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2010AMC10B真题与答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2018年5月5日上午11:02

2010AMC10B真题

答案解析请参考文末

Problem 1

What is $100(100-3)-(100\cdot100-3)$ ?

$\textbf{(A)}\ -20,000 \qquad \textbf{(B)}\ -10,000 \qquad \textbf{(C)}\ -297 \qquad \textbf{(D)}\ -6 \qquad \textbf{(E)}\ 0$

Problem 2

Makarla attended two meetings during her $9$ -hour work day. The first meeting took $45$ minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?

$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 35$

Problem 3

A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

Problem 4

For a real number $x$ , define $\heartsuit(x)$ to be the average of $x$ and $x^2$ . What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$ ?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$

Problem 5

A month with $31$ days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Problem 6

A circle is centered at $O$ , $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$ . What is the degree measure of $\angle CAB$ ?

$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$

Problem 7

A triangle has side lengths $10$ , $10$ , and $12$ . A rectangle has width $4$ and area equal to the area of the triangle. What is the perimeter of this rectangle?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 32 \qquad \textbf{(E)}\ 36$

Problem 8

A ticket to a school play cost $x$ dollars, where $x$ is a whole number. A group of 9th graders buys tickets costing a total of $\textdollar 48$ , and a group of 10th graders buys tickets costing a total of $\textdollar 64$ . How many values for $x$ are possible?

$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Problem 9

Lucky Larry's teacher asked him to substitute numbers for $a$ , $b$ , $c$ , $d$ , and $e$ in the expression $a-(b-(c-(d+e)))$ and evaluate the result. Larry ignored the parentheses but added and subtracted correctly and obtained the correct result by coincidence. The numbers Larry substituted for $a$ , $b$ , $c$ , and $d$ were $1$ , $2$ , $3$ , and $4$ , respectively. What number did Larry substitute for $e$ ?

$\textbf{(A)}\ -5 \qquad \textbf{(B)}\ -3 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 5$

Problem 10

Shelby drives her scooter at a speed of $30$ miles per hour if it is not raining, and $20$ miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of $16$ miles in $40$ minutes. How many minutes did she drive in the rain?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

Problem 11

A shopper plans to purchase an item that has a listed price greater than $\textdollar 100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\textdollar 30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds $\textdollar 100$ .
Let $x$ and $y$ be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or Coupon C. What is $y - x$ ?

$\textbf{(A)}\ 50 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 75 \qquad \textbf{(D)}\ 80 \qquad \textbf{(E)}\ 100$

Problem 12

At the beginning of the school year, $50\%$ of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", and $50\%$ answered "No." At the end of the school year, $70\%$ answered "Yes" and $30\%$ answered "No." Altogether, $x\%$ of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of $x$ ?

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$

Problem 13

What is the sum of all the solutions of $x = \left|2x-|60-2x|\right|$ ?

$\textbf{(A)}\ 32 \qquad \textbf{(B)}\ 60 \qquad \textbf{(C)}\ 92 \qquad \textbf{(D)}\ 120 \qquad \textbf{(E)}\ 124$

Problem 14

The average of the numbers $1, 2, 3,\cdots, 98, 99,$ and $x$ is $100x$ . What is $x$ ?

$\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}$

Problem 15

On a $50$ -question multiple choice math contest, students receive $4$ points for a correct answer, $0$ points for an answer left blank, and $-1$ point for an incorrect answer. Jesse’s total score on the contest was $99$ . What is the maximum number of questions that Jesse could have answered correctly?

$\textbf{(A)}\ 25 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 29 \qquad \textbf{(D)}\ 31 \qquad \textbf{(E)}\ 33$

Problem 16

A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?

$\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qquad \textbf{(E)}\ \dfrac{2\pi}{9}$

Problem 17

Every high school in the city of Euclid sent a team of $3$ students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed $37$ th and $64$ th, respectively. How many schools are in the city?

$\textbf{(A)}\ 22 \qquad \textbf{(B)}\ 23 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 25 \qquad \textbf{(E)}\ 26$

Problem 18

Positive integers $a$ , $b$ , and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$ . What is the probability that $abc + ab + a$ is divisible by $3$ ?

