# Difference between revisions of "1989 AIME Problems/Problem 11"

## Problem

A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let $D$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D\rfloor$? (For real $x$, $\lfloor x\rfloor$ is the greatest integer less than or equal to $x$.)

## Solution

Let the mode be $x$, which we let appear $n > 1$ times. We let the arithmetic mean be $M$, and the sum of the numbers $\neq x$ be $S$. Then \begin{align*} D &= \left|M-x\right| = \left|\frac{S+xn}{121}-x\right| = \left|\frac{S}{121}-\left(\frac{121-n}{121}\right)x\right| \end{align*} As $S$ is essentially independent of $x$, it follows that we wish to minimize or maximize $x$ (in other words, $x=1,1000$). Indeed, $D(x)$ is symmetric about $x = 500.5$; consider replacing all of numbers $x_i$ in the sample with $1001-x_i$, and the value of $D$ remains the same. So, without loss of generality, let $x=1$. Now, we would like to maximize the quantity

$\frac{S}{121}-\left(\frac{121-n}{121}\right)(1) = \frac{S+n}{121}-1$

$S$ contains $121-n$ numbers that may appear at most $n-1$ times. Therefore, to maximize $S$, we would have $1000$ appear $n-1$ times, $999$ appear $n-1$ times, and so forth. We can thereby represent $S$ as the sum of $n-1$ arithmetic series of $1000, 999, \ldots, 1001 - \left\lfloor \frac{121-n}{n-1} \right\rfloor$. We let $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor$, so

$S = (n-1)\left[\frac{k(1000 + 1001 - k)}{2}\right] + R(n)$

where $R(n)$ denotes the sum of the remaining $121-(n-1)k$ numbers, namely $R(n) = (121-(n-1)k)(1000-k)$.

At this point, we introduce the crude estimate[1] that $k=\frac{121-n}{n-1}$, so $R(n) = 0$ and \begin{align*}2S+2n &= (121-n)\left(2001-\frac{121-n}{n-1}\right)+2n = (120-(n-1))\left(2002-\frac{120}{n-1}\right)\end{align*} Expanding (ignoring the constants, as these do not affect which $n$ yields a maximum) and scaling, we wish to minimize the expression $5(n-1) + \frac{36}{n-1}$. By AM-GM, we have $5(n-1) + \frac{36}{n-1} \ge 2\sqrt{5(n-1) \cdot \frac{36}{n-1}}$, with equality coming when $5(n-1) = \frac{36}{n-1}$, so $n-1 \approx 3$. Substituting this result and some arithmetic gives an answer of $\boxed{947}$.

In less formal language, it quickly becomes clear after some trial and error that in our sample, there will be $n$ values equal to one and $n-1$ values each of $1000, 999, 998 \ldots$. It is fairly easy to find the maximum. Try $n=2$, which yields $924$, $n=3$, which yields $942$, $n=4$, which yields $947$, and $n=5$, which yields $944$. The maximum difference occurred at $n=4$, so the answer is $947$.

## Notes

• ^ In fact, when $n = 2,3,4,5$ (which some simple testing shows that the maximum will occur around), it turns out that $\frac{121-n}{n-1}$ is an integer anyway, so indeed $k = \left\lfloor \frac{121-n}{n-1} \right\rfloor = \frac{121-n}{n-1}$.