2015 AMC 10B Exam Problems

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Used with permission of the Mathematical Association of America.

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1.

What is the value of 2(2)2?2-(-2)^{-2}?

2 -2

116 \dfrac{1}{16}

74 \dfrac{7}{4}

94 \dfrac{9}{4}

6 6

Answer: C
Solution(s):

2(2)2=2122=214=74.\begin{align*} 2-(-2)^{-2} &= 2-\dfrac 12^2 \\ &= 2-\dfrac 14\\ &= \dfrac 74.\end{align*}

Thus, the correct answer is C.

2.

Marie does three equally time-consuming tasks in a row without taking breaks. She begins the first task at 1 ⁣: ⁣001\!:\!00 PM and finishes the second task at 2 ⁣: ⁣402\!:\!40 PM. When does she finish the third task?

3:10 PM

3:30 PM

4:00 PM

4:10 PM

4:30 PM

Answer: B
Solution(s):

The time it takes to do 22 tasks is 100100 minutes. Thus, it takes 5050 more minutes after 2:40,2:40, which is 3:30.3:30.

Thus, the correct answer is B.

3.

Kaashish has written down one integer two times and another integer three times. The sum of the five numbers is 100,100, and one of the numbers is 28.28. What is the other number?

8 8

11 11

14 14

15 15

18 18

Answer: A
Solution(s):

Kaashish either wrote 2828 three times or two times.

If he wrote it twice, then the sum of the other three numbers is: 100282=44,100-28\cdot 2 = 44, which isn't a multiple of 3.3. Therefore, this case wouldn't work.

If he wrote it three times, then the sum of the other two numbers is 100283=16.100-28\cdot 3 = 16. Therefore, the other number is 162=8.\frac {16}2 = 8.

Thus, the correct answer is A.

4.

Four siblings ordered an extra large pizza. Alex ate 15,\frac15, Beth 13,\frac13, and Cyril 14\frac14 of the pizza. Dan got the leftovers. What is the sequence of the siblings in decreasing order of the part of pizza they consumed?

Alex, Beth, Cyril, Dan

Beth, Cyril, Alex, Dan

Beth, Cyril, Dan, Alex

Beth, Dan, Cyril, Alex

Dan, Beth, Cyril, Alex

Answer: C
Solution(s):

Since 13>14>15,\frac 13 > \frac 14 > \frac 15, we know Beth ate more than Cyril and Cyril ate more than Alex. Thus, those three are in order.

The amount Dan ate is 1131415=1360.1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}. This is greater than 15\frac 15 and less than 14,\frac 14 , so Dan is in between Cyril and Alex. This makes the order Beth, Cyril, Dan, Alex.

Thus, the correct answer is C.

5.

David, Hikmet, Jack, Marta, Rand, and Todd were in a 1212-person race with 66 other people. Rand finished 66 places ahead of Hikmet. Marta finished 11 place behind Jack. David finished 22 places behind Hikmet. Jack finished 22 places behind Todd. Todd finished 11 place behind Rand. Marta finished in 66th place. Who finished in 88th place?

David

Hikmet

Jack

Rand

Todd

Answer: B
Solution(s):

Jack is one place in front of Marta who is in 6th place, so he is in 5th place.

Todd is two places in front of Marta who is in 5th place, so he is in 3rd place.

Rand is one place in front of Todd who is in 3rd place, so he is in 2nd place.

Himket is six places behind Rand who is in 2nd place, so he is in 8th place.

Thus, the correct answer is B.

6.

Mahdi practices exactly one sport each day of the week. He runs three days a week but never on two consecutive days. On Monday he plays basketball and two days later golf. He swims and plays tennis, but he never plays tennis the day after running or swimming. Which day of the week does Mahdi swim?

Sunday

Tuesday

Thursday

Friday

Saturday

Answer: E
Solution(s):

There are 44 days between Wednesday and Monday, so he can't fit all of his running days in that time without them being consecutive.

This means he runs on Tuesday. Then, all thats left is running, swimming, or tennis. He must play tennis on Thursday or else he plays tennis after swimming or running.