$\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}$

Problem 19

A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$ ?

$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Problem 20

Two circles lie outside regular hexagon $ABCDEF$ . The first is tangent to $\overline{AB}$ , and the second is tangent to $\overline{DE}$ . Both are tangent to lines $BC$ and $FA$ . What is the ratio of the area of the second circle to that of the first circle?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 108$

Problem 21

A palindrome between $1000$ and $10,000$ is chosen at random. What is the probability that it is divisible by $7$ ?

$\textbf{(A)}\ \dfrac{1}{10} \qquad \textbf{(B)}\ \dfrac{1}{9} \qquad \textbf{(C)}\ \dfrac{1}{7} \qquad \textbf{(D)}\ \dfrac{1}{6} \qquad \textbf{(E)}\ \dfrac{1}{5}$

Problem 22

Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?

$\textbf{(A)}\ 1930 \qquad \textbf{(B)}\ 1931 \qquad \textbf{(C)}\ 1932 \qquad \textbf{(D)}\ 1933 \qquad \textbf{(E)}\ 1934$

Problem 23

The entries in a $3 \times 3$ array include all the digits from $1$ through $9$ , arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$

Problem 24

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?

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

Problem 25

Let $a > 0$ , and let $P(x)$ be a polynomial with integer coefficients such that

$P(1) = P(3) = P(5) = P(7) = a$ , and
$P(2) = P(4) = P(6) = P(8) = -a$ .What is the smallest possible value of $a$ ?

$\textbf{(A)}\ 105 \qquad \textbf{(B)}\ 315 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 7! \qquad \textbf{(E)}\ 8!$

