2017 AMC 10A 考试题目

Scroll down and press Start to try the exam! Or, go to the printable PDF, answer key, or professional solutions curated by LIVE, by Po-Shen Loh.

All of the real AMC 8 and AMC 10 problems in our complete solution collection are used with official permission of the Mathematical Association of America (MAA).

Want to learn professionally through interactive video classes?

Learn LIVE

考试时间还剩下:

1:15:00

1.

What is the value of (2(2(2(2(2(2+1)+1)(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)?+1)+1)+1)+1)?

7070

9797

127127

159159

729729

Answer: C
Solution:

Simplifying yields (2(2(2(2(2(2+1)+1)+1)+1)+1)+1)=(2(2(2(2(2(3)+1)+1)+1)+1)+1)=(2(2(2(2(7)+1)+1)+1)+1)=(2(2(2(15)+1)+1)+1)=(2(2(31)+1)+1)=2(63)+1=127. \begin{align*} &(2(2(2(2(2(2+1)+1)\\ &+1)+1)+1)+1) \\ =& (2(2(2(2(2(3)+1)\\ &+1)+1)+1)+1)\\=&(2(2(2(2(7)+1)+1)+1)+1) \\=& (2(2(2(15) + 1) + 1) + 1) \\ =&(2(2(31) + 1) + 1) \\=& 2(63) + 1 \\=& 127. \end{align*}

Thus, C is the correct answer.

2.

Pablo buys popsicles for his friends. The store sells single popsicles for $1\$1 each, 3-popsicle boxes for $2\$2 each, and 5-popsicle boxes for $3.\$3. What is the greatest number of popsicles that Pablo can buy with $8?\$8?

88

1111

1212

1313

1515

Answer: D
Solution:

The $3 boxes give us the most popsicles per dollar, so we want to buy as many of those as possible.

We can buy two of those, getting 52=105 \cdot 2 = 10 popsicles with $8 - $6 = $2 remaining.

The $1 single popsicles are the worst deal, so Pablo should spend the rest of his money on the 33-popsicle box.

He then ends up with 10+3=1310 + 3 = 13 popsicles.

Thus, D is the correct answer.

3.

Tamara has three rows of two 66-feet by 22-feet flower beds in her garden. The beds are separated and also surrounded by 11-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?

7272

7878

9090

120120

150150

Answer: B
Solution:

We can see that the width of the garden is 26+31=15.2 \cdot 6 + 3 \cdot 1 = 15. We can also see that the height is 32+41=10. 3 \cdot 2 + 4 \cdot 1 = 10. The total area of the garden is therefore 1510=150.15 \cdot 10 = 150. The area of all the flower beds is 626=72. 6 \cdot 2 \cdot 6 = 72. Subtracting this from the area of the garden yields 15072=78,150 - 72 = 78, which is the area of the walkways.

Thus, B is the correct answer.

4.

Mia is "helping" her mom pick up 3030 toys that are strewn on the floor. Mia’s mom manages to put 33 toys into the toy box every 3030 seconds, but each time immediately after those 3030 seconds have elapsed, Mia takes 22 toys out of the box. How much time, in minutes, will it take Mia and her mom to put all 3030 toys into the box for the first time?

13.513.5

1414

14.514.5

1515

15.515.5

Answer: B
Solution:

Note that after 3030 seconds, there are 33 toys added and 22 removed, leaving a net total of +1+1 toys in the box.

We have to be careful towards the end, however, since it is possible for the box to have 3030 toys right after Mia's mom adds the toys and before Mia removes them.

After there are 2727 toys in the box, Mia's mom can add 3,3, leaving 3030 toys in the box.

It will take 273027 \cdot 30 seconds, plus another 3030 seconds, which gives us 2830÷60=14 28 \cdot 30 \div 60 = 14 minutes to get 3030 toys in the box.

Thus, B is the correct answer.

5.

The sum of two nonzero real numbers is 44 times their product. What is the sum of the reciprocals of the two numbers?

