Problem 1

What is the value of ?

Solution 1

Notice this is the term  in a recursive sequence, defined recursively as  Thus:

Solution 2

Starting to compute the inner expressions, we see the results are . This is always  less than a power of . The only admissible answer choice by this rule is thus .

Solution 3

Working our way from the innermost parenthesis outwards and directly computing, we have .

Solution 4

If you distribute this you get a sum of the powers of . The largest power of  in the series is , so the sum is

Problem 2

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

Solution

 boxes give us the most popsicles/dollar, so we want to buy as many of those as possible. After buying , we have  left. We cannot buy a third  box, so we opt for the  box instead (since it has a higher popsicles/dollar ratio than the  pack). We’re now out of money. We bought  popsicles, so the answer is .

Problem 3

Tamara has three rows of two 6-feet by 2-feet flower beds in her garden. The beds are separated and also surrounded by 1-foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?

Solution

Finding the area of the shaded walkway can be achieved by computing the total area of Tamara’s garden and then subtracting the combined area of her six flower beds.

Since the width of Tamara’s garden contains three margins, the total width is  feet.

Similarly, the height of Tamara’s garden is  feet.

Therefore, the total area of the garden is  square feet.

Finally, since the six flower beds each have an area of  square feet, the area we seek is , and our answer is

Problem 4

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

Solution

Every  seconds  toys are put in the box, so after  seconds there will be  toys in the box. Mia’s mom will then put  toys into to the box and we have our total amount of time to be  seconds, which equals  minutes.

Problem 5

The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?

Solution

Let the two real numbers be . We are given that

and dividing both sides by ,

Problem 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?

Solution

Rewriting the given statement: “if someone got all the multiple choice questions right on the upcoming exam then he or she would receive an A on the exam.” If that someone is Lewis the statement becomes: “if Lewis got all the multiple choice questions right, then he got an A on the exam.” The contrapositive: “If Lewis did not receive an A, then he got at least one of the multiple choice questions wrong (did not get all of them right)” must also be true leaving B as the correct answer. B is also equivalent to the contrapositive of the original statement, which implies that it must be true, so the answer is .

Problem 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?

Solution

Let  represent how far Jerry walked, and  represent how far Sylvia walked. Since the field is a square, and Jerry walked two sides of it, while Silvia walked the diagonal, we can simply define the side of the square field to be one, and find the distances they walked. Since Jerry walked two sides, Since Silvia walked the diagonal, she walked the hypotenuse of a 45, 45, 90 triangle with leg length 1. Thus,  We can then take

Problem 8

At a gathering of 30 people, there are 20 people who all know each other and 10 people who know no one. People who know each other a hug, and people who do not know each other shake hands. How many handshakes occur?

Solution 1

Each one of the ten people has to shake hands with all the  other people they don’t know. So . From there, we calculate how many handshakes occurred between the people who don’t know each other. This is simply counting how many ways to choose two people to shake hands, or . Thus the answer is .

Problem 9

Minnie rides on a flat road at  kilometers per hour (kph), downhill at  kph, and uphill at  kph. Penny rides on a flat road at  kph, downhill at  kph, and uphill at  kph. Minnie goes from town  to town , a distance of  km all uphill, then from town  to town , a distance of  km all downhill, and then back to town , a distance of  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 -km ride than it takes Penny?

Solution

The distance from town  to town  is  km uphill, and since Minnie rides uphill at a speed of  kph, it will take her  hours. Next, she will ride from town  to town , a distance of  km all downhill. Since Minnie rides downhill at a speed of  kph, it will take her half an hour. Finally, she rides from town  back to town , a flat distance of  km. Minnie rides on a flat road at  kph, so this will take her  hour. Her entire trip takes her  hours. Secondly, Penny will go from town  to town , a flat distance of  km. Since penny rides on a flat road at  kph, it will take her  of an hour. Next Penny will go from town  to town , which is uphill for Penny. Since penny rides at a speed of  kph uphill, and town  and  are  km apart, it will take her  hours. Finally, Penny goes from Town  back to town , a distance of  km downhill. Since Penny rides downhill at  kph, it will only take her  of an hour. In total, it takes her  hours, which simplifies to  hours and  minutes. Finally, Penny’s  Hour  Minute trip was  minutes less than Minnie’s  Hour  Minute Trip

Problem 10

Joy has  thin rods, one each of every integer length from  cm through  cm. She places the rods with lengths  cm,  cm, and  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?