Therefore, he has Friday, Saturday, and Sunday to run twice and swim once. Since running isn't done on consecutive days, Madhi cannot run on Saturday, leaving that day free for swimming.

Thus, the correct answer is E.

7.

Consider the operation "minus the reciprocal of," defined by ab=a1b.a\diamond b=a-\frac{1}{b}. What is ((12)3)(1(23))?((1\diamond2)\diamond3)-(1\diamond(2\diamond3))?

730 -\dfrac{7}{30}

16 -\dfrac{1}{6}

0 0

16 \dfrac{1}{6}

730 \dfrac{7}{30}

Answer: A
Solution(s):

((12)3)(1(23))=(1213)(153)=(1213)(135)=1625=730\begin{align*} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{align*}

Thus, the correct answer is A.

8.

The letter F shown below is rotated 9090^\circ clockwise around the origin, then reflected in the yy-axis, and then rotated a half turn around the origin. What is the final image?

Answer: E
Solution(s):

The rotation puts the F under the xx-axis with its lines going to the right.

Then, note that a half turn is the same as reflecting upon both axes in any order, so it undoes the reflection upon the yy-axis and reflects it upon the xx-axis. This means the last two turns just reflects it upon the xx-axis.

The reflection puts the F above the xx-axis with its lines going to the right.

Thus, the correct answer is E.

9.

The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius 33 and center (0,0)(0,0) that lies in the first quadrant, the portion of the circle with radius 32\tfrac{3}{2} and center (0,32)(0,\tfrac{3}{2}) that lies in the first quadrant, and the line segment from (0,0)(0,0) to (3,0).(3,0). What is the area of the shark's fin falcata?

4π5 \dfrac{4\pi}{5}

9π8 \dfrac{9\pi}{8}

4π3 \dfrac{4\pi}{3}

7π5 \dfrac{7\pi}{5}

3π2 \dfrac{3\pi}{2}

Answer: B
Solution(s):

The area of the portion of the circle of radius 33 in the first quadrant is equal to: 14π32=94π\dfrac14 \cdot \pi \cdot 3^2 = \dfrac94 \pi

Similarly, the are of the portion of the cicle with radius 32\frac32 centered at (0,32)(0,\frac 32) that lies in the first quadrant is equal to: 12π(32)2=98π\dfrac 12 \cdot \pi \cdot \left(\dfrac32\right)^2 = \dfrac98 \pi

The area of the shark's fin falcata is equal to the difference of these regions: 94π98π=98π\dfrac94 \pi - \dfrac 98 \pi = \dfrac98 \pi

Thus, the correct answer is B.

10.

What are the sign and units digit of the product of all the odd negative integers strictly greater than 2015?-2015?

It is a negative number ending with a 1.

It is a positive number ending with a 1.

It is a negative number ending with a 5.

It is a positive number ending with a 5.

It is a negative number ending with a 0.

Answer: C
Solution(s):

There are 10071007 odd numbers greater than 2015.-2015.

Our product is of an odd number of negative numbers, so the result is negative.

Also, we multiply by 5-5 in there, so the product is a multiple of 5,5, making it end in 55 or 0.0. None of our factors are even, so the product can't be even.

Therefore, the product must end in 5.5.

Thus, the correct answer is C.

11.

Among the positive integers less than 100,100, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?

899 \dfrac{8}{99}

25 \dfrac{2}{5}

920 \dfrac{9}{20}

12 \dfrac{1}{2}

916 \dfrac{9}{16}

Answer: B
Solution(s):

The only digits that are prime are 2,3,5,2,3,5, and 7.7.

Our number can either be one or two digits. There are 44 one digit numbers that have its digits being prime, and 1616 two digit numbers that have its digits being prime. This makes a total of 20.20.

Each of the one one digit numbers are prime. For the two digit numbers, if it ends in 22 or 5,5, then it isn't prime since it would be a multiple of 22 or 5.5.

Thus, we only need to check the numbers that end in 33 or 7.7.