11

22

44

88

1212

Answer: C
Solution:

Let xx and yy be the two numbers. We are given that x+y=4xy.x + y = 4xy.

Note that 1x+1y=x+yxy=4. \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{x + y}{xy} = 4.

Thus, C is the correct answer.

6.

Ms. Carroll promised that anyone who got all the multiple choice questions right on the upcoming exam would receive an A on the exam. Which one of these statements necessarily follows logically?

If Lewis did not receive an A, then he got all of the multiple choice questions wrong.

If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong.

If Lewis got at least one of the multiple choice questions wrong, then he did not receive an A.

If Lewis received an A, then he got all of the multiple choice questions right.

If Lewis received an A, then he got at least one of the multiple choice questions right.

Answer: B
Solution:

There is no stipulation on how to get an A other than that getting all the multiple choice right guarantees an A.

This means that it is possible to get an A without getting all the multiple choice questions right.

It is also possible to not get an A even if all but one of the multiple choice questions are answered correctly.

This rules out A, C, D, and E.

Thus, B is the correct answer.

7.

Jerry and Silvia wanted to go from the southwest corner of a square field to the northeast corner. Jerry walked due east and then due north to reach the goal, but Silvia headed northeast and reached the goal walking in a straight line. Which of the following is closest to how much shorter Silvia's trip was, compared to Jerry's trip?

30%30\%

40%40\%

50%50\%

60%60\%

70%70\%

Answer: A
Solution:

Let ss be the side length of the field. Then Jerry traveled 2s2s and Silvia traveled s2s\sqrt{2} from the Pythagorean theorem.

The desired value is 2ss22s=22221.42=.3=30%.\begin{align*} \dfrac{2s - s\sqrt{2}}{2s} &= \dfrac{2 - \sqrt{2}}{2} \\ &\approx \dfrac{2 - 1.4}{2} \\&= .3\\ &= 30\%. \end{align*}

Thus, A is the correct answer.

8.

At a gathering of 3030 people, there are 2020 people who all know each other and 1010 people who know no one. People who know each other hug, and people who do not know each other shake hands. How many handshakes occur within the group?

240240

245245

290290

480480

490490

Answer: B
Solution:

Each of the 1010 people shake hands with each of the 2020 people. This results in 1020=20010 \cdot 20 = 200 handshakes.

There are also (102)=45\binom{10}{2} = 45 handshakes within the 1010 people (every pair of people shake hands).

Therefore, the total number of handshakes is 200+45=245.200 + 45 = 245.

Thus, B is the correct answer.

9.

Minnie rides on a flat road at 2020 kilometers per hour (kph), downhill at 3030 kph, and uphill at 55 kph. Penny rides on a flat road at 3030 kph, downhill at 4040 kph, and uphill at 1010 kph. Minnie goes from town AA to town B,B, a distance of 1010 km all uphill, then from town BB to town C,C, a distance of 1515 km all downhill, and then back to town A,A, a distance of 2020 km on the flat. Penny goes the other way around using the same route. How many more minutes does it take Minnie to complete the 4545-km ride than it takes Penny?

4545

6060

6565

9090

9595

Answer: C
Solution:

It will take Minnie 10÷5=210 \div 5 = 2 hours to travel the uphill distance. It will take her 15÷30=1215 \div 30 = \frac{1}{2} hours to travel the downhill distance.

Finally, it will take her 20÷20=120 \div 20 = 1 hour to travel the flat. This will take her a total of 60(2+12+1)=6072=210 \begin{align*}60\left(2 + \frac{1}{2} + 1\right) &= 60 \cdot \frac{7}{2} \\&= 210 \end{align*} minutes.

It will take Penny 20÷30=2320 \div 30 = \frac{2}{3} hours to travel the flat. It will take her another 15÷10=3215 \div 10 = \frac{3}{2} hours to travel the uphill.