Solution

The triangle inequality generalizes to all polygons, so  and  to get . Now, we know that there are  numbers between  and  exclusive, but we must subtract  to account for the 2 lengths already used, which gives

Problem 11

The region consisting of all point in three-dimensional space within 3 units of line segment  has volume 216. What is the length ?

Solution 1

In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within  units of a point would be a sphere with radius . However, we need to find the region containing all points within 3 units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal ):

, where  is equal to the length of our line segment.

Solving, we find that .

Problem 12

Let  be a set of points  in the coordinate plane such that two of the three quantities  and  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

Solution

If the two equal values are  and , then . Also,  because 3 is the common value. Solving for , we get . Therefore the portion of the line  where  is part of . This is a ray with an endpoint of .

Similar to the process above, we assume that the two equal values are  and . Solving the equation  then . Also,  because 3 is the common value. Solving for , we get . Therefore the portion of the line  where  is also part of . This is another ray with the same endpoint as the above ray: .

If  and  are the two equal values, then . Solving the equation for , we get . Also  because  is one way to express the common value. Solving for , we get . Therefore the portion of the line  where  is part of  like the other two rays. The lowest possible value that can be achieved is also .

Since  is made up of three rays with common endpoint , the answer is

Problem 13

Define a sequence recursively by  and  the remainder when  is divided by  for all  Thus the sequence starts  What is

Solution

A pattern starts to emerge as the function is continued. The repeating pattern is  The problem asks for the sum of eight consecutive terms in the sequence. Because there are eight numbers in the repeating sequence, we just need to find the sum of the numbers in the sequence, which is

Problem 14

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

Solution

Let  = cost of movie ticket

Let  = cost of soda

We can create two equations:

Substituting we get:

 

which yields:

Now we can find s and we get:

Since we want to find what fraction of  did Roger pay for his movie ticket and soda, we add  and  to get:

Problem 15

Chloé chooses a real number uniformly at random from the interval . Independently, Laurent cooses a real number uniformly at random from the interval . What is the probability that Laurent’s number is greater than Chloé’s number?

Solution 1

Denote “winning” to mean “picking a greater number”. There is a  chance that Laurent chooses a number in the interval . In this case, Chloé cannot possibly win, since the maximum number she can pick is . Otherwise, if Laurent picks a number in the interval , with probability , then the two people are symmetric, and each has a  chance of winning. Then, the total probability is

Solution 2

We can use geometric probability to solve this. Suppose a point  lies in the -plane. Let  be Chloe’s number and  be Laurent’s number. Then obviously we want , which basically gives us a region above a line. We know that Chloe’s number is in the interval  and Laurent’s number is in the interval , so we can create a rectangle in the plane, whose length is  and whose width is . Drawing it out, we see that it is easier to find the probability that Chloe’s number is greater than Laurent’s number and subtract this probability from . The probability that Chloe’s number is larger than Laurent’s number is simply the area of the region under the line , which is . Instead of bashing this out we know that the rectangle has area . So the probability that Laurent has a smaller number is . Simplifying the expression yields  and so .

Problem 16

There are  horses, named Horse , Horse , . . . , Horse . They get their names from how many minutes it takes them to run one lap around a circular race track: Horse  runs one lap in exactly  minutes. At time  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 , in minutes, at which all  horses will again simultaneously be at the starting point is . Let  be the least time, in minutes, such that at least 5 of the horses are again at the starting point. What is the sum of the digits of

Solution 1

If we have horses, , then any number that is a multiple of the all those numbers is a time when all horses will meet at the starting point. The least of these numbers is the LCM. To minimize the LCM, we need the smallest primes, and we need to repeat them a lot. By inspection, we find that . Finally, .

Solution 2

We are trying to find the smallest number that has  one-digit divisors. We quickly consider  since it is the smallest number that has at least  divisors. Since  has  single-digit divisors, namely , , , , and , our answer is

Problem 17

Distinct points , , ,  lie on the circle  and have integer coordinates. The distances  and  are irrational numbers. What is the greatest possible value of the ratio ?