The possible numbers are: 23,27,33,37,23,27,33,37,53,57,73,77.53,57,73,77.

3333 and 7777 are multiples of 11,11, so they aren't prime.

Then, 2727 and 5757 are multiples of 33 so they aren't prime.

This leaves 23,27,53,23,27,53, and 7373 as the only two digit primes.

Therefore, there are 88 primes out of 20,20, making the probability 25.\dfrac 25.

Thus, the correct answer is B.

12.

For how many integers xx is the point (x,x)(x, -x) inside or on the circle of radius 1010 centered at (5,5)?(5, 5)?

11 11

12 12

13 13

14 14

15 15

Answer: A
Solution(s):

The distance between (5,5)(5,5) and (x,x)(x,-x) is (x5)2+(x+5)2\sqrt{(x-5)^2+(x+5)^2} =2x2+50.= \sqrt{2x^2+50}. We must find how many xx are such that that value is less than or equal to 10.10. Therefoe, 2x2+501002x^2 + 50 \leq 100 x225.x^2 \leq 25. 5x5.-5 \leq x \leq 5. Therefore, we have 1111 included tintegers.

Thus, the correct answer is A.

13.

The line 12x+5y=6012x+5y=60 forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

20 20

36017 \dfrac{360}{17}

1075 \dfrac{107}{5}

432 \dfrac{43}{2}

28113 \dfrac{281}{13}

Answer: E
Solution(s):

The triangle is a right triangle with legs of 1212 and 5.5. This makes the hypotenuse 13.13.

Two of the altitudes are then 1212 and 5.5. Also, for any side, A=bh2A = \frac{bh}2 where bb is the base and hh is the altitude.

The area is 1252=30,\frac {12\cdot 5}2 = 30, so the other altitude hh can be found with 30=13h2.30 = \frac{13h}2. Thus, this altitude is 6013.\frac{60}{13}.

Therefore, the sum is 12+5+6013=28113.12+5+\dfrac{60}{13} = \dfrac{281}{13} .

Thus, the correct answer is E.

14.

Let a,a, b,b, and cc be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation (xa)(xb)(x-a)(x-b)+(xb)(xc)=0?+(x-b)(x-c)=0?

15 15

15.5 15.5

16 16

16.5 16.5

17 17

Answer: D
Solution(s):

The equation is equal to (2x(a+c))(xb)(2x-(a+c))(x-b) =2(xb)(xa+c2).= 2(x-b)\left(x- \dfrac{a+c}2\right). This makes the roots equal to: b,a+c2b,\dfrac{a+c}2 and the sum is 2b+a+c2.\dfrac{2b+a+c}2.

Therefore, we want to maximize a,b,c,a,b,c, while making bb the highest.

As such, we can have b=9,a=8,c=7b=9,a=8,c=7 and get a sum: 9+7.5=16.59+7.5=16.5

Thus, the correct answer is D.

15.

The town of Hamlet has 33 people for each horse, 44 sheep for each cow, and 33 ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

41 41

47 47

59 59

61 61

66 66

Answer: B
Solution(s):

Let the number of horses be hh and let the number of cows be c.c.

Then, the number of people is 3h3h and the number of ducks is 9h.9h.

Also, the total number of sheep is 4c,4c, so the number of total number of animals or people is 13h+5c.13h+5c.

If we take the total and subtract 5c,5c, then we get a multiple of 13.13. Thus, any valid number is such that there is a multiple of 1313 that is lower than the number that has the same remainder when divided by 5.5.

For 41,41, the lowest multiple of 1313 that has the same remainder when divided by 55 is 26,26, so this is a valid solution.

For 47,47, the lowest multiple of 1313 that has the same remainder when divided by 55 is 52,52, so this isn't valid solution.

For 59,59, the lowest multiple of 1313 that has the same remainder when divided by 55 is 39,39, so this is a valid solution.

For 61,61, the lowest multiple of 1313 that has the same remainder when divided by 55 is 26,26, so this is a valid solution.

For 66,66, the lowest multiple of 1313 that has the same remainder when divided by 55 is 26,26, so this is a valid solution.