Finally, it will take her 10÷40=1410 \div 40 = \frac{1}{4} hours to travel the downhill. This is a total of 60(23+32+14)=602912=145\begin{align*} 60\left(\dfrac{2}{3} + \dfrac{3}{2} + \dfrac{1}{4}\right) &= 60 \cdot \dfrac{29}{12} \\&= 145 \end{align*} minutes. The trip takes Minnie 210145=65210 - 145 = 65 more minutes to travel than Penny.

Thus, C is the correct answer.

10.

Joy has 3030 thin rods, one each of every integer length from 11 cm through 3030 cm. She places the rods with lengths 33 cm, 77 cm, and 1515 cm on a table. She then wants to choose a fourth rod that she can put with these three to form a quadrilateral with positive area. How many of the remaining rods can she choose as the fourth rod?

1616

1717

1818

1919

2020

Answer: B
Solution:

Note that no one side can be greater than or equal to the sum of the other side lengths.

Let xx be the length fourth rod. Then we have that x<3+7+15 x \lt 3 + 7 + 15 and x+3+7>15 x + 3 + 7 \gt 15 Simplifying, we know that 5<x<25.5\lt x\lt 25. Counting the number of integers in this range, we are left with 2551=1925 - 5 - 1 = 19 values for x.x.

The rods with length 77 and 1515 are already being used, however, so xx cannot equal these.

This leaves 192=1719 - 2 = 17 viable solutions for x.x.

Thus, B is the correct solution.

11.

The region consisting of all points in three-dimensional space within 33 units of line segment AB\overline{AB} has volume 216π.216\pi. What is the length AB?AB?

66

1212

1818

2020

2424

Answer: D
Solution:

Recall that all the points at most a fixed distance rr away from a point form a sphere.

At the end points of this line segment, we can visualize two hemispheres being formed at each end.

All the points in the middle also have spheres forming around them, but they get merged into the ones right next to them.

This means that the middle section forms a cylinder with radius 3.3. The two hemispheres form a sphere with radius 3,3, and therefore a volume of 43π33=2743π=36π. \dfrac{4}{3} \pi 3^3 = 27 \cdot \dfrac{4}{3} \pi = 36 \pi. This means that the cylinder has a volume of 216π36π=180π.216 \pi - 36 \pi = 180 \pi. We know the base is 3π,3 \pi, so if hh is AB,AB, then the volume is π32h=180π9hπ=180πh=20. \begin{align*} \pi 3^2 h &= 180\pi \\9h \pi &= 180 \pi \\ h &= 20. \end{align*}

Thus, D is the correct answer.

12.

Let SS be a set of points (x,y)(x,y) in the coordinate plane such that two of the three quantities 3,x+2,3,x+2, and y4y-4 are equal and the third of the three quantities is no greater than this common value. Which of the following is a correct description for S?S?

a single point

two intersecting lines

three lines whose pairwise intersections are three distinct points

a triangle

three rays with a common endpoint

Answer: E
Solution:

Let us case on which of the values are equal. If 3=x+2,3 = x + 2, then x=1.x = 1. This also tells us that y43 y - 4 \leq 3 y7. y \leq 7. This describes a ray starting at (1,7)(1, 7) and extending in the negative yy direction.

Similarly, if 3=y4,3 = y - 4, then y=7y = 7 and x+23 x + 2 \leq 3 x1. x \leq 1. This also describes a ray starting at (1,7)(1, 7) but instead extending in the negative xx direction.

Finally, if x+2=y4,x + 2 = y - 4, then we have the line y=x+6.y = x + 6. Furthermore, we have that 3x+2 3 \leq x + 2 x1 x \geq 1 and 3y4 3 \leq y - 4 y7. y \geq 7. Note that if one if these conditions is met, the other is also necessarily true due to the equation of the line.

If y=7,y = 7, then x=1.x = 1. The other points are along the line, where y>7y \gt 7 and x>1.x \gt 1.

This describes another ray that starts at (1,7)(1, 7) and goes off in some third direction.

