2010 AMC 10B Problem 24

P21-25~8 minPrint

Sequences & Series Diophantine Equations Bound the quantity above/below or estimate to pin it down Organized listing / direct enumeration

A high school basketball game between the Raiders and 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?

Show solution without answering

Key insight

The 100-point limit leaves the Raiders few options, such as 1, 2, 4, 8 or 8, 12, 18, 27. Only 5, 10, 20, 40 fits a Wildcats total one lower.

Solution

Both teams scored the same amount, call it $a$ , in the first quarter.

Wildcats: $a,\ a + d,\ a + 2d,\ a + 3d$ , with $d$ a positive whole number. Total: $4a + 6d$ .
Raiders: $a,\ ar,\ ar^2,\ ar^3$ , with ratio $r$ bigger than $1$ . Their total is $1$ more than the Wildcats' total.

So the Raiders' total minus $1$ minus $4a$ must equal $6d$ : a positive multiple of $6$ . The Raiders' side has far fewer options, so list those first.

Step 1: list the possible Raiders' scores (total at most $100$ ).

Ratio $2$ : $a, 2a, 4a, 8a$ , total $15a$ . So $a$ is at most $6$ .
Ratio $3$ : total $a + 3a + 9a + 27a = 40a$ . So $a$ is $1$ or $2$ .
Ratio $4$ : total $85a$ , so $a = 1$ .
Ratio $5$ or more: even with $a = 1$ the total is at least $1 + 5 + 25 + 125 = 156$ . Too big.
A fraction ratio such as $\frac32$ : every quarter score must be whole, and the fourth score is $a \cdot \frac{27}{8}$ , so $a$ must be a multiple of $8$ . With $a = 8$ the scores are $8, 12, 18, 27$ , total $65$ ; $a = 16$ doubles that to $130$ , too big.
Other fractions fail too. With a bottom of $2$ , the next choice is $\frac52$ , and $a = 8$ gives $8, 20, 50, 125$ , too big. With a bottom of $3$ or more (like $\frac43$ ), the fourth score has a $27$ or more on the bottom, so $a$ is at least $27$ ; then all four scores are at least $27$ and the total is over $100$ .

Step 2: test each option. We need (Raiders' total) $- 1 - 4a$ to be a positive multiple of $6$ .

Raiders	$a$	Raiders' total	total $- 1 - 4a$	Multiple of $6$ ?
ratio $2$	$1$	$15$	$10$	no
ratio $2$	$2$	$30$	$21$	no
ratio $2$	$3$	$45$	$32$	no
ratio $2$	$4$	$60$	$43$	no
ratio $2$	$5$	$75$	$54$	yes, $d = 9$
ratio $2$	$6$	$90$	$65$	no
ratio $3$	$1$	$40$	$35$	no
ratio $3$	$2$	$80$	$71$	no
ratio $4$	$1$	$85$	$80$	no
ratio $\frac32$	$8$	$65$	$32$	no

Step 3: read off the game. Only $a = 5$ , ratio $2$ , $d = 9$ works.

Raiders: $5, 10, 20, 40$ , total $75$ .
Wildcats: $5, 14, 23, 32$ , total $74$ . Raiders win by $1$ .

The first half is the first two quarters: $5 + 10 + 5 + 14 = 34$ .

The answer is $\boxed{\textbf{(E)}\ 34}$ .

Why this works

A growing geometric sequence gets big very fast, so a limit of $100$ points leaves only a handful of possibilities, and the whole-number scores rule out most fraction ratios. Once the short list is written down, the arithmetic side is just a check of whether a number is a multiple of $6$ . When one side of a problem has few options, list that side first.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Let both teams score $a$ in the first quarter. Raiders: $a, ar, ar^2, ar^3$ with ratio $r > 1$ ; Wildcats: $a, a+d, a+2d, a+3d$ with $d > 0$ . The totals satisfy
$a(1 + r + r^2 + r^3) = 4a + 6d + 1 \le 100.$

Since all Raiders' scores are integers, if $r = \frac{p}{q}$ in lowest terms then $q^3 \mid a$ . Check the possible ratios under the $100$ -point cap:

$r = 2$ : total $15a \le 100$ , so $a \le 6$ .
$r = 3$ : total $40a$ , so $a \le 2$ .
$r = 4$ : total $85a$ , so $a = 1$ .
$r = \frac32$ : $a = 8k$ , total $65k$ , so $a = 8$ , total $65$ .
Any other ratio ( $5$ , $\frac43$ , $\frac52$ , ...) exceeds $100$ .

Now require $6d = (\text{Raiders total}) - 1 - 4a$ to be a positive multiple of $6$ :
- $r = 2$ : $11a - 1$ must be divisible by $6$ ; for $a = 1, \dots, 6$ the values are $10, 21, 32, 43, 54, 65$ , and only $a = 5$ works, with $d = 9$ .
- $r = 3$ : $36a - 1$ is never a multiple of $6$ .
- $r = 4$ , $a = 1$ : $80$ is not a multiple of $6$ .
- $r = \frac32$ , $a = 8$ : $65 - 1 - 32 = 32$ , not a multiple of $6$ .

So the Raiders scored $5, 10, 20, 40$ (total $75$ ) and the Wildcats $5, 14, 23, 32$ (total $74$ ). First-half points: $5 + 10 + 5 + 14 = 34$ .

The answer is $\boxed{\textbf{(E)}\ 34}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Number Theory, Chapter 7: Algebra With Integers
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)