Problem #116

116.

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 $100M$?

This problem is copyrighted by the American Mathematics Competitions.

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Instructions for entering answers:

  • Reduce fractions to lowest terms and enter in the form 7/9.
  • Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
  • Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
  • Exponents should be entered in the form 10^10.
  • If the problem is multiple choice, enter the appropriate (capital) letter.
  • Enter points with parentheses, like so: (4,5)
  • Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).

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