All three cases result in rays originating from (1,7)(1, 7) that all go in different directions.

Thus, E is the correct answer.

13.

Define a sequence recursively by F0=0, F1=1,F_{0}=0,~F_{1}=1, and Fn=F_{n}= the remainder when Fn1+Fn2F_{n-1}+F_{n-2} is divided by 3,3, for all n2.n\geq 2. Thus the sequence starts 0,1,1,2,0,2,.0,1,1,2,0,2,\ldots. What is F2017+F2018+F2019+F2020+F_{2017}+F_{2018}+F_{2019}+F_{2020}+F2021+F2022+F2023+F2024?F_{2021}+F_{2022}+F_{2023}+F_{2024}?

66

77

88

99

1010

Answer: D
Solution:

Let us list out the first few values to see if we can find a pattern in this sequence.

0,1,1,2,0,2,2,1,0,1, 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, \cdots

From this we can see that the pattern repeats every 88 terms.

The desired answer is the sum of 88 consecutive numbers, which is fixed. This sum is 0+1+1+2+0+2 0 + 1 + 1 + 2 + 0 + 2+2+1=9. + 2 + 1 = 9. Thus, D is the correct answer.

14.

Every week Roger pays for a movie ticket and a soda out of his allowance. Last week, Roger's allowance was AA dollars. The cost of his movie ticket was 20%20\% of the difference between AA and the cost of his soda, while the cost of his soda was 5%5\% of the difference between AA and the cost of his movie ticket. To the nearest whole percent, what fraction of AA did Roger pay for his movie ticket and soda?

9%9\%

19%19\%

22%22\%

23%23\%

25%25\%

Answer: D
Solution:

Let tt be the cost of the ticket and ss be the cost of the soda. Then we get the following equations. t=As5 t = \dfrac{A - s}{5} s=At20 s = \dfrac{A - t}{20}

Cross-multiplying the first equation gives us 5t=As.5t = A - s. Substituting in the expression for ss yields 5t=AAt20. 5t = A - \dfrac{A - t}{20}. Solving yields 5t=AAt20100t=20AA+t99t=19At=19A99. \begin{align*} 5t &= A - \dfrac{A - t}{20} \\ 100t &= 20A - A + t \\ 99t &= 19A \\ t &= \dfrac{19A}{99}. \end{align*}

This also gives us s=A19A9920=4A99. s = \dfrac{A - \frac{19A}{99}}{20} = \dfrac{4A}{99}.

Adding together the costs gives us 19A99+4A99=23A9923%. \dfrac{19A}{99} + \dfrac{4A}{99} = \dfrac{23A}{99} \approx 23 \%.

Thus, D is the correct answer.

15.

Chloe chooses a real number uniformly at random from the interval [0,2017].[0, 2017].

Independently, Laurent chooses a real number uniformly at random from the interval [0,4034].[0, 4034].

What is the probability that Laurent's number is greater than Chloe's number?

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

56\dfrac{5}{6}

78\dfrac{7}{8}

Answer: C
Solution:

If Laurent chooses a number in the interval (2017,4034],(2017, 4034], then there is no way that Chloe can have the greater number.

This means that Laurent has a 12\frac{1}{2} chance of automatically winning.

Otherwise, Laurent chooses a number in the interval [0,2017].[0, 2017]. The probability that she gets a greater number than Chloe is the same as Chloe getting a greater number then Laurent.

This means that Laurent has a 12\frac{1}{2} chance of getting a greater number (when working with real intervals, the probability of a tie is essentially 00 due to the infinite size of the intervals).

Laurent's total chance of getting a greater number is 121+1212=34. \dfrac{1}{2} \cdot 1 + \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{3}{4}.

Thus, C is the correct answer.

16.

There are 1010 horses, named Horse 1,1, Horse 2,2, . . . , Horse 10.10. They get their names from how many minutes it takes them to run one lap around a circular race track: Horse kk runs one lap in exactly kk minutes. At time 00 all the horses are together at the starting point on the track. The horses start running in the same direction, and they keep running around the circular track at their constant speeds.

