Difference between revisions of "1999 AHSME Problems/Problem 28"

Revision as of 11:24, 3 October 2023

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 1

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 $(\underbrace{-1,\dots,-1}_{21},1,1,\underbrace{2,\dots,2}_{19})$

Solution 2

As said in Solution 1, $x_i=0$ can be ignored, and only $x_i = -1, 1, 2$ need to be considered.

Minimum

To minimize $x_1^3 + \cdots + x_n^3$ , there are no $2$ s and maximize the number of $-1$ s.

Therefore, the number of $1$ s are $19 + \frac{99-19}{2} = 59$ , the number of $-1$ s are $\frac{99-19}{2} = 40$ .

$x_1^3 + \cdots + x_n^3 = 59 - 40 = 19$

Maximum

Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$ s

$4a+b+c=99$ $2a+b-c=19$

Multiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ . As we need to minimize the value of $c$ , $c = 21$ , $y = 2$ , $a = 19$

$x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133$

Therefore, $\frac Mm = \frac{133}{19}=\boxed{\textbf{(E) } 7}$ .

~isabelchen

@@ Line 41: / Line 41: @@
 ===Minimum===
-To minimize <math>x_1^3 + \cdots + x_n^3</math>, assume that there are no <math>2</math>s and maximize the number of <math>-1</math>s.
+To minimize <math>x_1^3 + \cdots + x_n^3</math>, there are no <math>2</math>s and maximize the number of <math>-1</math>s.
 Therefore, the number of <math>1</math>s are <math>19 + \frac{99-19}{2} = 59</math>, the number of <math>-1</math>s are <math>\frac{99-19}{2} = 40</math>.

1999 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
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