1999 AMC 12 Problem 28

Attempt Problem 28 of the 1999 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

28.

Let x1,x2,,xnx_1, x_2, \ldots, x_n be a sequence of integers such that

(i) for i=1,2,3,,n,i = 1, 2, 3, \ldots, n, 1xi2-1 \le x_i \le 2;

(ii) x1+x2++xn=19;x_1 + x_2 + \cdots + x_n = 19; and

(iii) x12+x22++xn2=99.x_1^2 + x_2^2 + \cdots + x_n^2 = 99.

Let mm and MM be the minimal and maximal possible values of x13+x23++xn3,x_1^3 + x_2^3 + \cdots + x_n^3, respectively. What is Mm?\dfrac{M}{m}?

33

44

55

66

77

Answer: E
Concepts:system of equationsoptimization
Difficulty rating: 2240
Solution:

Let a,b,ca, b, c be the numbers of 1-1s, 11s, and 22s. Then a+b+2c=19-a + b + 2c = 19 and a+b+4c=99,a + b + 4c = 99, giving a=40ca = 40 - c and b=593cb = 59 - 3c with 0c19.0 \le c \le 19.

The sum of cubes is a+b+8c=19+6c.-a + b + 8c = 19 + 6c. The minimum is at c=0c = 0 (value 1919) and the maximum at c=19c = 19 (value 133133), so Mm=13319=7.\dfrac{M}{m} = \dfrac{133}{19} = 7.

Thus, the correct answer is E.

← Problem 27#27
Full Exam