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$100(100-3)-(100\cdot{100}-3)=10000-300-10000+3=-300+3=\boxed{\textbf{(C)}\ -297}$ .
The percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday.The total time spent in meetings is $45 \text{ min} + 2*45\text{ min} = 2\text{ hours } 15 \text{ min} = 9/4 \text{ hours}$ Therefore, the percentage is $\frac{9/4 \text{ hours} }{9 \text{ hours}} = \frac{1}{4} = 25 \% = \boxed{\textbf{(C)} 25}$
After you draw $4$ socks, you can have one of each color, so (according to the pigeonhole principle), if you pull $\boxed{\textbf{(C)}\ 5}$ then you will be guaranteed a matching pair.
The average of two numbers, $a$ and $b$ , is defined as $\frac{a+b}{2}$ . Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$ . With that said, we need to find the sum when we plug, $1$ , $2$ and $3$ into that equation. So: $\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}$
In this month there are four weeks and three remaining days. As long as the last three days and the first four days have the same number of Mondays and Wednesdays, then it works. The number of days the month can start on is $\boxed{\textbf{(B)}\ 3}$
Assuming we do not already know an inscribed angle is always half of its central angle, we will try a different approach. Since $O$ is the center, $OC$ and $OA$ are radii and they are congruent. Thus, $\triangle COA$ is an isosceles triangle. Also, note that $\angle COB$ and $\angle COA$ are supplementary, then $\angle COA = 180 - 50 = 130^{\circ}$ . Since $\triangle COA$ is isosceles, then $\angle OCA \cong \angle OAC$ . They also sum to $50^{\circ}$ , so each angle is $\boxed{\textbf{(B)}\ 25}$ .
The triangle is isosceles. The height of the triangle is therefore given by $h = \sqrt{10^2 - ( \dfrac{12}{2})^2} = \sqrt{64} = 8$ Now, the area of the triangle is $\dfrac{bh}{2} = \dfrac{12*8}{2} = \dfrac{96}{2} = 48$ We have that the area of the rectangle is the same as the area of the triangle, namely $48$ . We also have the width of the rectangle: $4$ .The length of the rectangle therefore is: $l = \dfrac{48}{4} = 12$ The perimeter of the rectangle then becomes: $2l + 2w = 2*12 + 2*4 = 32$ The answer is: $\boxed{\textbf{(D)}\ 32}$
We see how many common integer factors $48$ and $64$ share. Of the factors of $48$ - $1, 2, 3, 4, 6, 8, 12, 16, 24, 48$ ; only $1, 2, 4, 8,$ and $16$ are factors of $64$ . So there are $\boxed{\textbf{(E)}\ 5}$ possibilities for the ticket price.
Simplify the expression $a-(b-(c-(d+e)))$ .So you get: $a-(b-(c-(d+e))) = a-(b-(c-d-e)) = a-(b-c+d+e)) = a-b+c-d-e$ Larry substituted $a, b, c, d$ with $1, 2, 3, 4$ respectively.We have to find the value of $e$ , such that $a-b+c-d-e = a-b-c-d+e$ (the same expression without parenthesis).Substituting and simplifying we get: $-2-e = -8+e \Rightarrow -2e = -6 \Rightarrow e=3$ So, Larry must have used the value $3$ for $e$ .Our answer is $3 \Rightarrow \boxed{\textbf{(D)}}$
We know that $d = vt$ Since we know that she drove both when it was raining and when it was not and that her total distance traveled is $16$ miles.We also know that she drove a total of $40$ minutes which is $\dfrac{2}{3}$ of an hour.We get the following system of equations, where $x$ is the time traveled when it was not raining and $y$ is the time traveled when it was raining: $\left\{\begin{array}{ccc} 30x + 20y & = & 16 \\x + y & = & \dfrac{2}{3} \end{array} \right.$ Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get: $-10y = -4 \Leftrightarrow y = \dfrac{2}{5}$ We know now that the time traveled in rain was $\dfrac{2}{5}$ of an hour, which is $\dfrac{2}{5}*60 = 24$ minutesSo, our answer is $\boxed{\textbf{(C)}\ 24}$
Let the listed price be $(100 + p)$ , where $p > 0$ Coupon A saves us: $0.15(100+p) = (0.15p + 15)$ Coupon B saves us: $30$ Coupon C saves us: $0.25p$ Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities: $A \geq B \Rightarrow 0.15p + 15 \geq 30 \Rightarrow p \geq 100$ $A \geq C \Rightarrow 0.15p + 15 \geq 0.25p \Rightarrow p \leq 150$ We see here that the greatest possible value for $p$ is $150$ , thus $y = 100 + 150 = 250$ and the smallest value for p is $100$ so $x = 100 + 100 = 200$ .The difference between $y$ and $x$ is $y - x = 250 - 200 = \boxed{\textbf{(A)}\ 50}$
The minimum possible value occurs when $20\%$ of the students who originally answered "No." answer "Yes." In this case, $x=20$ The maximum possible value occurs when $60\%$ of the students (which accounts for $30\%$ of the overall student population) who originally answered "Yes." answer "No." and the $100\%$ of the students (which accounts for $50\%$ of the overall student population) who originally answered "No." answer "Yes." In this case, $x=50+30=80$ Subtract $80-20$ to obtain an answer of $\boxed{\textbf{(D)}\ 60}$
We evaluate this in cases:Case 1 $x<30$ When $x<30$ we are going to have $60-2x>0$ . When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$ . Therefore we have $x=|2x-(60-2x)|$ . $x=|2x-60+2x|\implies x=|4x-60|$ Subcase 1 $30>x>15$ When $30>x>15$ we are going to have $4x-60>0$ . When this happens, we can express $|4x-60|$ as $4x-60$ . Therefore we get $x=4x-60\implies -3x=-60\implies x=20$ . We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$ . Therefore $20$ is one possible solution.Subcase 2 $x<15$ When $x<15$ we are going to have $4x-60<0$ , therefore $|4x-60|$ can be expressed in the form $60-4x$ . We have the equation $x=60-4x\implies 5x=60\implies x=12$ . Since $12$ is less than $15$ , $12$ is another possible solution. $x=|2x-|60-2x||$ Case 2 : $x>30$ When $x>30$ , $60-2x<0$ . When $x<0$ we can express this in the form $-x$ . Therefore we have $-(60-2x)=2x-60$ . This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have $x=|2x-(2x-60)|$ $x=|2x-2x+60|$ $x=|60|$
$x=60$

