Difference between revisions of "2010 AMC 12B Problems/Problem 19"

Revision as of 22:37, 25 October 2021

The following problem is from both the 2010 AMC 12B #19 and 2010 AMC 10B #24, so both problems redirect to this page.

Problem

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?

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 32 \qquad \textbf{(D)}\ 33 \qquad \textbf{(E)}\ 34$

Solution 1

Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know $r > 1$ because the sequence is said to be increasing. We also know that each of $a, ar, ar^2, ar^3$ is an integer. We start by showing that $r$ must also be an integer.

Suppose not, and say $r = m/n$ where $m>n>1$ , and $\gcd(m,n)=1$ . Then $n, n^2, n^3$ must all divide $a$ so $a=n^3k$ for some integer $k$ . Then $S_R = n^3k + n^2mk + nm^2k + m^3k < 100$ and we see that even if $k=1$ and $n=2$ , we get $m < 4$ , which means that the only option for $r$ is $r=3/2$ . A quick check shows that even this doesn't work. Thus $r$ must be an integer.

Let $a, a+d, a+2d, a+3d$ be the quarterly scores for the Wildcats. Let $S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d$ . Let $S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)$ . Then $S_R<100$ implies that $r<5$ , so $r\in \{2, 3, 4\}$ . The Raiders win by one point, so $a(1+r)(1+r^2) = 4a+6d+1.$

If $r=4$ we get $85a = 4a+6d+1$ which means $3(27a-2d) = 1$ , which is not possible with the given conditions.
If $r=3$ we get $40a = 4a+6d+1$ which means $6(6a-d) = 1$ , which is also not possible with the given conditions.
If $r=2$ we get $15a = 4a+6d+1$ which means $11a-6d = 1$ . Reducing modulo 6 we get $a \equiv 5\pmod{6}$ . Since $15a<100$ we get $a<7$ . Thus $a=5$ . It then follows that $d=9$ .

Then the quarterly scores for the Raiders are $5, 10, 20, 40$ , and those for the Wildcats are $5, 14, 23, 32$ . Also $S_R = 75 = S_W + 1$ . The total number of points scored by the two teams in the first half is $5+10+5+14=\boxed{\textbf{(E)}\ 34}$ .

Solution 2

Let $a,ar,ar^{2},ar^{3}$ be the quarterly scores for the Raiders. We know that the Raiders and Wildcats both scored the same number of points in the first quarter so let $a,a+d,a+2d,a+3d$ be the quarterly scores for the Wildcats. The sum of the Raiders scores is $a(1+r+r^{2}+r^{3})$ and the sum of the Wildcats scores is $4a+6d$ . Now we can narrow our search for the values of $a,d$ , and $r$ . Because points are always measured in positive integers, we can conclude that $a$ and $d$ are positive integers. We can also conclude that $r$ is a positive integer by writing down the equation:

$a(1+r+r^{2}+r^{3})=4a+6d+1$

Now we can start trying out some values of $r$ . We try $r=2$ , which gives

$15a=4a+6d+1$

$11a=6d+1$

We need the smallest multiple of $11$ (to satisfy the <100 condition) that is $\equiv 1 \pmod{6}$ . We see that this is $55$ , and therefore $a=5$ and $d=9$ .

So the Raiders' first two scores were $5$ and $10$ and the Wildcats' first two scores were $5$ and $14$ .

$5+10+5+14=34 \longrightarrow \boxed{\textbf{(E)}}$

Solution 3 (Quick Solve?)

It should become apparent that the geometric ratio is less than or equal to $2$ (try the first quarter $1$ , ratio $3$ ). Obviously, the ratio is more than $1$ , so we naively try the ratio of $2$

$1, 2, 4, 8$ - The Raiders

$1, ?, ?, ?$ - The Wildcats

To find the $?$ s, we need the sum of the raiders, which is $15$ . $15-1=14$ , and $\frac{14}2=7$ . So the last term is $7-1=6$ . But, now we can't have integrated 2nd and 3rd quarter scores, so we stop.

$2, 4, 8, 16$ - The Raiders, this also fails (try it out)

The same goes for $3, 6, 12, 24$ , and $4, 8, 16, 32$ . With $5, 10, 20, 40$ we get:

$5, 10, 20, 40$ - The Raiders

$5, ?, ?, ?$ - The wildcats

$5+10+20+40=75$ , $75-1=74$ , $\frac{74}2=37$ , $37-5=32$

$5, ?, ?, 32$ - The Wildcats.

Now, we can finish the arithmetic progression:

$5, 14, 23, 32$ - The Wildcats

Thus our answer is $5+10+5+14=34$

~ firebolt360

Video Solution

https://youtu.be/krRrPxRdgD0

~IceMatrix

Revision as of 22:37, 25 October 2021 (view source) Firebolt360 (talk \| contribs) (→‎Solution 2) ← Older edit		Revision as of 22:37, 25 October 2021 (view source) Firebolt360 (talk \| contribs) m (→‎Solution 3 (Quick Solve?)) Newer edit →
Line 41:		Line 41:

	<math>1, ?, ?, ?</math> - The Wildcats		<math>1, ?, ?, ?</math> - The Wildcats
		+
	To find the <math>?</math>s, we need the sum of the raiders, which is <math>15</math>. <math>15-1=14</math>, and <math>\frac{14}2=7</math>. So the last term is <math>7-1=6</math>. But, now we can't have integrated 2nd and 3rd quarter scores, so we stop.		To find the <math>?</math>s, we need the sum of the raiders, which is <math>15</math>. <math>15-1=14</math>, and <math>\frac{14}2=7</math>. So the last term is <math>7-1=6</math>. But, now we can't have integrated 2nd and 3rd quarter scores, so we stop.

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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
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