The least time S>0,S > 0, in minutes, at which all 1010 horses will again simultaneously be at the starting point is S=2520.S=2520. Let T>0T > 0 be the least time, in minutes, such that at least 55 of the horses are again at the starting point. What is the sum of the digits of T?T?

22

33

44

55

66

Answer: B
Solution:

The time it will take for 55 horses to meet again at the start is the least common multiple of their times.

We want to find the 55 numbers that share lots of prime factors and have small prime factors.

This is because to find the least common multiple, we choose the highest power of a prime that is present among all the numbers.

As such, we can choose Horses 1,2,3,4,1, 2, 3, 4, and 6.6. These have prime factors of 22 and 3,3, which the best we can do.

The least common multiple is 12.12. The sum of its digits is 1+2=3.1 + 2 = 3.

Thus, B is the correct answer.

17.

Distinct points P,P, Q,Q, R,R, SS lie on the circle x2+y2=25x^{2}+y^{2}=25 and have integer coordinates. The distances PQPQ and RSRS are irrational numbers.

What is the greatest possible value of the ratio PQRS?\dfrac{PQ}{RS}?

33

55

353\sqrt{5}

77

525\sqrt{2}

Answer: D
Solution:

Note that the only integer coordinate pairs on this circle are (±3,±4), (\pm 3, \pm 4),(±4,±3), (\pm 4, \pm 3),(±5,0), (\pm 5, 0), and (0,±5). (0, \pm 5).

To get the greatest possible ratio, we want to maximize PQPQ and minimize RS.RS.

We can see that the distance between any of these points is irrational as long as it is not a diameter.

There are only 22 logical candidates for the longest distance: (4,3)(-4, 3) and (3,4)(3, -4) or (4,3)(-4, 3) and (5,0).(5, 0).

Using the distance formula gives us the two distances as 98\sqrt{98} and 90.\sqrt{90}. The first is greater.

There is only one viable choice for the shortest distance: (3,4)(3, 4) and (4,3),(4, 3), which give us a distance of 2.\sqrt{2}.

The desired ratio is then 982=49=7. \dfrac{\sqrt{98}}{\sqrt{2}} = \sqrt{49} = 7.

Thus, D is the correct answer.

18.

Amelia has a coin that lands heads with probability 13,\frac{1}{3}, and Blaine has a coin that lands on heads with probability 25.\frac{2}{5}. Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. What is qp?q-p?

11

22

33

44

55

Answer: D
Solution:

Let xx be the probability that Amelia wins.

There is a 13\frac{1}{3} chance Amelia wins off her first flip.

If she gets a tails, we want Blaine to lose, which happens with a 35\frac{3}{5} chance.

The total probability of this case is 2335=25.\dfrac{2}{3} \cdot \dfrac{3}{5} = \dfrac{2}{5}.

The game then goes back to Amelia, who then again has a xx chance of winning.

Therefore, we get the following equation. x=13+25x x = \dfrac{1}{3} + \dfrac{2}{5}x35x=13\dfrac{3}{5}x = \dfrac{1}{3} x=59. x = \dfrac{5}{9}.

The difference between the denominator and numerator is 95=4.9 - 5 = 4.

Thus, D is the correct answer.

19.

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 55 chairs under these conditions?

1212

1616

2828

3232

4040

Answer: C
Solution:

If Alice sets on an edge, then the person next to her cannot be Bob or Carla. This means that it must be Derek or Eric.

WLOG, let the person be Eric. Then the person next to Eric has to be Bob or Carla. After that there are no more restrictions.

This gives us a total of 2222=16. 2 \cdot 2 \cdot 2 \cdot 2 = 16.

The first 22 is for both edges. The second 22 is for Derek or Eric. The third 22 is for Bob or Carla. The final 22 is just for the 22 people that are remaining.

