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1999 AHSME Problems/Problem 28: Difference between revisions

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


===m===
===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>.
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>.
Line 47: Line 47:
<math>x_1^3 + \cdots + x_n^3 = 59 - 40 = 19</math>
<math>x_1^3 + \cdots + x_n^3 = 59 - 40 = 19</math>


===M===
===Maximum===


Let <math>a =</math> number of <math>2</math>s, <math>b =</math> number of <math>1</math>s, <math>c =</math> number of <math>-1</math>s
Let <math>a =</math> number of <math>2</math>s, <math>b =</math> number of <math>1</math>s, <math>c =</math> number of <math>-1</math>s
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<cmath>2a+b-c=19</cmath>
<cmath>2a+b-c=19</cmath>


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


<math>x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133</math>
<math>x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133</math>

Latest revision as of 11:25, 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 $3c-b=61$, $3c=61+b$. As we need to minimize the value of $c$, $c = 21$, $b = 2$, $a = 19$

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

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

~isabelchen

See also

1999 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 27 		Followed by
Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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