Solution

Because , , , and  are integers there are only a few coordinates that actually satisfy the equation. The coordinates are  and  We want to maximize  and minimize  They also have to be the square root of something, because they are both irrational. The greatest value of  happens when it  and  are almost directly across from each other and are in different quadrants. For example, the endpoints of the segment could be  and  because the two points are almost across from each other. The least value of  is when the two endpoints are in the same quadrant and are very close to each other. This can occur when, for example,  is  and  is  They are in the same quadrant and no other point on the circle with integer coordinates is closer to the point  than  Using the distance formula, we get that  is  and that  is  

Problem 18

Amelia has a coin that lands heads with probability , and Blaine has a coin that lands on heads with probability . 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 , where  and  are relatively prime positive integers. What is ?

Solution

Let  be the probability Amelia wins. Note that , as if she gets to her turn again, she is back where she started with probability of winning . The chance she wins on her first turn is . The chance she makes it to her turn again is a combination of her failing to win the first turn –  and Blaine failing to win – . Multiplying gives us . Thus,Therefore, , so the answer is .

Solution 2

Let  be the probability Amelia wins. Note that This can be represented by an infinite geometric series,. Therefore, , so the answer is

Problem 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 5 chairs under these conditions?

Solution 1

For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E. We can split this problem up into two cases:

 A sits on an edge seat.

Then, since B and C can’t sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of .

 A does not sit in an edge seat.

In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can’t sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are  seatings in this case.

Adding up all the cases, we have .

Solution 2

Label the seats  through . The number of ways to seat Derek and Eric in the five seats with no restrictions is . The number of ways to seat Derek and Eric such that they sit next to each other is  (which can be figure out quickly), so the number of ways such that Derek and Eric don’t sit next to each other is . Note that once Derek and Eric are seated, there are three cases.

The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us  ways.

Another possible case is if Derek and Eric seat in seats  and  in some order. There are 2 possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are  ways to do this. So the second case gives us  total ways for the second case.

The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats is available. There are  ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot not sit in one of the two consecutive available seats without sitting next to Bob and Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us  ways.

So in total there are . So our answer is .

Problem 20

Let  equal the sum of the digits of positive integer . For example, . For a particular positive integer , . Which of the following could be the value of ?

Solution

Note that . This can be seen from the fact that . Thus, if , then , and thus . The only answer choice that is  is .

Problem 21

A square with side length  is inscribed in a right triangle with sides of length , , and  so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length  is inscribed in another right triangle with sides of length , , and  so that one side of the square lies on the hypotenuse of the triangle. What is ?

Solution

Analyze the first right triangle.

Note that  and  are similar, so . This can be written as . Solving, .

Now we analyze the second triangle.

Problem 22

Sides  and  of equilateral triangle  are tangent to a circle at points  and  respectively. What fraction of the area of  lies outside the circle?

Solution

Let the radius of the circle be , and let its center be . Since  and  are tangent to circle , then , so . Therefore, since  and  are equal to , then (pick your favorite method) . The area of the equilateral triangle is , and the area of the sector we are subtracting from it is . The area outside of the triangle is . Therefore, the answer is

Similarly,  and  are similar, so , and . Thus, . Solving for , we get . Thus, .

Problem 23

How many triangles with positive area have all their vertices at points  in the coordinate plane, where  and  are integers between  and , inclusive?

Solution

There are a total of  sets of three points. However, some of them form degenerate triangles (i.e., they have area of 0) if the three points are collinear. There are a total of 12 lines that go through 5 points (5 vertical, 5 horizontal, 2 diagonal), which contributes  degenerate triangles, 4 lines that go through exactly 4 points, which contributes  degenerate triangles, and 16 lines that go through exactly three points, which contributes  degenerate triangles. Subtracting these degenerate triangles, we get an answer of .

Problem 24

For certain real numbers , , and , the polynomialhas three distinct roots, and each root of  is also a root of the polynomialWhat is ?