Otherwise, let Alice be in the middle. Then the two people next to her have to be Derek and Eric. Bob and Carla are forced to be in the last 22 seats.

There are 33 choices for Alice sets. The side on which Derek is sat has 22 options, and then there are 22 options for where Bob and Carla go.

This gives us 322=12 3 \cdot 2 \cdot 2 = 12 configurations.

Therefore, there are a total of 12+16=2812 + 16 = 28 total seating arrangements.

Thus, C is the correct answer.

20.

Let S(n)S(n) equal the sum of the digits of positive integer n.n. For example, S(1507)=13.S(1507) = 13. For a particular positive integer n,n, S(n)=1274.S(n) = 1274.

Which of the following could be the value of S(n+1)?S(n+1)?

11

33

1212

12391239

12651265

Answer: D
Solution:

Recall that a number is divisible by 99 if and only if the sum of its digits is also divisible by 9.9.

This means that looking at S(n)S(n) mod 99 would also give us nn mod 9.9.

Let us prove this. If we add xx to nn without carrying, it is clear that the sum of the digits increases by xx and that nn itself increases by x.x.

This would increase both their values mod 99 by x.x.

Now, if it does carry, we would be subtracting 1010 from some digit and adding on 11 to the next digit.

This would keep the value mod 99 constant. We did, however, add xx in there, so the value mod 99 still increased by x.x.

These are the only two cases, and in both we have shown that the value mod 99 for both nn and S(n)S(n) increased by x.x.

Therefore, we have that nS(n)5(mod9). n \equiv S(n) \equiv 5 \pmod 9.

From this, we can see that n+16S(n+1)(mod9). n + 1 \equiv 6 \equiv S(n + 1) \pmod 9.

The only answer choice that leaves a remainder of 66 when divided by 99 is 1239.1239.

Thus, D is the correct answer.

21.

A square with side length xx is inscribed in a right triangle with sides of length 3,3, 4,4, and 55 so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length yy is inscribed in another right triangle with sides of length 3,3, 4,4, and 55 so that one side of the square lies on the hypotenuse of the triangle. What is xy?\dfrac{x}{y}?

1213\dfrac{12}{13}

3537\dfrac{35}{37}

11

3735\dfrac{37}{35}

1312\dfrac{13}{12}

Answer: D
Solution:

We can see that ABC\triangle ABC and FBE\triangle FBE are similar (angle-angle). This gives us BFFE=ABAC \dfrac{BF}{FE} = \dfrac{AB}{AC} 4xx=43. \dfrac{4 - x}{x} = \dfrac{4}{3}.

Cross-multiplying yields 123x=4x 12 - 3x = 4x x=127. x = \dfrac{12}{7}.

Here, we have that ABC,\triangle ABC, RBQ,\triangle RBQ, and STC\triangle STC are similar (angle-angle).

This means that RB=43yRB = \dfrac{4}{3}y and CS=34y.CS = \dfrac{3}{4}y. This gives us the equation 43y+34y+y=5 \dfrac{4}{3}y + \dfrac{3}{4}y + y = 5 3712y=5. \dfrac{37}{12}y = 5.

Finally, we get that y=6037.y = \dfrac{60}{37}. The desired ratio is 1276037=3735. \dfrac{\frac{12}{7}}{\frac{60}{37}} = \dfrac{37}{35}.

Thus, D is the correct answer.

22.

Sides AB\overline{AB} and AC\overline{AC} of equilateral triangle ABCABC are tangent to a circle at points BB and CC respectively. What fraction of the area of ABC\triangle ABC lies outside the circle?

43π2713\dfrac{4\sqrt{3}\pi}{27}-\dfrac{1}{3}

32π8\dfrac{\sqrt{3}}{2}-\dfrac{\pi}{8}

12\dfrac{1}{2}

323π9\sqrt{3}-\dfrac{2\sqrt{3}\pi}{9}

4343π27\dfrac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}

Answer: E
Solution:

Let the radius of the circle be r.r.

