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2018 AMC 10B Answer Key答案解析

Category: AMC美国数学竞赛, 国际竞赛, 福利干货, 竞赛资料 Date: 2018年2月23日下午7:17

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2018 AMC 10B Answer Key答案解析

翰林国际教育全网首发

Problem 1

答案：A
Solution 1：The area of the pan is $20\cdot18$ = $360$ . Since the area of each piece is $4$ , there are $\frac{360}{4} = 90$ pieces. Thus, the answer is $\boxed{A}$ .
Solution 2：By dividing the each of the dimensions by $2$ , we get a $10\times9$ grid which makes $90$ pieces. Thus, the answer is $\boxed{A}$ .

Problem 2

答案：D
Solution：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 $2$ 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 $\boxed{\textbf{(D) } 67}$ miles per hour in the third half hour.

Problem 3

答案：B
Solution 1：We have $\binom{4}{2}$ ways to choose the pairs, and we have $2$ ways for the values to be switched so $\frac{6}{2}=\boxed{3.}$ (harry1234)
Solution 2：We have four available numbers $(1, 2, 3, 4)$ . Because different permutations do not matter because they are all addition and multiplication, if we put $1$ on the first space, it is obvious there are $\boxed{3}$ possible outcomes $(2, 3, 4)$ .

Problem 4

答案：B
Solution 1：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 to equations to get $\frac{Z}{X} = 3$ . Then, multiply by the last equation to get $Z^2 = 144$ , giving $Z = 12$ . Following, $X = 4$ and $Y = 6$ . The final answer is $4 + 6 + 12 = 22$ . $\boxed{B}$
Solution 2：Simply use guess and check to find that the dimensions are $4$ by $6$ by $12$ . Therefore, the answer is $4 + 6 + 12 = 22$ . $\boxed{B}$

Problem 5：

答案：D
Solution 1：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 $\{4, 6, 8, 9\}$ , 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 $256 - 16 = 240$ . $\boxed{D}$
Solution 2 (Using Answer Choices)：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 permutations $\binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4} = 15$ . Using the answer choices, the only multiple of 15 is $\boxed{\textbf{(D) }240}$

Problem 6：

答案：D
Solution 1：Notice that the only two ways such that more than $2$ draws are required 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 $\boxed{D}$ .
Solution 2：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 probability of getting a $1$ is $\frac{1}{5}$ . It is necessary to pull either a $2$ or $3$ on the next draw and the probability 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 $\boxed {D}$ By: Soccer_JAMS

Problem 7：

答案：D
Solution 1 (Work using Answer Choices) ：Use the answer choices and calculate them. The one that works is D
Solution 2 (More Algebraic Approach)：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 $\bold{\boxed{\text{(D)19}}}$ . Solution by: Archimedes15; edited correct by kevinmathz

Problem 8：

答案：C
Solution 1：A staircase with $n$ steps contains $4 + 6 + 8 + ... + 2n + 2$ 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 $13 - 1 = \boxed {(C) 12}$
Solution 2：Layer $1$ : $4$ stepsLayer $1,2$ : $10$ stepsLayer $1,2,3$ : $18$ stepsLayer $1,2,3,4$ : $28$ stepsFrom 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 indeed $\boxed {(C) 12}$

By: Soccer_JAMS

Problem 9：

答案：D
Solution 1：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)... 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 $\textbf{(D)} \text{ 39}$ , and we are done. Written By: Archimedes15
Solution 2：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 $\textbf{(D)}$ By: Soccer_JAMS
Solution 3 (Simple Logic)：For the sums to have equal probability, the average sum of both sets of 7 dies has to be (6+1) x 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 $\textbf{(D)}$ By: epicmonster
Solution 4：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 $\textbf{(D)}$ . By: dajeff

Problem 10：

答案：E
Solution 1：Consider the cross-sectional plane and label its area $b$ . Note that the volume of the triangular prism that encloses the pyramid is $bh/2=3$ , and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is $bh/3$ , so the answer is $\boxed{2}$ . (AOPS12142015)
Solution 2：We can start by finding the total volume of the parallelepiped. It is $2 \cdot 3 \cdot 1 = 6$ , 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 the parallelepiped. 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 $\boxed{E}$ . Written by: Archimedes15NOTE: 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.

