Difference between revisions of "2010 AMC 10B Problems/Problem 24"

Revision as of 12:05, 18 December 2019

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 absurd.
If $r=3$ we get $40a = 4a+6d+1$ which means $6(6a-d) = 1$ , which is also absurd.
If $r=2$ we get $15a = 4a+6d+1$ which means $11a-6d = 1$ . Since $15a<100$ we get $a<7$ . Checking for the allowed values of $a$ by hand we get $a=5$ and $d=9$ .

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

Represent the teams' scores as: $(a, an, an^2, an^3)$ and $(a, a+m, a+2m, a+3m)$

We have $a+an+an^2+an^3=4a+6m+1$ Factoring out the $a$ from the left side of the equation, we can get $a(1+n+n^2+n^3)=4a+6m+1$ , or $a(n^4-1)/(n-1)=4a+6m+1$

Since both are increasing sequences, $n>1$ . We can check cases up to $n=4$ because when $n=5$ , we get $156a>100$ . When

$n=2, a=[1,6]$
$n=3, a=[1,2]$
$n=4, a=1$

Checking each of these cases individually back into the equation $a+an+an^2+an^3=4a+6m+1$ , we see that only when $a=5$ and $n=2$ , we get an integer value for $m$ , which is $9$ . The original question asks for the first half scores summed, so we must find $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}$

Note: This solution is incomplete because it doesn't consider fractional values of $r$ .

Solution 3

Note: This solution is incomplete because it doesn't consider fractional values of $n$ .

As above, represent the teams' scores as: $(a, an, an^2, an^3)$ and $(a, a+m, a+2m, a+3m)$ .

Note that the Wildcat's sum, $4a+6m$ , is even. Therefore, since the Wildcat's sum is one less than the Raiders, the Raider's team's score should be odd. But if all of $(a, an, an^2, an^3)$ are of the same parity, the sum will be even. If $a$ is even, then the rest of the scores will be even, so clearly $a$ is odd. Then, $n$ is even.

But if $n=4$ , only $a=1$ satisfies the requirement that the total score of each team is less than $100$ . We can test this out and see it doesn't work.

Therefore, $n=2$ . If we try $a=(1,3,5)$ , we quickly see only $a=5$ satisfies all of the conditions. Therefore, our team's scores are $(5,10,20,40)$ and $(5,14,23,32)$ , and the answer is $(5)+(10)+(5)+(14)=\boxed{\textbf{(E)}\ 34}$ .

Discussion

Couldn't we also say Raiders:1,2,4,8 and Wildcats:1,3,5,7 so the Raiders have won by one point? Is this false because in actual basketball you can only score 2 or more points? ---> Raiders would have a total of 15 points and the WildCats would have a total of 16 points.

Note: In example above, the Raiders score a total of 15 points while the Wildcats score 16. So it is the Wildcats who win by one point, not the Raiders.

@@ Line 5: / Line 5: @@
 == Solution 1 ==
+Let <math>a,ar,ar^{2},ar^{3}</math> be the quarterly scores for the Raiders. We know <math>r > 1</math> because the sequence is said to be increasing. We also know that each of <math>a, ar, ar^2, ar^3</math> is an integer. We start by showing that '''<math>r</math> must also be an integer.'''
+Suppose not, and say <math>r = m/n</math> where <math>m>n>1</math>, and <math>\gcd(m,n)=1</math>. Then <math>n, n^2, n^3</math> must all divide <math>a</math> so <math>a=n^3k</math> for some integer <math>k</math>. Then <math>S_R = n^3k + n^2mk + nm^2k + m^3k < 100</math> and we see that even if <math>k=1</math> and <math>n=2</math>, we get <math>m < 4</math>, which means that the only option for <math>r</math> is <math>r=3/2</math>. A quick check shows that even this doesn't work. Thus <math>r</math> must be an integer.
+Let <math>a, a+d, a+2d, a+3d</math> be the quarterly scores for the Wildcats. Let <math>S_W = a+(a+d) + (a+2d)+(a+3d) = 4a+6d</math>. Let <math>S_R = a+ar+ar^2+ar^3 = a(1+r)(1+r^2)</math>. Then <math>S_R<100</math> implies that <math>r<5</math>, so <math>r\in \{2, 3, 4\}</math>. The Raiders win by one point, so<cmath>a(1+r)(1+r^2) = 4a+6d+1.</cmath>
+*If <math>r=4</math> we get <math>85a = 4a+6d+1</math> which means <math>3(27a-2d) = 1</math>, which is absurd.
+*If <math>r=3</math> we get <math>40a = 4a+6d+1</math> which means <math>6(6a-d) = 1</math>, which is also absurd.
+*If <math>r=2</math> we get <math>15a = 4a+6d+1</math> which means <math>11a-6d = 1</math>. Since <math>15a<100</math> we get <math>a<7</math>. Checking for the allowed values of <math>a</math> by hand we get <math>a=5</math> and <math>d=9</math>.
+Then the quarterly scores for the Raiders are <math>5, 10, 20, 40</math>, and those for the Wildcats are <math>5, 14, 23, 32</math>. Also <math>S_R = 75 = S_W + 1</math>. The total number of points scored by the two teams in the first half is <math>5+10+5+14=\boxed{\textbf{(E)}\ 34}</math>.
+== Solution 2 ==
 Represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>
@@ Line 15: / Line 28: @@
 *<math>          n=4, a=1</math>
 Checking each of these cases individually back into the equation <math>a+an+an^2+an^3=4a+6m+1</math>, we see that only when <math>a=5</math> and <math>n=2</math>, we get an integer value for <math>m</math>, which is <math>9</math>. The original question asks for the first half scores summed, so we must find <math>(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}</math>
-*This solution is incomplete because it doesn't consider r could be a fraction.
-== Solution 2 ==
+*'''Note:''' This solution is incomplete because it doesn't consider fractional values of <math>r</math>.
+== Solution 3 ==
+*'''Note:''' This solution is incomplete because it doesn't consider fractional values of <math>n</math>.
 As above, represent the teams' scores as: <math>(a, an, an^2, an^3)</math> and <math>(a, a+m, a+2m, a+3m)</math>.
@@ Line 30: / Line 46: @@
 Is this false because in actual basketball you can only score 2 or more points?
 ---> Raiders would have a total of 15 points and the WildCats would have a total of 16 points.
+*'''Note:''' In example above, the Raiders score a total of 15 points while the Wildcats score 16. So it is the '''Wildcats''' who win by one point, not the '''Raiders'''.
 == See also ==
 {{AMC10 box|year=2010|ab=B|num-b=23|num-a=25}}
 {{MAA Notice}}

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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