We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{\textbf{(C)}\ 92}$
We must find the average of the numbers from $1$ to $99$ and $x$ in terms of $x$ . The sum of all these terms is $\frac{99(100)}{2}+x=99(50)+x$ . We must divide this by the total number of terms, which is $100$ . We get: $\frac{99(50)+x}{100}$ . This is equal to $100x$ , as stated in the problem. We have: $\frac{99(50)+x}{100}=100x$ . We can now cross multiply. This gives $100(100x)=99(50)+x \rightarrow 10000x=99(50)+x \rightarrow 9999x=99(50) \rightarrow 101x=50 \rightarrow x=\boxed{\text{(B)} \frac{50}{101}}$
Let $a$ be the amount of questions Jesse answered correctly, $b$ be the amount of questions Jesse left blank, and $c$ be the amount of questions Jesse answered incorrectly. Since there were $50$ questions on the contest, $a+b+c=50$ . Since his total score was $99$ , $4a-c=99$ . Also, $a+c\leq50 \Rightarrow c\leq50-a$ . We can substitute this inequality into the previous equation to obtain another inequality: $4a-(50-a)\leq99 \Rightarrow 5a\leq149 \Rightarrow a\leq \frac{149}5=29.8$ . Since $a$ is an integer, the maximum value for $a$ is $\boxed{\textbf{(C)}\ 29}$ .
The radius of the circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$ . Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}$ . We can see that the circle passes outside the square, but the square is NOT completely contained in the circle Therefore the picture will look something like this: $[asy] unitsize(5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3; real r=sqrt(1/3); pair O=(0,0); pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5); pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5); pair V=(0,0.5); path outer=Circle(O,r); draw(outer); draw(W--X--Y--Z--cycle); draw(O--A); draw(O--B); draw(V--O); pair[] ps={A,B,V,O}; dot(ps); label("$O$",O,SW); label("$\frac{\sqrt{3}}{3}$",O--B,SE); label("$A$",A,NW); label("$B$",B,NE); label("$X$",V,NW); label("$a$",B--V,S); label("$\frac12$",O--V,W); [/asy]$ Then we proceed to find: 4 * (area of sector marked off by the two radii - area of the triangle with sides on the square and the two radii).First we realize that the radius perpendicular to the side of the square between the two radii marking off the sector splits $AB$ in half. Let this half-length be $a$ . Also note that $OX=\frac12$ because it is half the sidelength of the square. Because this is a right triangle, we can use thePythagorean Theorem to solve for $a.$ $a^2+\left( \frac12 \right) ^2 = \left( \frac{\sqrt{3}}{3} \right) ^2$ Solving, $a= \frac{\sqrt{3}}{6}$ and $2a=\frac{\sqrt{3}}{3}$ . Since $AB=AO=BO$ , $\triangle AOB$ is an equilateral triangle and the central angle is $60^{\circ}$ . Therefore the sector has an area $\pi \left( \frac{\sqrt{3}}{3} \right) ^2 \left( \frac{60}{360} \right) = \frac{\pi}{18}$ .Now we turn to the triangle. Since it is equilateral, we can use the formula for the area of an equilateral triangle which is $\frac{s^2\sqrt{3}}{4} = \frac{\frac13 \sqrt{3}}{4} = \frac{\sqrt{3}}{12}$ Putting it together, we get the answer to be $4 \left( \frac{\pi}{18}-\frac{\sqrt{3}}{12} \right)= \boxed{\textbf{(B)}\ \frac{2\pi}{9}-\frac{\sqrt{3}}{3}}$
