1999 AMC 12 Problem 28

Question

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,

3,

\ldots,

n

;

(ii)

x_1 + x_2 + \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 + x_2^3 + \cdots + x_n^3,

respectively. What is

\dfrac{M}{m}?

3

4

5

6

7

Let

a, b, c

count the

-1

s,

1

s, and

2

s (zeros do not matter)

The sum of cubes is

-a + b + 8c = 19 + 6c;

optimize over the allowed range of

c

Let

a, b, c

be the numbers of

-1

s,

1

s, and

2

s. Then

-a + b + 2c = 19

and

a + b + 4c = 99,

giving

a = 40 - c

and

b = 59 - 3c

with

0 \le c \le 19.

The sum of cubes is

-a + b + 8c = 19 + 6c.

The minimum is at

c = 0

(value

19

) and the maximum at

c = 19

(value

133

), so

\dfrac{M}{m} = \dfrac{133}{19} = 7.

Thus, the correct answer is E.

Problem 28 in other years