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Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 2"

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The three biggest distinct positive integers that are <math>100</math> or less are <math>100</math>, <math>98</math>, and <math>96</math>. Thus, our answer is <math>\frac{100+98+96}{3} = \boxed{98}</math>.
 
The three biggest distinct positive integers that are <math>100</math> or less are <math>100</math>, <math>98</math>, and <math>96</math>. Thus, our answer is <math>\frac{100+98+96}{3} = \boxed{98}</math>.
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~Bradygho
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== Solution 2 ==
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Notice that the average of <math>3</math> numbers that form an arithmetic progression is the median of the numbers. Also, notice that to maximize the average of three numbers, we need to maximize the numbers themselves. Using these observations, it is obvious the answer is <math>\boxed{98}</math>.
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~Mathdreams
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==See also==
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#[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]]
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#[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 12:11, 13 July 2021

Problem

Three distinct even positive integers are chosen between $1$ and $100,$ inclusive. What is the largest possible average of these three integers?

Solution

The three biggest distinct positive integers that are $100$ or less are $100$, $98$, and $96$. Thus, our answer is $\frac{100+98+96}{3} = \boxed{98}$.

~Bradygho

Solution 2

Notice that the average of $3$ numbers that form an arithmetic progression is the median of the numbers. Also, notice that to maximize the average of three numbers, we need to maximize the numbers themselves. Using these observations, it is obvious the answer is $\boxed{98}$.

~Mathdreams

See also

  1. Other 2021 JMPSC Accuracy Problems
  2. 2021 JMPSC Accuracy Answer Key
  3. All JMPSC Problems and Solutions

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