Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 2"

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== Solution ==
 
== Solution ==
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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.
  
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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.
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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.
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Thus, the answer is
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<math>\{1,4,9,10\} </math> or <math>\{2,3,8,11\}</math>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 13:10, 23 December 2019

Problem

Let $S = \left \{a,b,c,d \right \}$ be a set of four positive integers. If pairs of distinct elements of $S$ are added, the following six sums are obtained: $5, 10, 11, 13, 14, 19.$ Determine the values of $a, b, c$, and $d.$ [Hint: there are two possibilities.]

Solution

The problem doesn't tell us that $a, b, c,$ and $d$ 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 $(4 * 3)/2 = 6$. If any of the elements $a, b, c,$ or $d$ 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 $a, b, c,$ and $d$ 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 $a < b < c < d$, it is obvious that $a + b = 5$ and $c + d = 19$, as those are the minimum and maximum pair-wise sums. We can do better though, because we know that $a + c = 10$ and $b + d = 14$, since those are the next smallest and next largest pair-wise sums.

With this knowledge, we can start considering cases. If $a = 1$ and $b = 4$, we can use the equations we found to determine that $c = 9$ and $d = 10$. Similarly, if $a = 2$ and $b = 3$, $c = 8$ and $d = 11$. Both of these solutions are valid, and are also the only solutions, because the only two possibilities for $a$ and $b$ were exhausted.

Thus, the answer is $\{1,4,9,10\}$ or $\{2,3,8,11\}$

See Also

2008 UNCO Math Contest II (ProblemsAnswer KeyResources)
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
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