Therefore, 4747 cannot possible be the total.

Thus, the correct answer is B.

16.

Al, Bill, and Cal will each randomly be assigned a whole number from 11 to 10,10, inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's?

91000 \dfrac{9}{1000}

190 \dfrac{1}{90}

180 \dfrac{1}{80}

172 \dfrac{1}{72}

2121 \dfrac{2}{121}

Answer: C
Solution(s):

Let their numbers be a,b,a,b, and c.c. Then, if bb is a multiple of aa and they aren't the same number, then b2a.b \geq 2a. Similarly, c2b.c \geq 2b.

Thus, 4ac10,4a \leq c \leq 10, making a2.5.a \leq 2.5. Thus, a=1a=1 or a=2.a=2. Also, b0.5c5,b \leq 0.5c \leq 5, creating an upper bound of b.b.

Now, casing:

If a=2,a=2, then b5b \leq 5 means b=4.b=4. Then, c=8.c=8. This makes exactly one case.

If a=1,a=1, then b5b \leq 5 means b=2,3,4b=2,3,4 or 5.5.

If b=2,b=2, then cc can be one of 4,6,8,104,6,8,10 making 44 cases.

If b=3,b=3, then cc can be one of 6,96,9 making 22 cases.

If b=4,b=4, then cc can is 88 making 11 case.

If b=5,b=5, then cc can is 1010 making 11 case.

Thus, a=1a=1 makes 88 cases.

There are then 99 total values of a,b,ca,b,c that work. The total number of pairs that are possible is 1098=720.10\cdot 9\cdot 8 = 720.

Therefore, the probability is 9720=180.\dfrac{9}{720} = \dfrac 1{80}.

Thus, the correct answer is C.

17.

The centers of the faces of the right rectangular prism shown below are joined to create an octahedron. What is the volume of this octahedron?

7512 \dfrac{75}{12}

10 10

12 12

102 10\sqrt2

15 15

Answer: B
Solution(s):

If we connect all the centers of the prism, we get an octohedron made of two different pyramids. The area of the base of the pyramid is half of the area of the rectangle it is parallel with. As such, the volume of each pyramid is 13Ah,\frac 13 Ah, so the combined area is: A(h1+h2)3.\dfrac{A(h_1+h_2)}{3}.

The sum of the height is the full side length, so this scales the volume down by a factor of 13.\frac 13.

Therefore, the ratio of the the volume of the prism to the octahedron is 16.\frac 16. The volume of the prism is 345=60,3\cdot 4\cdot 5=60, so the volume of the octohedron is 10.10.

Thus, the correct answer is B.

18.

Johann has 6464 fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?

32 32

40 40

48 48

56 56

64 64

Answer: D
Solution(s):

A coin ends as tails if and only if it has 33 flips that are tails, which happens with probability 18.\frac 18. Thus, the probability of any coin being heads is 78.\frac 78 .

As the probability that a given coin flip is 78,\frac 78, and there are 6464 coin flips in total, the expected number of coins that are now heads is: 6478=56.64 \cdot \dfrac 78 = 56.

Thus, the correct answer is D.

19.

In ABC,\triangle{ABC}, C=90\angle{C} = 90^{\circ} and AB=12.AB = 12. Squares ABXYABXY and ACWZACWZ are constructed outside of the triangle. The points X,Y,Z,X, Y, Z, and WW lie on a circle. What is the perimeter of the triangle?

12+93 12+9\sqrt{3}

18+63 18+6\sqrt{3}

12+122 12+12\sqrt{2}

30 30

32 32

Answer: C
Solution(s):

Consider the following diagram:

The points X,YX,Y are on the circle, so the center goes through its perpendicular bisector, which is the same as the perpendicular bisector of A,B.A,B.

The points W,ZW,Z are on the circle, so the center goes through its perpendicular bisector, which is the same as the perpendicular bisector of A,C.A,C.

Thus, the center is on the perpendicular bisector of A,BA,B and A,C,A,C, which is the circumcenter. The circumcenter of a right triangle is the midpoint of the hypotenuse, so it is the midpoint of AB.AB.

Then, let MM be the midpoint of AB.AB. This would make YM2=MA2+AY2YM^2 = MA^2+AY^2 =62+122= 6^2 + 12^2 =180.= 180.

Therefore, the radius is 180,\sqrt{180}, so MW=180.MW = \sqrt{180}.

Then, the distance from MM to WW is also equal to: (WC+BC2)2+(AC2)2\small \sqrt{\left(WC+ \dfrac{BC}2\right)^2 + \left(\dfrac{AC}2\right)^2} Which simplifies to equal AC2+ACBC+36.\sqrt{ AC^2 + AC\cdot BC + 36}.

Therefore, AC2+ACBC=12AC^2 + AC\cdot BC = 12=AC2+BC2. = AC^2+ BC^2. Since BC0,BC \neq 0, we have AC=BC,AC = BC, making an isoceles right triangle. Therefore AC=BC=62.AC = BC = 6\sqrt 2.

As such, the perimeter is 262+122\cdot 6 \sqrt 2 + 12 =122+12.= 12\sqrt 2 + 12.

Thus, the correct answer is C.

20.

Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?

6 6

9 9

12 12

18 18

24 24

Answer: A
Solution(s):

Suppose we have the unit cube where the points are ordered triples with coordinates that are 00 or 1.1.

Then, after an odd number of moves, the sum of the coordinates is odd. Therefore, the only valid last point is (1,1,1).(1,1,1).

Then, there are 33 choices for the first move and 22 for the second. These choices are all identical, so their next moves would be the same.

From here everything is forced to prevent us from going to (1,1,1)(1,1,1) until the last move, so there are just 66 moves.

Thus, the correct answer is A.

21.

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump.

Cozy goes two steps up with each jump (though if necessary, he will just jump the last step).

Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 55 steps left).

Suppose Dash takes 1919 fewer jumps than Cozy to reach the top of the staircase. Let ss denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of s?s?

9 9

11 11

12 12

13 13

15 15

Answer: D
Solution(s):

The amount Cozy jumps is s2.\left\lceil \frac s2 \right\rceil . The amount Dash jumps is s5.\left\lceil \frac s5 \right\rceil . Thus, s2s5=19. \left\lceil \frac s2 \right\rceil - \left\lceil \frac s5 \right\rceil = 19 . As such, we can case on whether ss is even or odd.

If it is even, then we have: s2s5=19. \dfrac s2 - \left\lceil \dfrac s5 \right\rceil = 19 . Which implies that, s5+1s219=s5s5\small\dfrac s5 +1\geq \dfrac s2 -19= \left\lceil \dfrac s5 \right\rceil \geq \dfrac s5 Therefore, 203s1019. 20\geq \dfrac {3s}{10} \geq 19.

Simplifying the inequality, we get 2003s=1903, \dfrac{200}3\geq s = \dfrac{190}3, so the only possible even ss are 66,64.66,64.

Both these cases satisfy the equation s2s5=19, \dfrac s2 - \left\lceil \dfrac s5 \right\rceil = 19 , so they are both valid answers.

If it is odd, then s2+12s5=19. \dfrac s2 + \dfrac 12- \left\lceil \dfrac s5 \right\rceil = 19 . Which implies that, s5+19.5s2=s5s5\small \dfrac s5 +19.5 \geq \dfrac s2= \left\lceil \dfrac s5 \right\rceil \geq \dfrac s5 Therefore, 19.53s10=18.5. 19.5\geq \dfrac {3s}{10} = 18.5. Simplifying the inequality, we get, 1953s=1853, \dfrac{195}3\geq s = \dfrac{185}3, so the only possible odd ss are 65,63.65,63.

Only 6363 satisfies s2+12s5=19, \dfrac s2 + \dfrac 12 - \left\lceil \dfrac s5 \right\rceil = 19 , so it is the only valid answer.

Therefore, the sum of the answers is 63+64+66=193,63+64+66=193, making the sum of the digits 1+9+3=13.1+9+3=13.

Thus, the correct answer is D.

22.

In the figure shown below, ABCDEABCDE is a regular pentagon and AG=1.AG=1. What is FG+JH+CD?FG + JH + CD?

3 3

1245 12-4\sqrt5

5+253 \dfrac{5+2\sqrt5}{3}

1+5 1+\sqrt5

11+11510 \dfrac{11+11\sqrt5}{10}

Answer: D
Solution(s):

Due to rotational symmetry, EBD=BDA=JHD,\small \angle EBD = \angle BDA = \angle JHD , so DJHDJH is isoceles. Thus, JH=DJ.JH = DJ . This is equal to AG=BG=1AG = BG = 1 by rotational symmetry.

Then, since FDBCFD || BC and BFDC,BF || DC , FBCDFBCD is a parallelogram, making DC=BF=BG+FG=1+FG.\begin{align*}DC &= BF \\&= BG + FG \\&= 1 + FG .\end{align*}

Thus, we need to just find 2(1+FG).2(1+ FG) .

Since AFGAJH,AFG \sim AJH , AGAH=FGJH.\dfrac{AG}{AH} = \dfrac{FG}{JH} . Therefore, 1FG+1=FG.\dfrac{1}{FG+1} = FG .

This means FG2+FG1=0,FG^2 + FG -1=0, so, FG=1+52.FG = \dfrac{-1 + \sqrt {5}}2.

This means 2(1+FG)=1+5.2(1+ FG) = 1 + \sqrt 5.

Thus, the correct answer is D.

23.

Let nn be a positive integer greater than 4 such that the decimal representation of n!n! ends in kk zeros and the decimal representation of (2n)!(2n)! ends in 3k3k zeros. Let ss denote the sum of the four least possible values of n.n. What is the sum of the digits of s?s?

7 7

8 8

9 9

10 10

11 11

Answer: B
Solution(s):

The number of zeros to end a number is equal to the power of 55 in the n!.n!.

Since we are trying to find the lowest numbers, we can inspect small numbers where n60,n \leq 60, so that the number of zeros to end n!n! is n5+n25\left\lfloor \dfrac n5 \right\rfloor + \left\lfloor \dfrac n{25} \right\rfloor and the number of zeros to end 2n!2n! is 2n5+2n25.\left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor .

This means 3(n5+n25) 3\left(\left\lfloor \dfrac n5 \right\rfloor + \left\lfloor \dfrac n{25} \right\rfloor\right) =2n5+2n25.= \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor . Then, we can inspect each value of n5+n25\left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor as follows:

Case 1: If n5+n25=0, \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =0, then n5=0\left\lfloor \frac n5 \right\rfloor =0 implying that 0n<40 \leq n < 4 which has no valid solutions.

Case 2: If n5+n25=1, \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =1, then n5=1\left\lfloor \frac n5 \right\rfloor =1 so 5n<10.5 \leq n < 10 .

Then, we also know 2n5+2n25=3. \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor =3. The only solutions are when 2n5=3,\left\lfloor \frac {2n}5 \right\rfloor = 3, so 152n<20,15 \leq 2n < 20, making 7.5n<10.7.5 \leq n < 10.

Thus, we have the solutions n=8,9.n = 8,9.

Case 3: If n5+n25=2, \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =2, then n5=2\left\lfloor \frac n5 \right\rfloor =2 so 10n<15.10 \leq n < 15 .

Then, we also know 2n5+2n25=6. \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor =6. The only solutions are when 2n5=5,\left\lfloor \frac {2n}5 \right\rfloor = 5, so 252n<30,25 \leq 2n < 30, making 12.5n<15.12.5 \leq n < 15.

Thus, we have the solutions n=13,14.n = 13,14.

Since n5+n25,\left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor , is non-decreasing, we found the four smallest solutions.

As such, their sum is 8+9+13+14=44,8+9+13+14=44, making the sum of the digits 8.8.

Thus, the correct answer is B.

24.

Aaron the ant walks on the coordinate plane according to the following rules.

