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 ===
  
 
== Solution ==
 
== Solution ==

Revision as of 23:21, 23 August 2019

Problem

The sum of the following seven numbers is exactly 19: $a_1 = 2.56$, $a_2 = 2.61$, $a_3 = 2.65$, $a_4 = 2.71$, $a_5 = 2.79$, $a_6 = 2.82$, $a_7 = 2.86$. It is desired to replace each $a_i$ by an integer approximation $A_i$, $1\le i \le 7$, so that the sum of the $A_i$'s is also 19 and so that $M$, the maximum of the "errors" $\| A_i-a_i\|$, the maximum absolute value of the difference, is as small as possible. For this minimum $M$, what is $100M$?

Explanation of the Question

Solution

If any of the approximations $A_i$ 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 $A_i$ 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 $a_i$, so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$, so the answer is $061$.

See also

1985 AIME (ProblemsAnswer KeyResources)
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
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