Let the $n$ be the number of schools, $3n$ be the number of contestants, and $x$ be Andrea's place. Since the number of participants divided by three is the number of schools, $n\geq\frac{64}3=21\frac13$ . Andrea received a higher score than her teammates, so $x\leq36$ . Since $36$ is the maximum possible median, then $2*36-1=71$ is the maximum possible number of participants. Therefore, $3n\leq71\Rightarrow n\leq\frac{71}3=23\frac23$ . This yields the compound inequality: $21\frac13\leq n\leq 23\frac23$ . Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, $n$ cannot be even. $\boxed{\textbf{(B)}\ 23}$ is the only other option.
First we factor $abc + ab + a$ as $a(bc + b + 1)$ , so in order for the number to be divisible by 3, either $a$ is divisible by $3$ , or $bc + b + 1$ is divisible by $3$ .We see that $a$ is divisible by $3$ with probability $\frac{1}{3}$ . We only need to calculate the probability that $bc + b + 1$ is divisible by $3$ .We need $bc + b + 1 \equiv 0\pmod 3$ or $b(c + 1) \equiv 2\pmod 3$ . Using some modular arithmetic, $b \equiv 2\pmod 3$ and $c \equiv 0\pmod 3$ or $b \equiv 1\pmod 3$ and $c \equiv 1\pmod 3$ . The both cases happen with probability $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$ so the total probability is $\frac{2}{9}$ .Then the answer is $\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}$ or $\boxed{E}$ .
The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$ Because $OA=4\sqrt{3} < \sqrt{156}, A$ must be in the interior of circle $O.$ $[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; real r=sqrt(156); pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147)); pair O=(0,0); pair X=(0,7sqrt(3)); path outer=Circle(O,r); draw(outer); draw(A--B--C--cycle); draw(O--X); draw(O--B); pair[] ps={A,B,C,O,X}; dot(ps); label("$A$",A,SE); label("$B$",B,NW); label("$C$",C,NE); label("$O$",O,S); label("$X$",X,N); label("$s$",A--C,SE); label("$\frac{s}{2}$",B--X,N); label("$\frac{s\sqrt{3}}{2}$",A--X,NE); label("$\sqrt{156}$",O--B,SW); label("$4\sqrt{3}$",A--O,E); [/asy]$ Let $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$ $\begin{align*} (\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\ 156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\ 0 &= s^2 + 12s - 108\\ 0 &= (s-6)(s+18)\\ s &= \boxed{\textbf{(B)}\ 6} \end{align*}$
A good diagram is very helpful. Also, BC and FA are LINES, not LINE SEGMENTS.The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle $GBA$ while the second circle is inscribed in $GKJ$ . From this, it's evident that the ratio of the blue area to the red area is equal to the ratio of the areas $\triangle GKJ$ to $\triangle GBA$ Since the ratio of areas is equal to the square of the ratio of lengths, we know our final answer is $\left(\frac{GK}{GB}\right)^2$ . From the diagram, we can see that this is $9^2=\boxed{\textbf{(D)}\ 81}$
The palindromes can be expressed as: $1000x+100y+10y+x$ (since it is a four digit palindrome, it must be of the form $xyyx$ , where x and y are integers from $[1, 9]$ and $[0, 9]$ , respectively.)We simplify this to $1001x+110y$ .Because the question asks for it to be divisible by 7,We express it as $1001x+110y \equiv 0 \pmod 7$ .Because $1001 \equiv 0 \pmod 7$ ,We can substitute $1001$ for $0$ We are left with $110y \equiv 0 \pmod 7$ Since $110 \equiv 5 \pmod 7$ we can simplify the $110$ in the expression to $5y \equiv 0 \pmod 7$ .In order for this to be true, $y \equiv 0 \pmod 7$ must also be true.Thus we solve:
$y \equiv 0 \pmod 7$

