
== See Also == | == See Also == | ||
Latest revision as of 12:10, 23 December 2019
Problem
Let 
be a set of four positive integers. If pairs of distinct elements of 
are added, the following six sums are obtained: 
Determine the values of 
, and 
[Hint: there are two possibilities.]
Solution
The problem doesn't tell us that 
and 
are distinct integers, but we can prove it easily. If we assume that these four positive integers are all distinct, then we will obtain six distinct pair-wise sums, since the number of ways to choose two objects out of four total where order does not matter is 
. If any of the elements or are equal to another, at least one of these distinct pair-wise sums would disappear, because at least two of the sums would be identical. However, we can see that the pair-wise sums we are given are all distinct, and that there are 6 of them, meaning that elements and are all distinct.
Before we move on to considering cases, it would help to consider what information might help save time. If we assume that 
, it is obvious that 
and 
, as those are the minimum and maximum pair-wise sums. We can do better though, because we know that 
and 
, since those are the next smallest and next largest pair-wise sums.
With this knowledge, we can start considering cases. If 
and 
, we can use the equations we found to determine that 
and 
. Similarly, if 
and 
, 
and 
. Both of these solutions are valid, and are also the only solutions, because the only two possibilities for 
and 
were exhausted.
Thus, the answer is
or 
See Also
| 2008 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