Solution 1

 must have four roots, three of which are roots of . Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of  and  are the same, we know that

where  is the fourth root of . Substituting  and expanding, we find that

Comparing coefficients with , we see that

Let’s solve for  and . Since , , so . Since , , and . Thus, we know that

Taking , we find that

Solution 4 (Slight guessing)

Let the roots of  be , , and . Let the roots of  be , , , and . From Vieta’s, we have:The fourth root is . Since , , and  are coon roots, we have:Let :Note that  This gives us a pretty good guess of

Problem 25

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

Solution 1

Let the three-digit number be :

If a number is divisible by , then the difference between the sums of alternating digits is a multiple of .

There are two cases:  and

We now proceed to break down the cases.

: .

:  , this case results in 110, 220, 330…990. There are two ways to arrange the digits in each of those numbers.

:  , this case results in 121, 231,… 891. There are  ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to  cases.

: , this case results in 242, 352,… 792. There are  ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to  cases.

: , this case results in 363, 473,…693. There are  ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to  cases.

: , this case results in 484 and 594. There are  ways to arrange the digits in all of those number except the first, and 3 ways ways for the first. This leads to  cases.

This case has  subcases.

: .

: , this cases results in 209, 308, …506. There are  ways to arrange each of those cases. This leads to  cases.

: , this cases results in 319, 418, …616. There are  ways to arrange each of those cases, except the last. This leads to  cases.

: , this cases results in 429, 528, …617. There are  ways to arrange each of those cases. This leads to  cases.

… If we continue this counting, we receive  subcases.

~Mathguy1492

Solution 2

We note that we only have to consider multiples of 11 and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of 11 has:

 All three digits are the same. By inspection, we find that there are no multiples of 11 here.

 Two of the digits are the same, and the third is different.

 There are 8 multiples of 11 without a zero that have this property: 121, 242, 363, 484, 616, 737, 858, 979. Each contributes 3 valid permutations, so there are  permutations in this subcase.

 There are 9 multiples of 11 with a zero that have this property: 110, 220, 330, 440, 550, 660, 770, 880, 990. Each one contributes 2 valid permutations (the first digit can’t be zero), so there are  permutations in this subcase.

 All the digits are different. Since there are  multiples of 11 between 100 and 999, there are  multiples of 11 remaining in this case. However, 8 of them contain a zero, namely 209, 308, 407, 506, 605, 704, 803, and 902. Each of those multiples of 11 contributes  valid permutations, but we overcounted by a factor of 2; every permutation of 209, for example, is also a permutation of 902. Therefore, there are . Therefore, there are  remaining multiples of 11 without a 0 in this case. Each one contributes  valid permutations, but once again, we overcounted by a factor of 2 (note that if a number ABC is a multiple of 11, then so is CBA). Therefore, there are  valid permutations in this subcase.

Adding up all the permutations from all the cases, we have .

Solution 3 (Shorter and Not Casework)

We can overcount and then subtract. We know there are 81 multiples of 11.

We can multiply be 6 for each permutation of these multiples. (Yes some multiples don’t have 6)

Now divide by 2, as if a number abc with digits a, b and c is a multiple of 11, then cab is also a multiple of 11 so we have counted the same permutations twice.

So basically we say that each multiple of 11 has it own 3 permutations (say abc has abc acb and bac where’s cab has cab cba and bca). We know that each multiple of 11 has 3 permutations at least as it cannot have 3 repeating digits.

Hence we have 243 permutations without subtracting for over count. Now note that we overcounted cases is which we have 0’s at the start of each number. So in theory we could just answer A and move on.

We can also solve it :/ We overcount cases where the Middle digit of the number is 0 and the last digit is 0.

Note that we assigned each multiple of 11 3 permutations.

The last digit is 0 gives 9 possibilities where we overcount by 1 permutation for each of 110,220, … , 990.

The middle digit is 0 gives 8 possibilities where we overcount by 1. 605, 704, 803, 902 and 506, 407, 308, 209

Subtracting 17 gives .

Now, we may ask if there is further overlap (I.e if two of abc and bac and acb were multiples of 11) Thankfully, using divisibility rules, this can never happen as taking the divisibility rule mod 11 and adding we get that 2a,2b or 2c is congruent to 0 mod 11. Since a,b,c are digits, this can never happen as none of them can equal 11 and they can’t equal 0 as they are the leading digit of a 3 digit number in each of the cases

~fuzz1