To find the area of the triangle outside of the circle, we can find the area of the triangle inside the circle and subtract it.

We get that BOC=120\angle BOC = 120^{\circ} since ABO\angle ABO and ACO\angle ACO are right angles.

This means that the area of sector OBCOBC is 120360πr2=πr23. \dfrac{120}{360} \cdot \pi r^2 = \dfrac{\pi r^2}{3}.

Now, we need to find the area of BOC.\triangle BOC. Using the formula for the area of a triangle with sine, we get the area to be 12sin(120)r2=r234. \dfrac{1}{2} \sin (120^{\circ}) \cdot r^2 = \dfrac{r^2\sqrt{3}}{4}.

Then the area of the triangle inside the circle is πr23r234=r2(4π33)12. \dfrac{\pi r^2}{3} - \dfrac{r^2\sqrt{3}}{4} = \dfrac{r^2(4\pi - 3\sqrt{3})}{12}.

The area of ABC\triangle ABC is (r3)234=3r234. \dfrac{(r\sqrt{3})^2\sqrt{3}}{4} = \dfrac{3r^2\sqrt{3}}{4}.

The desired fraction is then 1r2(4π33)123r234=14π3393=14π327+13=434π327.\begin{align*} 1 - \dfrac{\frac{r^2(4\pi - 3\sqrt{3})}{12}}{\frac{3r^2\sqrt{3}}{4}} &= 1 - \dfrac{4\pi - 3\sqrt{3}}{9\sqrt{3}} \\&= 1 - \dfrac{4\pi\sqrt{3}}{27} + \dfrac{1}{3} \\&= \dfrac{4}{3} - \dfrac{4\pi\sqrt{3}}{27}. \end{align*}

Thus, E is the correct answer.

23.

How many triangles with positive area have all their vertices at points (i,j)(i,j) in the coordinate plane, where ii and jj are integers between 11 and 5,5, inclusive?

21282128

21482148

21602160

22002200

23002300

Answer: B
Solution:

We can use complementary counting to find the total number of triangles and subtract out the ones that don't work.

There are a total of 52=255^2 = 25 points, so there are (253)=2300\binom{25}{3} = 2300 possible triangles.

Note that the only way a triangle doesn't work is if all the 33 points are in a straight line.

There are 55 rows, 55 columns, and 22 long diagonals. Each of these 1212 lines have 55 points, which means they contribute 12(53)=1210=120 12 \cdot \binom{5}{3} = 12 \cdot 10 = 120 degenerate triangles.

There are also the diagonal lines with 44 points, such as (0,1)(0, 1) to (4,5).(4, 5). There are 44 of these lines, so they have 4(43)=44=16 4 \cdot \binom{4}{3} = 4 \cdot 4 = 16 degenerate triangles.

Similarly, they are 44 diagonal lines with 33 points. These give us 41=44 \cdot 1 = 4 extra triangles that don't work.

Now, we have to look at the lines with slopes of 12,2,12,\dfrac{1}{2}, 2, -\dfrac{1}{2}, and 2.-2.

There are 33 such lines for each slope, and they all have 33 points on them. Therefore, they contribute 431=12 4 \cdot 3 \cdot 1 = 12 more triangles to discount.

The total number of working triangles is then 230012016412 2300 - 120 - 16 - 4 - 12 =2148.= 2148. Thus, B is the correct answer.

24.

For certain real numbers a,a, b,b, and c,c, the polynomial g(x)=x3+ax2+x+10g(x) = x^3 + ax^2 + x + 10has three distinct roots, and each root of g(x)g(x) is also a root of the polynomial f(x)=x4+x3+bx2+100x+c.f(x) = x^4 + x^3 + bx^2 + 100x + c.What is f(1)?f(1)?

