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Difference between revisions of "1985 AIME Problems/Problem 8"

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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>061</math>.
 
== See also ==
 
== See also ==
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* [[1985 AIME Problems/Problem 7 | Previous problem]]
+
{{AIME box|year=1985|num-b=7|num-a=9}}
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* [[1985 AIME Problems/Problem 9 | Next problem]]
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* [[AIME Problems and Solutions]]
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* [[1985 AIME Problems]]
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* [[American Invitational Mathematics Examination]]
 +
* [[Mathematics competition resources]]
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Revision as of 13:33, 6 May 2007

Problem

The sum of the following seven numbers is exactly 19: $a_1 = 2.56$, $a_2 = 2.61$, $a_3 = 2.65$, $a_4 = 2.71$, $a_5 = 2.79$, $a_6 = 2.81$, $a_7 = 2.86$. 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 $100M$?

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 (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
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