2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 1 Easy

Kate bakes a 20-inch by 18-inch pan of cornbread. The cornbread is cut into pieces that measure 2 inches by 2 inches. How many pieces of cornbread does the pan contain? (2018 AMC 10B Problem, Question#1)

  • A.

    90

  • B.

    100

  • C.

    180

  • D.

    200

  • E.

    360

Answer:A

The area of the pan is 20\times 18=360. Since the area of each piece is 4, there are \frac{360}{4}=90 pieces.

Thus, the answer is \rm A.

By dividing each of the dimensions by 2, we get a 10\times 9 grid which makes 90 pieces. Thus, the answer is \rm A.

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Problem 2 Easy

Sam drove 96 miles in 90 minutes. His average speed during the first 30 minutes was 60 mph (miles per hour), and his average speed during the second 30 minutes was 65 mph. What was his average speed, in mph, during the last 30 minutes? (2018 AMC 10B Problem, Question#1)

  • A.

    64

  • B.

    65

  • C.

    66

  • D.

    67

  • E.

    68

Answer:D

Let Sam drive at exactly 60 mph in the first half hour, 65 mph in the second half hour, and x mph in the third half hour.

Due to rt=d, and that 30 min is half an hour, he covered  60 \cdot \frac{1}{2}=30 miles in the first 30 mins.

Similarly, he covered \frac{65}{2} miles in the 2\rm nd half hour period.

The problem states that Sam drove 96 miles in 90 min, so that means that he must have covered 96-\left(30+ \frac{65}{2}\right)=33 \frac{1}{2} miles in the third half hour period.

rt=d, so x \cdot \frac{1}{2}=33 \frac{1}{2}.

Therefore,Sam was driving (\rm D) \ 67 miles per hour in the third half hour.

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Problem 3 Easy

In the expression (            \times            )+(            \times            ) each blank is to be filled in with one of the digits 1, 2, 3, or 4, with each digit being used once. How many different values can be obtained? (2018 AMC 10B Problem, Question#3)

  • A.

    2

  • B.

    3

  • C.

    4

  • D.

    6

  • E.

    24

Answer:B

We have \left( \begin{array}{l} {4}\\{2} \end{array}\right) ways to choose the pairs, and we have 2 ways for the values to be switched so \frac{6}{2}= 3.

We have four available numbers (1,2,3,4).Because different permutations do not matter because they are all addtion and multiplication, if we put 1 on the first space, it is obvious there are possible outcomes (2,3,4).

There are 4! ways to arrange the numbers and 2!2!2! overcounts per way due to commutativity.

Therefore, the answer is \dfrac{4!}{2!2!2!}=3.

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Problem 4 Easy

A three-dimensional rectangular box with dimensions X, Y, and Z has faces whose surface areas are 24, 24, 48, 48, 72, and 72 square units. What is X+Y+Z? (2018 AMC 10B Problem, Question#4)

  • A.

    18

  • B.

    22

  • C.

    24

  • D.

    30

  • E.

    36

Answer:B

Let X be the length of the shortest dimension and Z be the length of the longest dimension.Thus，XY=24，YZ=72,~and XZ=48. Divide the first two equations to get \frac{Z}{X}=3.Then, mutiply by the last equation to get Z^2=144,~giving Z=12. Following，X=4 and Y=6. The final answer is 4+6+12=22.

Simply use guess and check to find that the dimensions are 4 by 6 by 12.Therefore, the answer is 4+6+12=22.

If you find the GCD of 24，48~and 72 you get your first number，12. After this, do 48\div12 and 72\div12 to get 4 and 6, the other 2 numbers.When you add up your 3 numbers, you get 22 which is \rm B.

Since the surface areas of the faces are the product of two of the dimensions．

Therefore，XY=24，XZ=48, and YZ=72.You can multiply XY\times XZ\times YZ, which simplifies to XYZ^2=24\times48\times72 which means that the volume XYZ equals \sqrt{24*48*72}= \sqrt{24^{2}*12^{2}}=24*12=288. The individual dimensions，X，Y, and Z can be found by doing  \frac{XYZ}{XY}, \frac{XYZ}{YZ}, and \frac{XYZ}{XZ}, which yields Z=12, Y=6, and X=4. Adding this up, we have that X+Y+Z=22 which is \rm B.

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Problem 5 Easy

How many subsets of \left\{2,3,4,5,6,7,8,9\right\} contain at least one prime number? (2018 AMC 10B Problem, Question#5)

  • A.

    128

  • B.

    192

  • C.

    224

  • D.

    240

  • E.

    256

Answer:D

Consider finding the number of subsets that do not contain any primes.There are four primes in the set：2, 3, 5, and 7.This means that the number of subsets without any primes is the number of subsets of \left\{4, 6, 8,9\right\},  which is just 2^4=16. The number of subsets with at least one prime is the number of subsets minus the number of subsets without any primes.The number of subsets is 2^8=256. Thus, the answer is 256-16=(\rm D)~240.

Well, there are 4 composite numbers, and you can list them in a 1 number format, a 2 number, 3 number, and a 4 number format. Now,  we can use combinations.  \left( \begin{array}{l} {4}\\{1} \end{array}\right)+\left( \begin{array}{l} {4}\\{2} \end{array}\right)+\left( \begin{array}{l} {4}\\{3} \end{array}\right)+\left( \begin{array}{l} {4}\\{4} \end{array}\right)=15. Using the answer choices, the only multiple of 15 is (\rm D)~240.

Subsets of \left\{2,3,4,5,6,7,8,9\right\} indude a singe digit up to all eight numbers.Therefore,we must add the combinations of all possible subsets and subtract from each of the subsets fomed by the composite numbers.

Hence：

\left( \begin{array}{l} {8}\\{1} \end{array}\right)-\left( \begin{array}{l} {4}\\{1} \end{array}\right)+\left( \begin{array}{l} {8}\\{2} \end{array}\right)-\left( \begin{array}{l} {4}\\{2} \end{array}\right)+\left( \begin{array}{l} {8}\\{3} \end{array}\right)-\left( \begin{array}{l} {4}\\{3} \end{array}\right)+\left( \begin{array}{l} {8}\\{4} \end{array}\right)-1+\left( \begin{array}{l} {8}\\{5} \end{array}\right)+\left( \begin{array}{l} {8}\\{6} \end{array}\right)+\left( \begin{array}{l} {8}\\{7} \end{array}\right)+1=(\rm D)~240.

