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.

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 $$398$$

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:

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