Which has two solutions: $0$ and $7$

There are thus two options for $y$ out of the 10, so the answer is $2/10 = \boxed{\textbf{(E)}\ \frac15}$
We can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.Each candy has three choices; it can go in any of the three bags.Since there are seven candies, that makes the total distributions $3^7=2187$ To find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.The number of distributions such that the red bag is empty is equal to $2^7$ , since it's equivalent to distributing the $7$ candies into $2$ bags.We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also $2^7$ .The case where both the red and the blue bags are empty (all $7$ candies are in the white bag) are included in both of the above calculations, and this case has only $1$ distribution.The total overcount is $2^7+2^7-1=2^8-1$ The final answer will be $\text{total}-\text{overcount}=2187-(2^8-1) = 2187-256+1=1931+1=\boxed{\textbf{(C)}\ 1932}$
The upper-left corner must contain the entry 1, and similarly the lower-right corner must contain the entry 9. Consider the entries 2 and 3 -- they may either both lie in the first row, both lie in the first column, or lie in the two squares neighboring 1. By symmetry (which we will take into account by a factor of 2 in the end), we may assume that 2 lies in the cell to the right of 1 and that 3 lies either in the cell to the right of 2 or in the cell below 1: $\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\\hline && \\\hline &&\\\hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 1 & 2 & \\\hline 3&& \\\hline &&\\\hline \end{array}$ Similarly, the entries 7 and 8 may either both lie in the last row, or lie in the two squares neighboring 9. This gives the following cases: $\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\\hline && \\\hline 7&8&9\\\hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 1 & 2 & \\\hline 3&& \\\hline 7&8&9\\\hline \end{array} \times 2 \qquad \begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\\hline &&7 \\\hline &8&9\\\hline \end{array}\times 2 \qquad \begin{array}{|c|c|c|} \hline 1 & 2 & \\\hline 3&&7 \\\hline &8&9\\\hline \end{array}\times 2,$ where the notation $\times 2$ denotes two possible cases, either by switching a row and column or by switching the 7 and 8. Finally, there are respectively 1, 2, 2, 6 ways to complete these four cases. This gives a total of $2\cdot\left(1+2\times2+2\times2+2\times6\right)=\boxed{\textbf{(D)}\ 42}$ possible ways to fill the diagram.NotesIn fact, there is a general formula (coming from the fields of combinatorics and representation theory) to answer problems of this form; it is known as the hook-length formula.
Represent the teams' scores as: and We have Factoring out the from the left side of the equation, we can get , or Since both are increasing sequences, . We can check cases up to because when , we get . 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 $(a)+(an)+(a)+(a+m)=(5)+(10)+(5)+(5+9)=\boxed{\textbf{(E)}\ 34}$
We observe that because $P(1) = P(3) = P(5) = P(7) = a$ , if we define a new polynomial $R(x)$ such that $R(x) = P(x) - a$ , $R(x)$ has roots when $P(x) = a$ ; namely, when $x=1,3,5,7$ .Thus since $R(x)$ has roots when $x=1,3,5,7$ , we can factor the product $(x-1)(x-3)(x-5)(x-7)$ out of $R(x)$ to obtain a new polynomial $Q(x)$ such that $(x-1)(x-3)(x-5)(x-7)(Q(x)) = R(x) = P(x) - a$ .Then, plugging in values of $2,4,6,8,$ we get $P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2) = -15Q(2) = -2a$ $P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4) = 9Q(4) = -2a$ $P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6) = -15Q(6) = -2a$ $P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8) = 105Q(8) = -2a$ $-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$ Thus, the least value of $a$ must be the $lcm(15,9,15,105)$ . Solving, we receive $315$ , so our answer is $\boxed{\textbf{(B)}\ 315}$ .To complete the solution, we can let $a = 315$ , and then try to find $Q(x)$ . We know from the above calculation that $Q(2)=42, Q(4)=-70, Q(6)=42$ , and $Q(8)=-6$ . Then we can let $Q(x) = T(x)(x-2)(x-6)+42$ , getting $T(4)=28, T(8)=-4$ . Let $T(x)=L(x)(x-8)-4$ , then $L(4)=-8$ . Therefore, it is possible to choose $T(x) = -8(x-8)-4 = -8x + 60$ , so the goal is accomplished. As a reference, the polynomial we get is $P(x) = (x-1)(x-3)(x-5)(x-7)((-8x + 60)(x-2)(x-6)+42) + 315$ $= -8 x^7+252 x^6-3248 x^5+22050 x^4-84392 x^3+179928 x^2-194592 x+80325$