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Problem 6 Easy

A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required? (2018 AMC 10B Problem, Question#6)

  • A.

    \dfrac{1}{15}

  • B.

    \frac{1}{10}

  • C.

    \frac{1}{6}

  • D.

    \frac{1}{5}

  • E.

    \frac{1}{4}

Answer:D

Notice that the only four ways such that no less than 2 draws are requred are 1, 2; 1, 3; 2, 1;  and 3, 1. Notice that each of those cases has a  \frac{1}{5}\cdot \frac{1}{4} chance, so the answer is \frac{1}{5}\cdot \frac{1}{4}\cdot 4= \frac{1}{5}, or \rm D.

Notice that only the first two draws are important, it doesn’t matter what number we get third because no matter what combination of 3 numbers is picked, the sum will always be greater than 5. Also, note that it is necessary to draw a 1 in order to have 3 draws, otherwise 5 will be attainable in two or less draws. So the probabilty of getting a 1 is \frac{1}{5}. It is necessary to pull either a 2 or 3 on the next draw and the probabilty of that is \frac{1}{2}.But,  the order of the draws can be switched so we get \frac{1}{5}\cdot \frac{1}{2}\cdot 2= \frac{1}{5}, or \rm D.

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Problem 7 Easy

In the figure below, N congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let A be the combined area of the small semicircles and B be the area of the region inside the large semicircle but outside the small semicircles. The ratio A:B is 1:18. What is N? (2018 AMC 10B Problem, Question#7)

  • A.

    16

  • B.

    17

  • C.

    18

  • D.

    19

  • E.

    36

Answer:D

Use the answer choices and calculate them. The one that works is (\rm D)~19.

Let the number of semicircles be n and let the radius of each semicircle to be r. To find the total area of all of the small semicircles, we have n\cdot\frac{{\pi}\cdot r^2}{2}.Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be n\cdot r. So, the area of the larger semicircle is \frac{\pi\cdot n^2\cdot r^2}{2} . Now that we have found the area of both A and B, we can find the ratio. \frac{A}{B}=\frac{1}{18}, so part-to-whole ratio is 1:19. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get \frac{1}{n}. This is equal to \frac{1}{19}, By setting them equal, we find that n=19. This is our answer, which corresponds to choice (\rm D)~19.

Each small semicircle is \frac{1}{N^2} of the large semicircle. Since N small semicircles make \frac{1}{19} of the large one, \frac{N}{N^2}=\frac{1}{19}. Solving this, we get 19.

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Problem 8 Easy

Sara makes a staircase out of toothpicks as shown:

This is a 3-step staircase and uses 18 toothpicks. How many steps would be in a staircase that used 180 toothpicks? (2018 AMC 10B Problem, Question#8)

  • A.

    10

  • B.

    11

  • C.

    12

  • D.

    24

  • E.

    30

Answer:C

A staircase with n steps contains 4+6+8+\cdots+2n+2 toothpicks. This can be rewritten as (n+1)(n+2)-2.

So, (n+1)(n+2)-2=180,

So, (n+1)(n+2)=182,

Inspection could tell us that 13*14=182 , so the answer is 13-1=\rm (C)~12.

Layer 1:4 steps,

Layer 1, 2:10  steps,

Layer 1, 2, 3:18 steps,

Layer 1, 2, 3, 4:28 steps,

From inspection, we can see that with each increase in layer the difference in toothpicks between the current layer and the previous increases by 2 . Using this pattern:

4, 10, 18, 28, 40, 54, 70, 88, 108, 130, 154, 180,

From this we see that the solution is \rm (C)~12.

We can find a function that gives us the number of toothpicks for every layer. Using finite difference, we know that the degree must be 2 and the leading coefficient is 1. The function is f(n)=n^2+3n where n is the layer and f(n) is the number of toothpicks. We have to solve for n when n^2+3n=180\Rightarrow n^2+3n-180=0. Factor to get (n-12)(n+15). The roots are 12 and -15. Clearly -15 is impossible so the answer is \rm (C)~12.

Notice that the number of toothpicks can be found by adding all the horizontal and all the vertical toothpicks. We can see that for the case of 3 steps, there are 2(3+3+2+1)=18 toothpicks.Thus, the equation is 2S+2(1+2+3\cdots+S)=180 with S being the number of steps. Solving, we get S=12, or \rm (C)~12.

If you are trying to look for a pattern, you can see that the first column is made of 4 toothpicks. The second one is made from 2 squares: 3 toothpicks for the first square and 4 for the second. The third one is made up of 3 squares: 3 toothpicks for the first and second one, and 4 for the third one. The pattern continues like that. So for the first one, you have 0 "3 toothpick squares" and 1 "4 toothpick squares". The second is 1 to 1. The third is 2:1. And the amount of three toothpick squares increase by one every column.The list is as follow for the number of toothpicks used\cdots 4, 4+3, 4+6, 4+9, and so on. 4, 7, 10, 13, 16, 19, \cdots

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Problem 9 Easy

The faces of each of 7 standard dice are labeled with the integers from 1 to 6. Let P be the probability that when all 7 dice are rolled, the sum of the numbers on the top faces is 10. What other sum occurs with the same probability P? (2018 AMC 10B Problem, Question#9)

  • A.

    13

  • B.

    26

  • C.

    32

  • D.

    39

  • E.

    42

Answer:D

It can be seen that the probability of rolling the smallest number possible is the same as the probability of rolling the largest number possible, the probability of rolling the second smallest number possible is the same as the probability of rolling the second largest number possible, and so on. This is because the number of ways to add a certain number of ones to an assortment of 7 ones is the same as the number of ways to take away a certain number of ones from an assortment of 7~6s. So, we can match up the values to find the sum with the same probability as 10. We can start by noticing that 7 is the smallest possible roll and 42 is the largest possible role. The pairs with the same probability are as follows: (7, 42), (8, 41), (9, 40), (10, 39), (11, 38)\cdots

However, we need to find the number that matches up with 10. So, we can stop at (10, 39) and deduce that the sum with equal probability as 10 is 39. So, the correct answer is \rm (D)~39, and we are done.

Let's call the unknown value x . By symmetry, we realize that the difference between 10 and the minimum value of the rolls is equal to the difference between the maximum and x. So, 10-7=42-x, x=39 and our answer is \rm (D)~39.

For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) \times 7= 49. Since having 10 is similar to not having 10, you just subtract 10 from the expected total sum. 49 -10 = 39 so the answer is  \rm (D)~39.

The expected value of the sums of the die rolls is 3.5*7=24.5, and since the probabilities should be distributed symmetrically on both sides of 24.5, the answer is 24.5+24.5-10=39, which is \rm (D)~39.

Calculating the probability of getting a sum of 10 is also easy. There are 3 cases:

Case 1: \left\{1,1,1,1,1,1,4\right\},

\frac{7!}{6!}=7 cases,

Case 2: \left\{1,1,1,1,1,2,3\right\},

\frac{7!}{5!}=6*7=42 cases,

Case 3: \left\{1,1,1,1,2,2,2\right\},

\frac{7!}{4!3!}=5*7=35 cases,

The probability is \frac{84}{6^7}=\frac{14}{6^6}.

Calculating 6^6:

6^6=(6^3)^2=216^2=46656,

Therefore, the probability is \frac{14}{46656}=\frac{7}{23328}.

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Problem 10 Easy

In the rectangular parallelepiped shown, AB=3, BC=1 , and CG=2. Point M is the midpoint of \overline{FG}. What is the volume of the rectangular pyramid with base BCHE and apex M? (2018 AMC 10B Problem, Question#10)

  • A.

    1

  • B.

    \frac{4}{3}

  • C.

    \frac{3}{2}

  • D.

    \frac{5}{3}

  • E.

    2

Answer:E

Consider the cross-sectional plane and label its area b. Note that the volume of the triangular prism that encloses the pyramid is \frac{bh}{2}=3, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is \frac{bh}{3}, so the answer is 2.

We can start by finding the total volume of the parallelepiped. It is 2\cdot3\cdot2=6, because a rectangular parallelepiped is a rectangular prism. Next, we can consider the wedge-shaped section made when the plane BCHE cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is \frac{1}{2}\cdot 2\cdot 3=3. Since BC is given to be 1, we have that FM is \frac{1}{2}. Using the formula for the volume of a triangular pyramid, we have V=\frac{1}{3}\cdot \frac{1}{2}\cdot 3=\frac{1}{2}. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume \frac{1}{2} as well.The original wedge we considered in the last step has volume 3, because it is half of the volume of theparallelepiped. We can subtract out the parts we found to have 3-\frac{1}{2}\cdot 2=2. Thus, the volume of the figure we are trying to find is 2. This means that the correct answer choice is \rm E.

For those who think that it isn't a rectangular prism, please read the problem. It says"rectangular parallelepiped." If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.

If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is \frac{1}{3}Bh, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is \frac{1}{3}\left(3\times 2/2\times \frac{1}{2}\right)=\frac{1}{2}. We can obtain the answer by subtracting twice this value from the diagonal half prism,or \left(\frac{1}{2}\times 3\times 2\times 1\right)-\left(2\times \frac{1}{2}\right)=2.

You can calculate the volume of the rectangular pyramid by using the formula,\frac{Ah}{3} . A is the area of the base,BCHE , and is equal to BC*BE. The height,h , is equal to the height of triangle FBE drawn from F to BE.

BE=\sqrt{BF^2+EF^2}=\sqrt{13} Area of BCHE=BC*BE=\sqrt{13},

h=\frac{2*\text{Area of}~FBE}{BE} (since Area=\frac{1}{2}bh).

Area of FBE=\frac{1}{2}*FB*FE=3,

h=\frac{2*3}{\sqrt{13}}=\frac{6}{\sqrt{13}},

Volume of pyramid =\frac{1}{3}*\sqrt{13}*\frac{6}{\sqrt{13}}=2, Answer is 2.

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Problem 11 Easy

Which of the following expressions is never a prime number when p is a prime number? (2018 AMC 10B Problem, Question#11)

  • A.

    p^2+16

  • B.

    p^2+24

  • C.

    p^2+26

  • D.

    p^2+46

  • E.

    p^2+96

Answer:C

Because squares of a non-multiple of 3 is always 1 mod 3, the only expression is always a multiple of 3 is \rm(C)~p^2+26. This is excluding when p=0 mod 3, which only occurs when p=3, then p^2+26=35 which is still composite.

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Problem 12 Easy

Line segment \overline{AB} is a diameter of a circle with AB=24. Point C, not equal to A or B , lies on the circle. As point C moves around the circle, the centroid (center of mass) of \triangle ABC traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve? (2018 AMC 10B Problem, Question#12)

  • A.

    25

  • B.

    38

  • C.

    50

  • D.

    63

  • E.

    75

Answer:C

Let A=(-12,0), B=(12,0). Therefore, C lies on the circle with equation x^2+y^2=144. Let it have coordinates (x, y). Since we know the centroid of a triangle with vertices with coordinates of (x_1,y_1), (x_2,y_2), (x_3,y_3) is \left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right), the centroid of \triangle ABC is \left(\frac{x}{3},\frac{y}{3}\right). Because x^2+y^2=144, we know that \left(\frac{x}{3}\right)^2+\left(\frac{y}{3}\right)^2=16, so the curve is a circle centered at the origin. Therefore, its area is 16\pi\approx\rm (C)~50.

We know the centroid of a triangle splits the medians into segments of ratio 2:1, and the median of the triangle that goes to the center of the circle is the radius (length 12), so the length from the centroid of the triangle to the center of the circle is always \frac{1}{3}\cdot 12=4 . The area of a circle with radius 4 is 16\pi, or around \rm(C)~50.

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Problem 13 Medium

How many of the first 2018 numbers in the sequence 101, 1001, 10001, 100001, \cdots are divisible by 101? (2018 AMC 10B Problem, Question#13)

  • A.

    253

  • B.

    504

  • C.

    505

  • D.

    506

  • E.

    1009

Answer:C

The number 10^n+1 is divisible by 101 if and only if 10^n\equiv -1 (mod 101). We note that (10, 10^2, 10^3, 10^4)\equiv(10,-1, -10, 1) (mod 101), so the powers of 10 are 4-periodic mod 101. It follows that 10^n\equiv -1 (mod 101) if and only if  n\equiv 2 (mod 4). In the given list, 10^2+1, 10^3+1, 10^4+1, \cdots, 10^{2019}+1, the desired exponents are 2, 6, 10, \cdots, 2018, and there are \frac{2020}{4}=\rm(C)~505 numbers in that list.

Note that 10^{2k}+1 for some odd k will suffice mod 101 . Each 2k\in\left\{2, 6,10,\cdots,2018\right\}, so the answer is \rm (C)~505.

If we divide each number by 101, we see a pattern occuring in every 4 numbers. 101, 1000001, 10000000001, \cdots. We divide 2018 by 4 to get 504 with 2 left over. Looking at our pattern of four numbers from above, the first number is divisible by 101. This means that the first of the 2 left over will be divisible by 101, so our answer is \rm(C)~505.

Note that 909 is divisible by 101, and thus 9999 is too. We know that 101 is divisible and 1001 isn't so let us start from 10001. We subtract 9999 to get 2. Likewise from 100001 we subtract, but we instead subtract 9999 times 10 or 99990 to get 11. We do it again and multiply the \rm 9's by 10 to get 101. Following the same knowledge, we can use mod 101 to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is 0, -9, -99(2), 11, 0, \cdots. Thus it repeats every four. Consider the sequence after the \rm 1st term and we have 2017 numbers. Divide 2017 by four to get 504 remainder 1. Thus the answer is 504 plus the \rm 1st term or \rm (C)~505.

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Problem 14 Medium

A list of 2018 positive integers has a unique mode, which occurs exactly 10 times. What is the least number of distinct values that can occur in the list? (2018 AMC 10B Problem, Question#14)

  • A.

    202

  • B.

    223

  • C.

    224

  • D.

    225

  • E.

    234

Answer:D

To minimize the number of distinct values, we want to maximize the number of times they appear. So, we could have 223 numbers appear 9 times, 1 number appear once, and the mode appear 10 times, giving us a total of 223+1+1 =\rm (D)~225.

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Problem 15 Medium

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point A in the figure on the right. The box has base length w and height h. What is the area of the sheet of wrapping paper? (2018 AMC 10B Problem, Question#15)

  • A.

    2(w+h)^2

  • B.

    \frac{(w+h)^2}{2}

  • C.

    2w^2+4wh

  • D.

    2w^2

  • E.

    w^2h

Answer:A

Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is h. The area of the rectangle that is w by h is wh . The combined figure of the two triangles with base h is a square with h as its diagonal. Using the Pythagorean Theorem, each side of this square is \sqrt\frac{h^2}{2}. Thus, the area is the side length squared which is \frac{h^2}{2}. Similarly, the combined figure of the two triangles with base w is a square with area \frac{w^2}{2}. Adding all of these together, we get \frac{h^2}{2}+\frac{w^2}{2}+wh . Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting 4\left(\frac{w^2}{2}+\frac{h^2}{2}+wh\right)=2(w^2+h^2+2wh)= (\text{A})~2(w+h)^2 .

The sheet of paper is made out of the surface area of the box plus the sum of the four triangles. The surface area is 2w^2+2wh+2wh which equals 2w^2+4wh .The four triangles each have a height and a base of h, so they each have an area of \frac{h^2}{2}. There are four of them, so multiplied by four is 2h^2.Together, paper's area is 2w^2+4wh+2h^2. This can be factored and written as (\text{A})~2(w+h)^2.

The sheet of paper is made out of 4 squares. Each square has a side length of \frac{w}{\sqrt{2}}+\frac{h}{\sqrt{2}}, which we get from the pythagorean theorem (a 45-45-90 triangle's legs is the hypotenuse divided by \sqrt{2}).Thus, to find the area of the entire paper, we square our side length and multiply by 4.

So, 4\cdot\left(\frac{w}{\sqrt{2}}+\frac{h}{\sqrt{2}}\right)^2\Rightarrow 2(w+h)^2 which is the answer.

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Problem 16 Medium

Let a_1, a_2, \cdots, a_{2018} be a strictly increasing sequence of positive integers such that a_1+a_2+\cdots+a_{2018}=2018^{2018},  What is the remainder when a_1^3+a_2^3+\cdots+a_{2018}^3 is divided by 6? (2018 AMC 10B Problem, Question#16)

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:E

One could simply list out all the residues to the third power \rm mod~6. (Edit: Euler's totient theorem is not a valid approach to showing that they are all congruent \rm mod~6. This is due to the fact that a_k need not be relatively prime to 6.)Therefore the answer is congruent to 2018^{2018}\equiv2^{2018} (\rm mod \ 6)=(E) \ 4.

Note that

(a_1+a_2+\cdots+a_{2018})^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(a_1+a_2+\cdots+a_{2018}-a_1)+3a_2^2(a_1+a_2+\cdots+a_{2018}-a_2)+\cdots+3a_{2018}^2(a_1+a_2+\cdots+a_{2018}-a_{2018})+6\sum _{i\neq j\neq k}^{2018}{a_ia_ja_k},

Note that

a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(a_1+a_2+\cdots+a_{2018}-a_1)+3a_2^2(a_1+a_2+\cdots+a_{2018}-a_2)+\cdots+3a_{2018}^2(a_1+a_2+\cdots+a_{2018}-a_{2018})+6\sum _{i\neq j\neq k}^{2018}{a_ia_ja_k}\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018^{2018}-a_1)+3a_2^2(2018^{2018}-a_2)+\cdots+3a_{2018}^2(2018^{2018}-a_{2018})\equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)(\rm mod ~6),

Therefore,-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv(2018^{2018})^3\equiv(2^{2018})^3\equiv4^3\equiv4(\rm mod ~6).

Thus, a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1(\rm mod ~3). However, since cubing preserves parity, and the sum of the individual terms is even, the some of the cubes is also even, and our answer is \rm (\rm E)~4.

We first note that 1^3+2^3+\cdots=(1+2+\cdots)^2. So what we are trying to find is what (2018^{2018})^2=(2018^{4036}) \ \rm mod \ 6. We start by noting that 2018 is congruent to 2 \ \rm mod \ 6. So we are trying to find (2^{4036}) \ \rm mod \ 6 . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of 2 and see that 2^1 is 2 \ \rm mod \ 6,2^2  is 4 \ \rm mod \ 6 , 2^3 is 2 \ \rm mod \ 6, 2^4 is 4 \ \rm mod \ 6 ,  and so on \cdots So we see that since (2^{4036}) has an even power, it must be congruent to 4 \ \rm mod \ 6 , thus giving our answer \rm (E) \ 4.

You can prove this pattern using mods. But I thought this was easier.

Assume a_1, a_2, \cdots a_{2017} are multiples of 6 and find 2018^{2018}(\rm mod \ 6) (which happens to be 4). Then a_1^3+\cdots+a_{2018}^3 is congruent to 64(\rm mod \ 6) or just 4 .

First note that each a_i^3\equiv a_i(\rm mod \ 3) by Fermat's Little Theorem. This implies that a_1^3+\cdots+a_{2018}^3\equiv a_1+\cdots+a_{2018}\equiv 2^{2018}\equiv 1(\rm mod \ 3). Also, all  a_i^2\equiv a_i(\rm mod \ 2), hence a_i^3\equiv(a_i)(a_i^2)\equiv a_i^2\equiv a_i(\rm mod \ 2) by Fermat's Little, Theorem.Thus, a_1^3+\cdots+a_{2018}^3\equiv 2^{2018}\equiv 0(\rm mod \ 2). Now set x=a_1^3+\cdots +a_{2018}^3 . Then, we have the congruences x\equiv 0(\rm mod \ 2) and x\equiv 1(\rm mod \ 3) . By the Chinese Remainder Theorem, a solution must exist, and indeed solving the congruence we get that x\equiv 4(\rm mod \ 6). Thus, the answer is \rm (E) \ 4.

Link Problem
Problem 17 Medium

In rectangle PQRS, PQ=8 and QR=6. Points A and B lie on \overline {PQ}, points C and D lie on \overline {QR}, points E and F lie on \overline {RS}, and points G and H lie on \overline {SP} so that AP=BQ<4 and the convex octagon ABCDEFGH is equilateral. The length of a side of this octagon can be expressed in the form k+m\sqrt n, where k, m, and n are integers and n is not divisible by the square of any prime. What is k+m+n? (2018 AMC 10B Problem, Question#17)

  • A.

    1

  • B.

    7

  • C.

    21

  • D.

    92

  • E.

    106

Answer:B

Let AP=BQ+x. Then AB=8-2x.

Now notice that since CD=8-2x we have QC=DR=x-1.

Thus by the Pythagorean Theorem we have x^2+(x-1)^2=(8-2x)^2 which becomes 2x^2-30x+63=0\Rightarrow x=\frac{15-3\sqrt{11}}{2}.

Our answer is 8-(15-3\sqrt{11})=3\sqrt{11}-7\Rightarrow \rm (B) \ 7.

Denote the length of the equilateral octagon as x. The length of \overline{BQ} can be expressed as \frac{8-x}{2}. By the Pythagorean Theorem, we find that: \left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\Rightarrow\overline{CQ}=\sqrt{x^2-\left(\dfrac{8-x}{2}\right)^2},

Since \overline{CQ}=\overline{DR} , we can say that x+2\sqrt{x^2-\left(\dfrac{8-x}{2}\right)^2}=6\Rightarrow x=-7\pm3\sqrt{11} . We can discard the negative solution, so k+m+n=-7+3+11=\rm (B) \ 7.

Let the octagon's side length be x. Then PH=\frac{6-x}{2} and PA=\frac{8-x}{2}. By the Pythagorean theorem, PH^2+PA^2=HA^2, so \left(\frac{6-x}{2}\right)^2+\left(\frac{8-x}{2}\right)^2=x^2. Solving this, we get one positive solution, x=-7\pm3\sqrt{11}, so k+m+n=-7+3+11=\rm (B) \ 7.

Link Problem
Problem 18 Medium

Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of hisor her sibling. How many seating arrangements are possible for this trip? (2018 AMC 10B Problem, Question#18)

  • A.

    60

  • B.

    72

  • C.

    92

  • D.

    96

  • E.

    120

Answer:D

We can begin to put this into cases. Let's call the pairs a, b and c, and assume that a member of pair a is sitting in the leftmost seat of the second row. We can have the following cases then.

Case 1: Second Row: a \ b \ c Third Row: b \ c \ a

Case 2: Second Row: a \ c \ b Third Row: c \ b \ a

Case 3: Second Row: a \ b \ c Third Row: c \ a \ b

Case 4: Second Row: a \ c \ b Third Row: b \ a \ c

For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has 2\cdot 2\cdot 2=8 possibilities. Since there are four cases, when pair a has someone in the leftmost seat of the second row, there are 32 ways to rearrange it. However, someone from either pair a, b, or c could be sitting in the leftmost seat of the second row. So, we have to multiply it by 3 to get our answer of 32\cdot 2=96. So, the correct answer is 96.

Lets call the siblings A_1, A_2, B_1, B_2, C_1, and C_2. We can split our problem into 2 cases:There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.

Starting off with the first case, we see that there are 3!=6 ways to arrange the A, B, C. Then, we have to choose which sibling sits. There are 2 choices for each set of siblings meaning we have 2^3=8 ways to arrange that. So, there are 48 ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are 2 ways to place the A sibling and each placement yields only 1 possibility. So, our first case has 48\cdot 2=96 possibilities.

In our second case, there are 3 ways to choose which set of siblings will be in the same row, two ways to choose which set of sibling will sit in between them and 2 ways to choose whether it is the brother orsister. So, there 3*2*2=18 ways to arrange the first row. In the second row, however, we see thatit is impossible to make everything work out. So, there are 0 possibilities for this case.Thus, there are 96+0=96 possibilities for this trip.

Call the siblings A_1, A_2, B_1, B_2, C_1, and C_2.

There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that A_1 takes it (a \circ is an empty seat):

\begin{array}{l} {A_{1}\circ\circ}\\{\circ\circ\circ} \end{array}

Then there are 4 choices for the second seat ( B_1, B_2, C_1, or C_2 ). Like before, it doesn't matter whotakes the seat, so WLOG suppose it is B_1:

The last seat in the first row cannot be A_2 because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be C_1 or C_2. Let's say WLOG that it is C_1. There are two ways to create a second row:

A_1B_1C_1
B_2C_2A_2
A_1B_1C_1
C_2A_2B_2

Therefore, there are 6\cdot4\cdot2\cdot2=96 possible seating arrangements.

Note that there is a free S_2, S_3 action on the set of allowed seating arrangements: any brother-sister pair can be switched, and the 3 pairs can be permuted among each other in any way. Hence the answer must be a multiple of the order of S_2, S_3, which is 2^3\cdot6=48. The only answer choice satisfying this is 96.

To finish the solution, with a bit of work we see that there are only two orbits of seating arrangements: the orbit of 123/231 and the orbit of 123/312.

So the answer is indeed 96.

Link Problem
Problem 19 Hard

Joey, Chloe and their daughter Zoe all have the same birthday. Joey is 1 year older than Chloe, and Zoe is exactly 1 year old today. Today is the first of the 9 birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age? (2018 AMC 10B Problem, Question#19)

  • A.

    7

  • B.

    8

  • C.

    9

  • D.

    10

  • E.

    11

Answer:E

Let Joey's age be j, Chloe's age be c, and we know that Zoe's age is 1. We know that there must be 9 values k\in\mathbf{Z} such that c+k=a(1+k) where a is an integer.Therefore, c-1+(1+k)=a(1+k) and c-1=(1+k)(a-1). Therefore, we know that, as there are 9 solutions for k, there must be 9 solutions for c-1. We know that this must be a perfect square. Testing perfect squares, we see that c-1=36, so c=37. Therefore, .j=38. Now, since j-1=37, by similar logic, 37=(1+k)(a-1), so k=36 and Joey will be 38+36=74 and the sum of the digits is 11.

Here's a different way of saying the above solution:

If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has 9 factors. Therefore, the difference between Chloe and Zoe's age is 36, so Chloe is 37, and Joey is 38. The common factor that will divide both of their ages is 37, so Joey will be 74. 7 + 4=11.

Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y),

Let C+n denote Chloe's age, J+n denote Joey's age, and Z+n denote Zoe's age, where n is the number of years from now. We are told that C+n is a multiple of Z+n exactly nine times. Because Z+n is 1 at n=0 and will increase until greater than C-Z, it will hit every natural number less than C-Z, including every factor of C-Z. For C+n to be an integral multiple of Z+n, the difference C-Z must also be a multiple of Z, which happens iff Z is a factor of C-Z. Therefore,C-Z has nine factors. The smallest number that has nine positive factors is 2^23^2=36 (we want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's). We also know Z=1 and J=C+1. Thus, C-Z=36J-Z=37 By our above logic, the next time J-Z is a multiple of Z+n will occur when Z+n is a factor of J-Z. Because 37 is prime, the next time this happens is at Z+n=37, when J+n=74. 7+4=11.

Link Problem
Problem 20 Hard

A function f is defined recursively by f(1)=f(2)=1 and f(n)=f(n-1)-f(n-2)+n for all integers n\geqslant3. What is f(2018)? (2018 AMC 10B Problem, Question#20)

  • A.

    2016

  • B.

    2017

  • C.

    2018

  • D.

    2019

  • E.

    2020

Answer:B

f(n)=f(n-1)-f(n-2)+n

=(f (n-2)-f(n-3)+n-1)-f(n-2)+n

=2n-1-f(n-3)

=2n-1-(2(n-3)-1-f(n-6))

=f (n-6)+6

Thus, f (2018)=2016+f(2)=2017.

Start out by listing some terms of the sequence. f(1)=1, f(2)=1, f(3)=3, f(4)=6, f(5)=8, f(6)=8, f(7)=7, f(8)=7, f(9)=9, f(10)=12, f(11)=14, f(12)=14, f(13)=13, f(14)=13, f(15)=15\cdots. Notice that f(n)=n whenever n is an odd muttple of 3, and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. The largest odd multiple of 3 smaller than 2018 is 2013, so we have f(2013)=2013, f(2014)=2016, f(2015)=2018, f(2016)=2018, f(2017)=2017, f(2018)= 2017.

Writing out the first few values, we get:1, 1, 3, 6, 8, 8, 7, 7, 9, 12, 14, 14, 13, 13, 15, 18, 20, 20, 19, 19\cdots . Examining,we see that every number x where x≡1 (\rm mod \ 6) has f(x)=x, f(x+1)=f(x)=x, and f(x-1)=f(x-2)=x+1. The greatest number that's 1 (\rm mod \ 6) and less 2018 is 2017,so we have f(2017)=f(2018)=2017.

Link Problem
Problem 21 Hard

Mary chose an even 4-digit number n. She wrote down all the divisors of n in increasing order from left to right: 1, 2,\cdots, \frac{n}{2}, n. At some moment Mary wrote 323 as a divisor of n. What is the smallest possible value of the next divisor written to the right of 323? (2018 AMC 10B Problem, Question#21)

  • A.

    324

  • B.

    330

  • C.

    340

  • D.

    361

  • E.

    646

Answer:C

Prime factorizing 323 gives you 17\cdot 19. The desired answer needs to be a multiple of 17 or 19 , because if it is not a multiple of 17 or 19, the LCM, or the least possible value for n, will not be more than 4 digits. Looking at the answer choices 340, is the smallest number divisible by 17 or 19. Checking, we can see that n would be 6460.

Let the next largest divisor be k. Suppose gcd (k,323)=1. Then,as 323|n,k|n, therefore, 323·k|n. However, because k>323, 323k>323\cdot 324>9999.

Therefore, gcd (k,323)>1. Note that 323=17\cdot 19. Therefore, the smallest the gcd can be is 17 and our answer is 323 +17=340.

Link Problem
Problem 22 Hard

Real numbers x and y are chosen independently and uniformly at random from the interval [0,1]. Which of the following numbers is closest to the probability that x, y, and 1 are the side lengths of an obtuse triangle? (2018 AMC 10B Problem, Question#22)

  • A.

    0.21

  • B.

    0.25

  • C.

    0.29

  • D.

    0.50

  • E.

    0.79

Answer:C

The Pythagorean inequality tells us that in an obtuse triangle，a^2+b^2<c^2. the  triangle inequality tells us that a+b>c. So, we have two inequalities: x^2+y^2<1, x+y>1 The first equation is \frac{1}{4} of a circle with radius 1, and the second equation is a line from (0,1) to (1,0), So, the area is  \frac{\pi}{4}- \frac{1}{2} which is approximately 0.29.

