# 1999 AHSME Problems/Problem 28

## Problem

Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that (i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$; and (iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$. Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$, respectively. Then $\frac Mm =$

$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$

## Solution

Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$.

Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$, $b$ equal to $1$, and $c$ equal to $2$.

The conditions (ii) and (iii) simplify to:
(ii) $2c + b - a = 19$
(iii) $4c + a + b = 99$
and we want to find the maximum and minimum of $2^3c + 1^3b + (-1)^3a = 8c + b - a$ over all non-negative solutions of the above two equations.

Subtracting twice (ii) from (iii) we get $b = 3a-61$. By entering that into one of the two equations and simplifying we get $c=40-a$.

Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$.

As all three variables must be non-negative integers, we have $3a-61 \geq 0 \Rightarrow a \geq 21$ and $40-a\geq 0 \Rightarrow a \leq 40$.

For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a = 259 - 6a$. Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$. Their values are $M=133$ and $m=19$, and therefore $\frac Mm = \boxed{7}$.

### Note

The minimum is achieved for $$(\underbrace{-1,\dots,-1}_{40},\underbrace{1,\dots,1}_{59})$$ The maximum is achieved for $$(\underbrace{-1,\dots,-1}_{21},1,1,\underbrace{2,\dots,2}_{19})$$