2017 AMC 10 A

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

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as, .Thus: .

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 ruleis thus.

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

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

＄ 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 .

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 .

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.

Though Mia's mom places toys every seconds, Mia takes out toys right after. Therefore, after seconds, the two have collectively placed toy into the box. Therefore by minutes, the two would have placed toys into the box. Therefore, at minutes, the two would have placed toys into the box. Though Mia may take toys out right after, the number of toys in the box first reaches by minutes.

Let the two real numbers be , . We are given that , and dividing both sides by , .

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Note: we can easily verify that this is the correct answer; for example, and work, and the sum of their reciprocals is .

Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction. See for yourself. And by looking into fractions, we immediately see that and would fit the rule.

Notice that from the information given above, .

Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or .

We can solve this by substituting .

Our answer is simplya .

Theretve, he answeris .

Let the two numbers be and , respectively. We wish to find . Note that . We are given that .

Subsituting, we have

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

Note that Answer choice is the contrapositive of the given statement. Weknow that

contrapositive is always true if the given statement is true.

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 , , triangle with leg length . Thus, . We can then

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 .

We can also use complementary counting. First of all, handshakes or hugs occur. Then, if we can find the number of hugs, then we can subtract it from to find the handshakes. Hugs only happen between the people who know each other, so there are hugs. .

We can focus on how many handshakes the people get.

The person gets handshakes.

gets .

And the receives handshakes.

We can write this as the sum of an arithmetic sequence.

. Therefore, the answer is .

The distance from town to town is uphill, and since Minnie rides uphill at a speed of , it will take her hours. Next, she will ride from town to town , a distance of all downhill. Since Minnie rides downhill at a speed of , it will take her half an hour. Finally, she rides from town back to town , a flat distance of . Minnie rides on a flat road at , so this will take her hour. Her entire trip takes her hours. Secondly, Penny will go from town to town , a flat distance of . Since Penny rides on a flat road at , 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 uphill, and town and are apart, it will take her hours. Finally, Penny goes from Town back to town , a distance of downhill. Since Penny rides downhill at , 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.

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 lengths already used that are between those numbers, which gives .

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 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 .

Because this is just a cylinder and "half spheres", and the radius is , the volume of the half spheres is . Since we also know that the volume of this whole thing is , we do to get as the area of the cylinder. Thus the height is over the base, or , .

If the two equal values are and , then . Also, because 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 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 2 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

A patten starts to emerge as the function is continued. The repeating patten 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 .

Let cost of movie ticket,

Let cost of soda,

We can create two equations:

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Substituting we get:

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which yields:

A.

Now we can find s and we get:

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

add and to get:

%.

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 probabilty is .

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 .

Scale down by to get that Chloe picks from and Laurent picks from . There are an infinite number of cases for the number that Chloe picks, but they are all centered around the average of . Therefore, Laurent has a range of to to pick from, on average, which is a length of out of a total length of . Therefore, the probability is .

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 . To minimize the , we need the smallest primes, and we need to repeat them a lot. By inspection, we find that . Finally, .

We are trying to find the smallest number that has onedigit divisors. Therefore we try to find the for smaller digits, such as , , , or . We quickly consider since it is the smallest number that is the of , , and . Since has singledigit divisors, namely , , , , and , our answer is .

Because , , , and are lattice points, 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 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. Another possible pair could be and . To find out which segment is longer, we have to compare the distances from their endpoints to a diameter (which must be the longest possible segment). The closest diameter would be from to . The distance between and is shorter than the distance between and Therefore, the segment from to is the longest attainable. 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 . .

Let be the probability Amelia wins. Note that chance she wins on her first turn chance she gets to her turn again, 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 .

Let be the probability Amelia wins. Note that chance she wins on her frst turn chance she gets to her second turn chance she gets to her third turn . This can be represented by an infinite geometric series. . Therefre , so the answer is .

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

Case : sits on an edge seat.

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

Case : does not sit in an edge seat.

In this case, then only two people that can sit next to are and , and there are two ways to permute them, and this also handles the restriction that can't sit next to . Then, there are two ways to arrange and , the remaining people. However, there are three initial seats that can sit in, so there are seatings in this case. Adding up all the cases, we have .

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 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 ouranswer is .

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