Note that the obtuse angle in the triangle has to be opposite the side that is always length 1. This is because the largest angle is always opposite the largest side, and if 2 sides of the triangle were 1, the last side would have to be greater than 1 to make an obtuse triangle. Using this observation, we can set up a law of cosines where the angle is opposite 1: 1^2=x^2+y^2-2xy\cos(\theta), where x and y are the sides that go from [0,1] and \theta is the angle opposite the side of length 1. By isolating \cos(\theta), we get: \frac{1-x^2-y^2}{-2xy}=\cos(\theta).

For \theta to be obtuse, \cos(\theta) must be negative. Therefore, \frac{1-x^2-y^2}{-2xy} is negative. Since x and y must be positive, -2xy must be negative, so we must make 1-x^2-y^2 positive. From here, we can set up the inequality x^2+y^2<1 Additionally, to satisfy the definition of a triangle, we need: x+y>1 The solution should be the overlap between the two equations in the \rm 1st quadrant.

By observing that x^2+y^2<1 is the equation for a circle, the amount that is in the \rm 1st quadrant is \frac{\pi}{4}. The line can also be seen as a chord that goes from (0,1) to (1,0). By cutting off the triangle of area \frac{1}{2} that is not part of the overlap, we get \frac{\pi}{4}-\frac{1}{2}\approx0.29.

Link Problem
Problem 23 Hard

How many ordered pairs (a,b) of positive integers satisfy the equation a\cdot b+63=20\cdot lcm(a,b)+12\cdot gcd(a,b), where gcd(a,b) denotes the greatest common divisor of a and b, and lcm(a,b) denotes their least common multiple? (2018 AMC 10B Problem, Question#23)

  • A.

    0

  • B.

    2

  • C.

    4

  • D.

    6

  • E.

    8

Answer:B

Let x=lcm(a,b), and y=gcd(a,b). Therefore, a\cdot b= lcm(a,b)\cdot gcd(a,b)=x\cdot y.Thus, the equation becomes

x\cdot y+63=20x+12yx\cdot y-20x-12y+63=0, Using Simon's Favorite Factoring Trick, we rewrite this equation as (x-12)(y-20)-240+63=0, (x-12)(y-20)=177, From here we can already see that this is a quadratic, and thus must have 2 solutions. But, let's continue, to see if one of the solutions is extraneous.

Since 177=3\cdot 59 and x>y, we have x- 12=59 and y-20=3, or x-12=177 and y - 20=1. This gives us the solutions (71,23) and (189, 21). Since the Greatest Common Denominator must be a divisor of the Lowest Common Multiple, the first pair does not work. Assume a>b. We must have a=21\cdot 9 and b=21, and we could then have a<b, so  there are 2 solutions.

Link Problem
Problem 24 Hard

Let ABCDEF be a regular hexagon with side length 1. Denote by X, Y, and Z the midpoints of sides \overline{AB},  \overline{CD},  and \overline{EF},  respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of \triangle ACE and \triangle XYZ? (2018 AMC 10B Problem, Question#24)

  • A.

    \frac{3}{8}\sqrt{3}

  • B.

    \frac{7}{16}\sqrt{3}

  • C.

    \frac{15}{32}\sqrt{3}

  • D.

    \frac{1}{2}\sqrt{3}

  • E.

    \frac{9}{16}\sqrt{3}

Answer:C

The desired area (hexagon MPNQOR) consists of an equilateral triangle (\triangle MNO) and three right triangles (\triangle MPN, \triangle NQO, and \triangle ORM)

Notice that \overline{AD} (not shown) and \overline{BC} are parallel. \overline{XY} divides transversals \overline{AB} and \overline{CD} into a 1:1 ratio.Thus,  it must also divide transversal \overline{AC} and transversal \overline{CO} into a 1:1 ratio.By Symmetry, the same applies for \overline{CE} and \overline{EA} as well as \overline{EM} and \overline{AN} in \triangle ACE, we see that \frac{\left[ MNO \right]}{\left[ ACE \right]}= \frac{1}{4} and \frac{\left[ MPN \right]}{\left[ ACE \right]}= \frac{1}{8}. Our desired area becomes (\frac{1}{4}+3 \cdot \frac{1}{8})\cdot \frac{(\sqrt{3})^{2}\cdot \sqrt{3}}{4}= \frac{15}{32}\sqrt{3}.

Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of 3 isosceles trapezoids (AXFZ , XBCY, and ZYED), and 3 right triangles, with one right angle on each of X, Y, and Z. Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is 1, and the other base is \frac{3}{2} (It is halfway in between the side and the longest diagonal, which has length 2) with a height of \frac{\sqrt{3}}{4} (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of \frac{5\sqrt{3}}{16} for a total area of \frac{15\sqrt{3}}{16}  (Alternatively, we could have calculated the area of hexagon ABCDEF and subtracted the area of \triangle XYZ, which, as we showed before, had a side length of \frac{3}{2}). Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on X, is similar to the triangle with a base of YC=\frac{1}{2}. Using similar triangles we calculate the base to be \frac{1}{4} and the height to be \frac{\sqrt{3}}{4} giving us an area of \frac{\sqrt{3}}{32} per triangle, and a total area of 3 \frac{\sqrt{3}}{32}, Adding the two areas together, we get \frac{15 \sqrt{3}}{16}+ \frac{3 \sqrt{3}}{32}= \frac{33 \sqrt{3}}{32}.  Finding the total area, we get 6 \cdot 1^{2}\cdot \frac{\sqrt{3}}{4}= \frac{3 \sqrt{3}}{2}. Taking the complement, we get \frac{3 \sqrt{3}}{2}- \frac{33 \sqrt{3}}{32}= \frac{15 \sqrt{3}}{32}=\frac{15}{32}\sqrt{3}.

