Difference between revisions of "1985 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
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 = 2.82</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 = 2.82</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>? | ||
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+ | == Explanation of the Question == | ||
+ | |||
+ | Note: please read the explanation AFTER YOU HAVE TRIED reading the problem but couldn't understand. | ||
+ | |||
+ | For the question. Let's say that you have determined 7-tuple <math>(A_1,A_2,A_3,A_4,A_5,A_6,A_7)</math>. Then you get the absolute values of the <math>7</math> differences. Namely, | ||
+ | <cmath>|A_1-a_1|, |A_2-a_2|, |A_3-a_3|, |A_4-a_4|, |A_5-a_5|, |A_6-a_6|, |A_7-a_7|</cmath> | ||
+ | Then <math>M</math> is the greatest of the <math>7</math> absolute values. So basically you are asked to find the 7-tuple <math>(A_1,A_2,A_3,A_4,A_5,A_6,A_7)</math> with the smallest <math>M</math>, and the rest would just be a piece of cake. | ||
== 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>. | + | 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>\boxed{061}</math>. |
== See also == | == See also == | ||
{{AIME box|year=1985|num-b=7|num-a=9}} | {{AIME box|year=1985|num-b=7|num-a=9}} |
Latest revision as of 02:17, 16 February 2021
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 ?
Explanation of the Question
Note: please read the explanation AFTER YOU HAVE TRIED reading the problem but couldn't understand.
For the question. Let's say that you have determined 7-tuple . Then you get the absolute values of the differences. Namely, Then is the greatest of the absolute values. So basically you are asked to find the 7-tuple with the smallest , and the rest would just be a piece of cake.
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 .
See also
1985 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |