### 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 $$2-(-2)^{-2}?$$

$$-2$$

$$\dfrac{1}{16}$$

$$\dfrac{7}{4}$$

$$\dfrac{9}{4}$$

$$6$$

###### Solution(s):

\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 $$1\!:\!00$$ PM and finishes the second task at $$2\!:\!40$$ PM. When does she finish the third task?

3:10 PM

3:30 PM

4:00 PM

4:10 PM

4:30 PM

###### Solution(s):

The time it takes to do $$2$$ tasks is $$100$$ minutes. Thus, it takes $$50$$ more minutes after $$2:40,$$ which is $$3:30.$$

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 $$100,$$ and one of the numbers is $$28.$$ What is the other number?

$$8$$

$$11$$

$$14$$

$$15$$

$$18$$

###### Solution(s):

Kaashish either wrote $$28$$ three times or two times.

If he wrote it twice, then the sum of the other three numbers is: $100-28\cdot 2 = 44,$ which isn't a multiple of $$3.$$ Therefore, this case wouldn't work.

If he wrote it three times, then the sum of the other two numbers is $100-28\cdot 3 = 16.$ Therefore, the other number is $$\frac {16}2 = 8.$$

Thus, the correct answer is A.

4.

Four siblings ordered an extra large pizza. Alex ate $$\frac15,$$ Beth $$\frac13,$$ and Cyril $$\frac14$$ 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

###### Solution(s):

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

The amount Dan ate is $1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}.$ This is greater than $$\frac 15$$ and less than $$\frac 14 ,$$ 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 $$12$$-person race with $$6$$ other people. Rand finished $$6$$ places ahead of Hikmet. Marta finished $$1$$ place behind Jack. David finished $$2$$ places behind Hikmet. Jack finished $$2$$ places behind Todd. Todd finished $$1$$ place behind Rand. Marta finished in $$6$$th place. Who finished in $$8$$th place?

David

Hikmet

Jack

Rand

Todd

###### 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

###### Solution(s):

There are $$4$$ 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 $$a\diamond b=a-\frac{1}{b}.$$ What is $((1\diamond2)\diamond3)-(1\diamond(2\diamond3))?$

$$-\dfrac{7}{30}$$

$$-\dfrac{1}{6}$$

$$0$$

$$\dfrac{1}{6}$$

$$\dfrac{7}{30}$$

###### Solution(s):

\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 $$90^\circ$$ clockwise around the origin, then reflected in the $$y$$-axis, and then rotated a half turn around the origin. What is the final image?

###### Solution(s):

The rotation puts the F under the $$x$$-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 $$y$$-axis and reflects it upon the $$x$$-axis. This means the last two turns just reflects it upon the $$x$$-axis.

The reflection puts the F above the $$x$$-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 $$3$$ and center $$(0,0)$$ that lies in the first quadrant, the portion of the circle with radius $$\tfrac{3}{2}$$ and center $$(0,\tfrac{3}{2})$$ that lies in the first quadrant, and the line segment from $$(0,0)$$ to $$(3,0).$$ What is the area of the shark's fin falcata?

$$\dfrac{4\pi}{5}$$

$$\dfrac{9\pi}{8}$$

$$\dfrac{4\pi}{3}$$

$$\dfrac{7\pi}{5}$$

$$\dfrac{3\pi}{2}$$

###### Solution(s):

The area of the portion of the circle of radius $$3$$ in the first quadrant is equal to: $\dfrac14 \cdot \pi \cdot 3^2 = \dfrac94 \pi$

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

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

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 $$-2015?$$

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.

###### Solution(s):

There are $$1007$$ odd numbers greater than $$-2015.$$

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

Also, we multiply by $$-5$$ in there, so the product is a multiple of $$5,$$ making it end in $$5$$ or $$0.$$ None of our factors are even, so the product can't be even.

Therefore, the product must end in $$5.$$

Thus, the correct answer is C.

11.

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

$$\dfrac{8}{99}$$

$$\dfrac{2}{5}$$

$$\dfrac{9}{20}$$

$$\dfrac{1}{2}$$

$$\dfrac{9}{16}$$

###### Solution(s):

The only digits that are prime are $$2,3,5,$$ and $$7.$$

Our number can either be one or two digits. There are $$4$$ one digit numbers that have its digits being prime, and $$16$$ two digit numbers that have its digits being prime. This makes a total of $$20.$$

Each of the one one digit numbers are prime. For the two digit numbers, if it ends in $$2$$ or $$5,$$ then it isn't prime since it would be a multiple of $$2$$ or $$5.$$

Thus, we only need to check the numbers that end in $$3$$ or $$7.$$

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

$$33$$ and $$77$$ are multiples of $$11,$$ so they aren't prime.

Then, $$27$$ and $$57$$ are multiples of $$3$$ so they aren't prime.

This leaves $$23,27,53,$$ and $$73$$ as the only two digit primes.

Therefore, there are $$8$$ primes out of $$20,$$ making the probability $$\dfrac 25.$$

Thus, the correct answer is B.

12.

For how many integers $$x$$ is the point $$(x, -x)$$ inside or on the circle of radius $$10$$ centered at $$(5, 5)?$$

$$11$$

$$12$$

$$13$$

$$14$$

$$15$$

###### Solution(s):

The distance between $$(5,5)$$ and $$(x,-x)$$ is $\sqrt{(x-5)^2+(x+5)^2}$$= \sqrt{2x^2+50}.$ We must find how many $$x$$ are such that that value is less than or equal to $$10.$$ Therefoe, $2x^2 + 50 \leq 100$ $x^2 \leq 25.$ $-5 \leq x \leq 5.$ Therefore, we have $$11$$ included tintegers.

Thus, the correct answer is A.

13.

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

$$20$$

$$\dfrac{360}{17}$$

$$\dfrac{107}{5}$$

$$\dfrac{43}{2}$$

$$\dfrac{281}{13}$$

###### Solution(s):

The triangle is a right triangle with legs of $$12$$ and $$5.$$ This makes the hypotenuse $$13.$$

Two of the altitudes are then $$12$$ and $$5.$$ Also, for any side, $$A = \frac{bh}2$$ where $$b$$ is the base and $$h$$ is the altitude.

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

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

Thus, the correct answer is E.

14.

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

$$15$$

$$15.5$$

$$16$$

$$16.5$$

$$17$$

###### Solution(s):

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

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

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

Thus, the correct answer is D.

15.

The town of Hamlet has $$3$$ people for each horse, $$4$$ sheep for each cow, and $$3$$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

$$41$$

$$47$$

$$59$$

$$61$$

$$66$$

###### Solution(s):

Let the number of horses be $$h$$ and let the number of cows be $$c.$$

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

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

If we take the total and subtract $$5c,$$ then we get a multiple of $$13.$$ Thus, any valid number is such that there is a multiple of $$13$$ that is lower than the number that has the same remainder when divided by $$5.$$

For $$41,$$ the lowest multiple of $$13$$ that has the same remainder when divided by $$5$$ is $$26,$$ so this is a valid solution.

For $$47,$$ the lowest multiple of $$13$$ that has the same remainder when divided by $$5$$ is $$52,$$ so this isn't valid solution.

For $$59,$$ the lowest multiple of $$13$$ that has the same remainder when divided by $$5$$ is $$39,$$ so this is a valid solution.

For $$61,$$ the lowest multiple of $$13$$ that has the same remainder when divided by $$5$$ is $$26,$$ so this is a valid solution.

For $$66,$$ the lowest multiple of $$13$$ that has the same remainder when divided by $$5$$ is $$26,$$ so this is a valid solution.

Therefore, $$47$$ 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 $$1$$ to $$10,$$ 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?

$$\dfrac{9}{1000}$$

$$\dfrac{1}{90}$$

$$\dfrac{1}{80}$$

$$\dfrac{1}{72}$$

$$\dfrac{2}{121}$$

###### Solution(s):

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

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

Now, casing:

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

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

If $$b=2,$$ then $$c$$ can be one of $$4,6,8,10$$ making $$4$$ cases.

If $$b=3,$$ then $$c$$ can be one of $$6,9$$ making $$2$$ cases.

If $$b=4,$$ then $$c$$ can is $$8$$ making $$1$$ case.

If $$b=5,$$ then $$c$$ can is $$10$$ making $$1$$ case.

Thus, $$a=1$$ makes $$8$$ cases.

There are then $$9$$ total values of $$a,b,c$$ that work. The total number of pairs that are possible is $$10\cdot 9\cdot 8 = 720.$$

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

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?

$$\dfrac{75}{12}$$

$$10$$

$$12$$

$$10\sqrt2$$

$$15$$

###### 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 $$\frac 13 Ah,$$ so the combined area is: $\dfrac{A(h_1+h_2)}{3}.$

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

Therefore, the ratio of the the volume of the prism to the octahedron is $$\frac 16.$$ The volume of the prism is $$3\cdot 4\cdot 5=60,$$ so the volume of the octohedron is $$10.$$

Thus, the correct answer is B.

18.

Johann has $$64$$ 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?

$$32$$

$$40$$

$$48$$

$$56$$

$$64$$

###### Solution(s):

A coin ends as tails if and only if it has $$3$$ flips that are tails, which happens with probability $$\frac 18.$$ Thus, the probability of any coin being heads is $$\frac 78 .$$

As the probability that a given coin flip is $$\frac 78,$$ and there are $$64$$ coin flips in total, the expected number of coins that are now heads is: $64 \cdot \dfrac 78 = 56.$

Thus, the correct answer is D.

19.

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

$$12+9\sqrt{3}$$

$$18+6\sqrt{3}$$

$$12+12\sqrt{2}$$

$$30$$

$$32$$

###### Solution(s):

Consider the following diagram:

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

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

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

Then, let $$M$$ be the midpoint of $$AB.$$ This would make $YM^2 = MA^2+AY^2$$= 6^2 + 12^2$$= 180.$

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

Then, the distance from $$M$$ to $$W$$ is also equal to: $\small \sqrt{\left(WC+ \dfrac{BC}2\right)^2 + \left(\dfrac{AC}2\right)^2}$ Which simplifies to equal $\sqrt{ AC^2 + AC\cdot BC + 36}.$

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

As such, the perimeter is $2\cdot 6 \sqrt 2 + 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?

$$6$$

$$9$$

$$12$$

$$18$$

$$24$$

###### Solution(s):

Suppose we have the unit cube where the points are ordered triples with coordinates that are $$0$$ or $$1.$$

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

Then, there are $$3$$ choices for the first move and $$2$$ 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 $$(1,1,1)$$ until the last move, so there are just $$6$$ 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 $$5$$ steps left).

Suppose Dash takes $$19$$ fewer jumps than Cozy to reach the top of the staircase. Let $$s$$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $$s?$$

$$9$$

$$11$$

$$12$$

$$13$$

$$15$$

###### Solution(s):

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

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

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

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

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

Only $$63$$ satisfies $\dfrac s2 + \dfrac 12 - \left\lceil \dfrac s5 \right\rceil = 19 ,$ so it is the only valid answer.

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

Thus, the correct answer is D.

22.

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

$$3$$

$$12-4\sqrt5$$

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

$$1+\sqrt5$$

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

###### Solution(s):

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

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

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

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

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

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

Thus, the correct answer is D.

23.

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

$$7$$

$$8$$

$$9$$

$$10$$

$$11$$

###### Solution(s):

The number of zeros to end a number is equal to the power of $$5$$ in the $$n!.$$

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

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

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

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

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

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

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

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

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

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

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

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 $$p_0=(0,0)$$ facing to the east and walks one unit, arriving at $$p_1=(1,0).$$

For $$n=1,2,3,\dots,$$ right after arriving at the point $$p_n,$$ if Aaron can turn $$90^\circ$$ left and walk one unit to an unvisited point $$p_{n+1},$$ he does that. Otherwise, he walks one unit straight ahead to reach $$p_{n+1}.$$ Thus the sequence of points continues $p_2=(1,1),$$p_3=(0,1),$ $p_4=(-1,1),$$p_5=(-1,0),$$\vdots$ and so on in a counterclockwise spiral pattern. What is $$p_{2015}?$$

$$(-22,-13)$$

$$(-13,-22)$$

$$(-13,22)$$

$$(13,-22)$$

$$(22,-13)$$

###### Solution(s):

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

Thus, he always makes a square length $$s$$ in as few points as possible, which would be $$(s+1)^2 .$$ Thus, after $$45^2 = 2025$$ points, he would make a square of side length $$44.$$ This would have corners at the points $(\pm 22,\pm 22).$

Since it is a counter-clockwise spiral, the square the $$2025$$th point is at $$(22,-22).$$ Note that $$p_0$$ is one of the points, so $p_{2024} = (22,-22).$

Then, since we have to go back $$9$$ points and it is a counter-clockwise spiral, we subtract $$9$$ from the $$x$$-value, making $p_{2015} = (13,-22).$

Thus, the correct answer is D.

25.

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

$$4$$

$$10$$

$$12$$

$$21$$

$$26$$

###### Solution(s):

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

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

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

We case on $$a$$ as follows:

If $$a = 3,$$ we have $3bc = 2(3b+3c+bc),$ so $bc-6b-6c = 0.$ Thus, $(b-6)(c-6)=36,$ which has $$5$$ solutions since $$36$$ has $$9$$ factors, making $$5$$ of them less than or equal to $$6$$ and thus possible values for $$b.$$

If $$a = 4,$$ we have $4bc = 2(4b+4c+bc),$ so $bc-4b-4c = 0.$ Thus, $(b-4)(c-4)=16,$ which has $$3$$ solutions since $$16$$ has $$5$$ factors, making $$3$$ of them less than or equal to $$4$$ and thus possible values for $$b.$$

If $$a = 5,$$ we have $5bc = 2(5b+5c+bc),$ so $3bc-10b-10c = 0.$ Thus, $(3b-10)(3c-10)=100,$ which has $$1$$ solution since we can only get $$(a,b,c) = (5,5,10)$$ since $$b \geq a.$$

If $$a = 6,$$ we have $6bc = 2(6b+6c+bc),$ so $bc-3b-3c = 0.$ Thus, $(b-3)(c-3)=9,$ which has $$1$$ solution since we can only get $$(a,b,c) = (6,6,6)$$ since $$b \geq a.$$ Thus, we have $$10$$ solutions.

Thus, the correct answer is B.