He starts at the origin p0=(0,0)p_0=(0,0) facing to the east and walks one unit, arriving at p1=(1,0).p_1=(1,0).

For n=1,2,3,,n=1,2,3,\dots, right after arriving at the point pn,p_n, if Aaron can turn 9090^\circ left and walk one unit to an unvisited point pn+1,p_{n+1}, he does that. Otherwise, he walks one unit straight ahead to reach pn+1.p_{n+1}. Thus the sequence of points continues p2=(1,1),p_2=(1,1), p3=(0,1),p_3=(0,1), p4=(1,1),p_4=(-1,1),p5=(1,0), p_5=(-1,0), \vdots and so on in a counterclockwise spiral pattern. What is p2015?p_{2015}?

(22,13) (-22,-13)

(13,22) (-13,-22)

(13,22) (-13,22)

(13,22) (13,-22)

(22,13) (22,-13)

Answer: D
Solution(s):

When walking around, Aaron always walks in a counter-clockwise spiral centered at (0,0).(0,0).

Thus, he always makes a square length ss in as few points as possible, which would be (s+1)2.(s+1)^2 . Thus, after 452=202545^2 = 2025 points, he would make a square of side length 44.44. This would have corners at the points (±22,±22).(\pm 22,\pm 22).

Since it is a counter-clockwise spiral, the square the 20252025th point is at (22,22).(22,-22). Note that p0p_0 is one of the points, so p2024=(22,22).p_{2024} = (22,-22).

Then, since we have to go back 99 points and it is a counter-clockwise spiral, we subtract 99 from the xx-value, making p2015=(13,22).p_{2015} = (13,-22).

Thus, the correct answer is D.

25.

A rectangular box measures a×b×c,a \times b \times c, where a,a, b,b, and cc are integers and 1abc.1\leq a \leq b \leq c. The volume and the surface area of the box are numerically equal. How many ordered triples (a,b,c)(a,b,c) are possible?

4 4

10 10

12 12

21 21

26 26

Answer: B
Solution(s):

Our problem statment is equivalent to the number of solutions for abc=2(ab+bc+ac).abc = 2(ab+bc+ac).

Then, abc=2(ab+bc+ac)6bc,\small abc = 2(ab+bc+ac) \leq 6bc, so a6.a \leq 6.

Also, abc=2(ab+bc+ac)>2bc,\small abc = 2(ab+bc+ac) > 2bc, so a>2,a > 2, so 2<a6.2 < a \leq 6.

We case on aa as follows:

If a=3,a = 3, we have 3bc=2(3b+3c+bc),3bc = 2(3b+3c+bc), so bc6b6c=0.bc-6b-6c = 0. Thus, (b6)(c6)=36,(b-6)(c-6)=36, which has 55 solutions since 3636 has 99 factors, making 55 of them less than or equal to 66 and thus possible values for b.b.

If a=4,a = 4, we have 4bc=2(4b+4c+bc),4bc = 2(4b+4c+bc), so bc4b4c=0.bc-4b-4c = 0. Thus, (b4)(c4)=16,(b-4)(c-4)=16, which has 33 solutions since 1616 has 55 factors, making 33 of them less than or equal to 44 and thus possible values for b.b.

If a=5,a = 5, we have 5bc=2(5b+5c+bc),5bc = 2(5b+5c+bc), so 3bc10b10c=0.3bc-10b-10c = 0. Thus, (3b10)(3c10)=100,(3b-10)(3c-10)=100, which has 11 solution since we can only get (a,b,c)=(5,5,10)(a,b,c) = (5,5,10) since ba.b \geq a.

If a=6,a = 6, we have 6bc=2(6b+6c+bc),6bc = 2(6b+6c+bc), so bc3b3c=0.bc-3b-3c = 0. Thus, (b3)(c3)=9,(b-3)(c-3)=9, which has 11 solution since we can only get (a,b,c)=(6,6,6)(a,b,c) = (6,6,6) since ba.b \geq a. Thus, we have 1010 solutions.

Thus, the correct answer is B.