2010 AMC 12B Problem 19

Question

A high school basketball game between the Raiders and the Wildcats was tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than

100

points. What was the total number of points scored by the two teams in the first half?

30

31

32

33

34

Answer: E

Concepts:geometric sequence arithmetic sequence bounding to limit cases

Difficulty rating: 2180

Small hint:

Let the Raiders score

a, ar, ar^2, ar^3

and the Wildcats

a, a+d, a+2d, a+3d,

tied at

a

Big hint:

The totals differ by

1

and both stay under

100,

which forces small values

Solution:

Let the Raiders score

a, ar, ar^2, ar^3

(increasing geometric,

r\gt1

) and the Wildcats

a, a+d, a+2d, a+3d

(increasing arithmetic), tied in the first quarter at

a.

Write

r=\frac{m}{n}

in lowest terms, with

m\gt n.

Since

ar^3

is an integer,

a

is divisible by

n^3;

put

a=A n^3.

The Raiders’ total is

R=A(n^3+mn^2+m^2n+m^3).

Since

m^3\lt R\le100,

we have

m\le4.

The pair

(4,3)

already makes the parenthesized sum

175,

so the only possible coprime pairs

(m,n)

are

(2,1),(3,1),(3,2),(4,1).

The corresponding base values of

\frac{R}{A}

are

15,40,65,85.

For

(4,1)

and

(3,2),

the bound forces

A=1,

but

R-1=4a+6d

gives a nonintegral

d.

For

(3,1),

it would give

36A-1=6d,

which is impossible modulo

6.

For

(m,n)=(2,1),

we have

R=15A

and

a=A,

so

11A-1=6d.

Thus

A\equiv5\pmod6,

and

R\le100

forces

A=5.

Then

d=9,

giving Raiders scores

5,10,20,40

and Wildcats scores

5,14,23,32.

The Raiders win

75

to

74.

The first-half total is

(5+10)+(5+14)=34.

Thus, the correct answer is E.