Difference between revisions of "1985 AIME Problems/Problem 8"
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The sum of the following seven numbers is exactly 19: <math>a_1 = 2.56</math>, <math>a_2 = 2.61</math>, <math>a_3 = 2.65</math>, <math>a_4 = 2.71</math>, <math>a_5 = 2.79</math>, <math>a_6 = 2.81</math>, <math>a_7 = 2.86</math>. It is desired to replace each <math>a_i</math> by an [[integer]] approximation <math>A_i</math>, <math>1\le i \le 7</math>, so that the sum of the <math>A_i</math>'s is also 19 and so that <math>M</math>, the [[maximum]] of the "errors" <math>\| A_i-a_i\|</math>, the maximum [[absolute value]] of the difference, is as small as possible. For this minimum <math>M</math>, what is <math>100M</math>? | The sum of the following seven numbers is exactly 19: <math>a_1 = 2.56</math>, <math>a_2 = 2.61</math>, <math>a_3 = 2.65</math>, <math>a_4 = 2.71</math>, <math>a_5 = 2.79</math>, <math>a_6 = 2.81</math>, <math>a_7 = 2.86</math>. It is desired to replace each <math>a_i</math> by an [[integer]] approximation <math>A_i</math>, <math>1\le i \le 7</math>, so that the sum of the <math>A_i</math>'s is also 19 and so that <math>M</math>, the [[maximum]] of the "errors" <math>\| A_i-a_i\|</math>, the maximum [[absolute value]] of the difference, is as small as possible. For this minimum <math>M</math>, what is <math>100M</math>? | ||
== Solution == | == Solution == | ||
| − | + | If any of the approximations <math>A_i</math> is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the <math>A_i</math> are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the <math>a_i</math>, so our approximations are <math>A_1 = A_2 = 2</math> and <math>A_3 = A_4 = A_5 = A_6 = A_7 = 3</math> and the largest error is <math>|A_2 - a_2| = 0.61</math>, so the answer is <math>061</math>. | |
== See also == | == See also == | ||
* [[1985 AIME Problems/Problem 7 | Previous problem]] | * [[1985 AIME Problems/Problem 7 | Previous problem]] | ||
* [[1985 AIME Problems/Problem 9 | Next problem]] | * [[1985 AIME Problems/Problem 9 | Next problem]] | ||
* [[1985 AIME Problems]] | * [[1985 AIME Problems]] | ||
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| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 18:25, 1 December 2006
Problem
The sum of the following seven numbers is exactly 19: 
, 
, 
, 
, 
, 
, 
. It is desired to replace each 
by an integer approximation 
, 
, so that the sum of the 's is also 19 and so that 
, the maximum of the "errors" 
, the maximum absolute value of the difference, is as small as possible. For this minimum , what is 
?
Solution
If any of the approximations is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1. However, if all of the are 2 or 3, the largest error will be less than 1. So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3. Then there must be five 3s and two 2s. It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the , so our approximations are 
and 
and the largest error is 
, so the answer is 
.