Notice, the area of the convex hexagon formed through the intersection of the 2 triangles can be found by finding the area of the triangle formed by the midpoints of the sides and subtracting the smaller triangles that are formed by the region inside this triangle but outside the other triangle. First, let's find the area of the area of the triangle formed by the midpoint of the sides. Notice, this is an equilateral triangle, thus all we need is to find the length of its side. To do this, we look at the isosceles trapezoid outside this triangle but inside the outer hexagon. Since the interior angle of a regular hexagon is 120^\circ and the trapezoid is isosceles, we know that the angle opposite is 60^\circ, and thus the side length of this triangle is 1+2\left(\frac{1}{2}\cos(60^\circ)\right)=1+\frac{1}{2}=\frac{3}{2}. So the area of this triangle is \frac{\sqrt{3}}{4}s^2=\frac{9\sqrt{3}}{16} Now let's find the area of the smaller triangles. Notice, triangle ACE cuts off smaller isosceles triangles from the outer hexagon. The base of these isosceles triangles is perpendicular to the base of the isosceles trapezoid mentioned before, thus we can use trigonometric ratios to find the base and height of these smaller triangles, which are all congruent due to the rotational symmetry of a regular hexagon. The area is then \frac{1}{2}\left(\frac{1}{2}\cos(60^\circ)\right)\left(\frac{1}{2}\sin(60^\circ)\right)=\frac{\sqrt{3}}{32} and the sum of the areas is 3\cdot\frac{\sqrt{3}}{32}=\frac{3\sqrt{3}}{32} Therefore, the area of the convex hexagon is \frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\frac{15\sqrt{3}}{32}.

Dividing \triangle MNO into two right trianges congruent to  \triangle PMN, we see that [MPNQOR]=\frac{5}{8}[ACE], Because [ACE]=\frac{1}{2}[ABCDEF], we have \left[ MPNQOR \right]= \frac{5}{16}\left[ ABCDEF \right] , From here, you should be able to tell that the answer will have a factor of 5, and \frac{15}{32}\sqrt{3} is the only answer that has a factor of 5. However, if you want to actually calculate the area, you would calculate [ABCDEF] to be 6\cdot \frac{\sqrt{3}}{2\cdot 2}=\frac{3\sqrt{3}}{2}, so [MPNQOR]=\frac{5}{16}\cdot\frac{3\sqrt{3}}{2}=\frac{15}{32}\sqrt{3}.

Link Problem
Problem 25 Hard

Let \left\lfloor {x} \right\rfloor denote the greatest integer less than or equal to x. How many real numbers x satisfy the equation x^{2}+10,000\left\lfloor {x} \right\rfloor=10,000x? (2018 AMC 10B Problem, Question#25)

  • A.

    197

  • B.

    198

  • C.

    199

  • D.

    200

  • E.

    201

Answer:C

This rewrites itself to x^2=10000\left\{x\right\}, Graphing y=10000\left\{x\right\} and y=x^2 we see that the former is a set of line segments with slope 10000 from 0 to 1 wth a hole at x=1, then 1 to 2 with a hole at x=2 etc. Here is a graph of y=x^2 and y=16\left\{x\right\} for visualization.

Now notice that when x=\pm100 then graph has a hole at (\pm100,10000) which the equation y=x^2 passes through and then continues upwards. Thus our set of possible solutions is bounded by (-100,100). We can see that y=x^2 intersects each of the lines once and there are 99-(-99)+1=199  lines for an answer of 199.

Note: from the graph we can clearly see there are 4 solution on the negative side and only 2 on the positive side. So the solution really should be from -100 to 98, which still counts to 199. A couple of the alternative solutions also seem to have the same flaw.

Same as the first soution，x2=10000\left\{x\right\}.

We can wrte x as \left\lfloor {x} \right\rfloor+\left\{x\right\} , Expanding everything, we get a quadratic in x in terms of \left\lfloor {x} \right\rfloor:\left\{x\right\}^2+(2 \left\lfloor {x} \right\rfloor-10,000)\left\{x\right\}+ \left\lfloor {x} \right\rfloor^2=0, We use the quadratic formula to solve for  \left\{x\right\}:\left\{ x \right\} = \frac{-2 \left\lfloor {x} \right\rfloor +10000 \pm \sqrt{(-2 \left\lfloor {x} \right\rfloor +10000^{2}-4 \left\lfloor {x} \right\rfloor ^{2})}}{2},

Since 0\leqslant\left\{x\right\}<1, we get an inequalty which we can then solve. After simplfying a lot, we get that \left\lfloor {x} \right\rfloor^2+2 \left\lfloor {x} \right\rfloor-9999<0. Solving over the integers, -101< \left\lfloor {x} \right\rfloor<99, and since \left\lfloor {x} \right\rfloor is an integer, there are 199 solutions. Each vaue of \left\lfloor {x} \right\rfloor shoud correspond to one value of x,  so we are done.

Let x=a+k where a is the integer part of x and k is the fractional part of x. We can then rewrite the problem below：

(a+k)^2+10000a=10000(a+k),

From here,  we get

(a+k)^2+10000a=10000a+10000k,

Solving for a+k=x,

(a+k)^2=10000k,

x=a+k= \pm 100 \sqrt{k},

Because 0\leqslant k<1,  we know that a+k cannot be less than or equal to -100 nor greater than or equal to100. Therefore： -99\leqslant x\leqslant 99.

There are 199 elements in this range,  so the answer is 199.

Notice the given equation is equvilent to \left(\left\lfloor {x} \right\rfloor+\left\{x\right\}\right)^2=10000\left\{x\right\}, Now we now that \left\{x\right\}<1 so plugging in 1 for \left\{x\right\} we can find the upper and lower bounds for the values.

(\left\lfloor {x} \right\rfloor+1)^2=10000(1),

(\left\lfloor {x} \right\rfloor+1)=\pm 100,

\left\lfloor {x} \right\rfloor=99, -101.

And just ike Solution 2, we see that -101<\left\lfloor {x} \right\rfloor<99, and since \left\lfloor {x} \right\rfloor is an integer, there are 199 solutions. Each value of \left\lfloor {x} \right\rfloor should correspond to one value of x, so we are done.

First, we can let \left\{x\right\}=b, \left\lfloor {x} \right\rfloor=a. we know that a+b=x by definition, We can rearrange the equation to obtain x^2=10^4(x-a). By taking square root on both side, we obtain x=\pm 100\sqrt{b} (because x-a=b),  We know since b is the fractional part of x, it must be tat 0\leqslant b<1, Thus, x may take any value in the inteval -100<x<100, Hehnce, we know that there are 199 potential values for \left\lfloor {x} \right\rfloor in that  range and we are done.

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