9009-9009

8008-8008

7007-7007

6006-6006

5005-5005

Answer: C
Solution:

We know that f(x)f(x) has 44 roots, 33 of which are the roots of g(x).g(x). This means that we can express as f(x)f(x) as f(x)=g(x)(xr), f(x) = g(x)(x - r), for some complex number rr that is the other root of f(x).f(x).

Plugging in g(x),g(x), we get f(x)f(x) equals: (x3+ax2+x+10)(xr) (x^3 + ax^2 + x + 10)(x - r) =x4+(ar)x3+(1ar)x2 = x^4 + (a - r)x^3 + (1 - ar)x^2 +(10r)x10r.+ (10 - r)x - 10r.

Comparing coefficients, we get 10r=100 10 - r = 100 r=90. r = -90. We also know that ar=1 a - r = 1 a=89. a = -89.

Finally, we have that f(1)f(1) equals: 14+(ar)13+(1ar)121^4 + (a - r)1^3 + (1 - ar)1^2 +(10r)110r+ (10 - r)1 - 10r =1+(89+90)+(18990)= 1+ (-89 + 90) + (1 - 89 \cdot 90) +(10+90)+1090+ (10 + 90) + 10 \cdot 90 =1+18009+100+900= 1 + 1 - 8009 + 100 + 900 =7007.= -7007.

Thus, C is the correct answer.

25.

How many integers between 100100 and 999,999, inclusive, have the property that some permutation of its digits is a multiple of 1111 between 100100 and 999?999? For example, both 121121 and 211211 have this property.

226226

243243

270270

469469

486486

Answer: A
Solution:

We can analyze all the multiple so 1111 and see how many permutations each of them contribute. We can do this by casing on the number of unique digits in the number.

Case 1:1: all the digits are the same

This cannot happen. We can see this by the divisibility rule for 11,11, which says that the sum of the first and last digit minus the middle digit must be divisible by 11.11.

If all the digits are the sum, then the above expression evaluates to that digit, which cannot be divisible by 11.11.

Case 2:2: two of the digits are the same

We can split this up into the numbers that have the digit 00 and those that don't.

There are 88 multiples of 1111 that do not have the digit 0:0: 121,242,363,484,616,737,858, 121, 242, 363, 484, 616, 737, 858, and 979.979.

Each of these numbers contributes 33 permutations, so this scenario has 83=248 \cdot 3 = 24 numbers.

There are 99 multiples of 1111 that have the digit 0:0:110,220,330,440,550,660,770, 110, 220, 330, 440, 550, 660, 770, 880,880, and 990.990.

For these numbers, 00 cannot be the hundreds digit, so each of them only contributes 22 permutations, for a total of 92=18.9 \cdot 2 = 18.

Case 3:3: all the digits are different

There are a total of 8181 multiples of 1111 between 100100 and 999.999. The number of these with all different digits is 8189=64. 81 - 8 - 9 = 64. As in case 2,2, we have to specially account for the numbers with 00 as a digit. There are 8:8: 209,308,407,506,605,704,803, 209, 308, 407, 506, 605, 704, 803, and 902.902.

Each of these gives us 22=42 \cdot 2 = 4 permutations, but we overcount by a factor of 22 since flipping the first and last digits creates another number already in the set.

Therefore, these numbers provide a total of 84÷2=16 8 \cdot 4 \div 2 = 16 unique permutations.

There are now 648=5664 - 8 = 56 multiples of 1111 that we need to account for.

We know that each of these provides 3!=63! = 6 permutations. As above, however, note that flipping the first and last digit of any number in this set produces another number in this set.

We can see this by using the divisibility rule for 11.11. If ABCABC is divisible by 11,11, then we have that A+CBA + C - B is divisible by 11.11.

This means that C+ABC + A - B is divisible by 11,11, which means that CBACBA is also divisible by 11.11.

Therefore, these numbers contribute 566÷2=168 56 \cdot 6 \div 2 = 168 more permutations.

Over all the cases, we have a total of 24+18+16+168=226 24 + 18 + 16 + 168 = 226 numbers.

Thus, A is the correct answer.