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Given that the mean, median, range and the only mode of 200 integers are also 200. If $A$ is the largest integer among those 200 integers, find the maximum value of $A$.

I have asked some of my friends and colleagues to solve this problem, but no one give me a light.

Attempt:

Assuming first that all the numbers are $200$. To maximize $A$, but satisfies all the criterion given, we need to make $A$ ascending while descending the value of other numbers. Logically, $100, 200, 200, \cdots, 300$ still satisfies. Maybe we have$A_{\text{max}} = 300$? I don't know how to approach it clearly.

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If $A\gt 400$ then, since the range is $200$, all the integers would exceed $200$, making the mean exceed $200$, so that is not possible

If $A=400$ then, since the range is $200$, all the integers would be at least $200$ and at least one is strictly greater, making the mean exceed $200$, so that is not possible

If $A=399$ then, since the range is $200$, the minimum would be $199$. Since the median is $200$, no more than $99$ of the integers can be $199$, with mean at least $\frac{1}{200}(99 \times 199 + 100 \times 200 +1 \times 399)=200.5$, making the mean exceed $200$, so that is not possible

If $A=398$ then, since the range is $200$, the minimum would be $198$. There is a solution with

  • $198$ appearing $99$ times,
  • $200$ appearing $100$ times, and
  • $398$ appearing $1$ time,

with mean $\frac{1}{200}(99 \times 198 + 100 \times 200 +1 \times 398)=200$, and clearly the range, median and mode are all $200$ too, so that is possible

So the maximum possible value of $A$ is $398$

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The distribution consisting of $\{196, \underbrace{198, \ldots, 198}_{96}, \underbrace{200, \ldots 200}_{102}, 396 \}$

Has mean = mode = range = $200$ and $A_{max} = 396$.

Here's the histogram, with the two "singletons" barely visible:

enter image description here

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  • $\begingroup$ @JackMoody: I posted my solution within 10 seconds of Henry. $\endgroup$ Jan 16 '19 at 17:10
  • $\begingroup$ For those coming along and seeing this later: the issue @JackMoody identified has been fixed. $\endgroup$ Jan 16 '19 at 17:50
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    $\begingroup$ @MichaelLugo shouldn't the max value be 398 as stated in the solution by @Henry? $\endgroup$
    – Jack Moody
    Jan 16 '19 at 17:55
  • $\begingroup$ @JackMoody: $398$ is just an upper bound. $\endgroup$ Jan 16 '19 at 17:56

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