Difference between revisions of "2005 AMC 12B Problems/Problem 20"
Arthy00009 (talk | contribs) m (→Solution) |
|||
Line 15: | Line 15: | ||
== Solution == | == Solution == | ||
The sum of the set is <math>-7-5-3-2+2+4+6+13=8</math>, so if we could have the sum in each set of parenthesis be <math>4</math> then the minimum value would be <math>2(4^2)=32</math>. Considering the set of four terms containing <math>13</math>, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be <math>13-7-5-3=-2</math>, and with two odd terms then its minimum value is <math>13-7+2-2=6</math>, so we cannot achieve two sums of <math>4</math>. The closest we could have to <math>4</math> and <math>4</math> is <math>3</math> and <math>5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}</math>. | The sum of the set is <math>-7-5-3-2+2+4+6+13=8</math>, so if we could have the sum in each set of parenthesis be <math>4</math> then the minimum value would be <math>2(4^2)=32</math>. Considering the set of four terms containing <math>13</math>, this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be <math>13-7-5-3=-2</math>, and with two odd terms then its minimum value is <math>13-7+2-2=6</math>, so we cannot achieve two sums of <math>4</math>. The closest we could have to <math>4</math> and <math>4</math> is <math>3</math> and <math>5</math>, which can be achieved through <math>13-7-5+2</math> and <math>6-3-2+4</math>. So the minimum possible value is <math>3^2+5^2=34\Rightarrow\boxed{\mathrm{C}}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | We start off with trying to get a <math>2(4^2) = 32</math> solution. Trying out the values for a bit leads to us getting <math>34</math> with <math>(-3, -2, 2, 6)</math> in 1 set for a total of 9 and the other numbers giving a total of 25. | ||
+ | |||
+ | We can add 7 to each number. We are now trying to get a set with 32. | ||
+ | |||
+ | <cmath>(0, 2, 4, 5, 9, 11, 13, 20)</cmath> | ||
+ | |||
+ | Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers <math>a,b,c,d</math>, the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26). For the 2 odds and evens, we can divide them into two cases: has 20 or doesn't have 20. | ||
+ | |||
+ | Case 1: <math>a,b,c,d</math> doesn't have 20. | ||
+ | |||
+ | The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need. | ||
+ | |||
+ | Case 2: <math>a,b,c,d</math> has 20. | ||
+ | |||
+ | The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need. | ||
+ | |||
+ | We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is <math>\boxed{C}</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2005|ab=B|num-b=19|num-a=21}} | {{AMC12 box|year=2005|ab=B|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:27, 6 November 2022
Contents
[hide]Problem
Let and be distinct elements in the set
What is the minimum possible value of
Solution
The sum of the set is , so if we could have the sum in each set of parenthesis be then the minimum value would be . Considering the set of four terms containing , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be , and with two odd terms then its minimum value is , so we cannot achieve two sums of . The closest we could have to and is and , which can be achieved through and . So the minimum possible value is .
Solution 2
We start off with trying to get a solution. Trying out the values for a bit leads to us getting with in 1 set for a total of 9 and the other numbers giving a total of 25.
We can add 7 to each number. We are now trying to get a set with 32.
Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers , the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26). For the 2 odds and evens, we can divide them into two cases: has 20 or doesn't have 20.
Case 1: doesn't have 20.
The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.
Case 2: has 20.
The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.
We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is .
See also
2005 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.