Problem 11：

答案：C
Solution 1：Because squares of a non-multiple of 3 is always $1\mod 3$ , the only expression is always a multiple of $3$ is $\boxed{\textbf{(C) } p^2+26}$ . This is excluding when $p=0\mod3$ , which only occurs when $p=3$ , then $p^2+26=35$ which is still composite.
Solution 2 (Bad Solution): We proceed with guess and check: $5^2+16=41 \qquad 7^2+24=73 \qquad 5^2+46=71 \qquad 19^2+96=457$ . Clearly only $\boxed{(\textbf{C})}$ is our only option left. (franchester)
Solution 3：From Fermat's Little Theorom, $p^2 \equiv 1\pmod 3$ if $p$ is coprime with $3$ . So for any $n\equiv2\pmod3$ , $p^2+n \equiv 0\pmod 3$ - divisible by 3, so not a prime. The only choice $\equiv2\pmod3$ is $\boxed{(\textbf{C})}$
Solution 4：Primes can only be $1$ or $-1\mod 6$ . Therefore, the square of a prime can only be $1\mod 6$ . $p^2+26$ then must be $3\mod 6$ , so it is always divisible by $3$ . Therefore, the answer is $\boxed{\text{(C)}}$ .

Problem 12：

答案：C
Solution 1：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 $16\pi\approx \boxed{\text{(C) }50}$ . -tdeng

Solution 2 (no coordinates)：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 $\dfrac{1}{3} \cdot 12 = 4$ . The area of a circle with radius $4$ is $16\pi$ , or around $\boxed{\textbf{(C)} \text{ 50}}$ . -That_Crazy_Book_Nerd

Problem 13

答案：C
Solution 1：Note that $10^{2k}+1$ for some odd $k$ will suffice $\mod {101}$ . Each $2k \in \{2,4,6,\dots,2018\}$ , so the answer is $\boxed{\textbf{(C) } 505}$ (AOPS12142015)
Solution 2 ：If we divide each number by $101$ , we see a pattern occuring in every 4 numbers. $101, 1000001, 10000000001, \dots$ . We divide $2018$ by $4$ to get $504$ with $2$ left over. One divisible number will be in the $2$ left over, so out answer is $\boxed{\textbf{(C) } 505}$ .
Solution 3：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 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 ${0,-9,-99 ( 2),11, 0, ...}$ . Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide $2017$ by four to get $504$ remainder $1$ . Thus the answer is $504$ plus the 1st term or $\boxed{\textbf{(C) } 505}$ . -googleghosh

Problem 14

答案：D
Solution：To minimize the number of 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 $223 + 1 + 1$ = $\boxed{\textbf{(D) } 225}$

Problem 15

答案：A
Solution：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 $\frac{w^2}{2} + \frac{h^2}{2} + wh$ . Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting $4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \boxed{\textbf{(A) } 2(w+h)^2} \qquad$ .

Problem 16

答案：E
Solution 1：By Euler's Totient Theorem, $n^{3}\equiv n \pmod{6}$ Alternatively, one could simply list out all the residues to the third power $\mod 6$ Therefore the answer is congruent to $2018^{2018}\equiv 2^{2018} \pmod{6} = \boxed{ (E)4}$
Solution 2 (not very good one)： Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\prod_{i\neq j\neq k}^{2018} a_ia_ja_k$ Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\prod_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2(2018-a_1)+3a_2^2(2018-a_2)+\cdots+3a_{2018}^2(2018-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ Therefore, $-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}$ . Thus, $a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3$ . 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 $\boxed{\text{(E) }4}$
Solution 3：We first note that $1^3+2^3+...=(1+2+...)^2$ . So what we are trying to find is what $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6$ . We start by noting that $2018$ is congruent to $2$ mod $6$ . So we are trying to find $\left(2^{4036}\right)$ 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 $2$ mod $6$ , $2^2$ is $4$ mod $6$ , $2^3$ is $2$ mod $6$ , $2^4$ is $4$ mod $6$ , and so on... So we see that since $\left(2^{4036}\right)$ has an even power, it must be congruent to $4$ mod $6$ , thus giving our answer $\boxed{\text{(E) }4}$ . You can prove this pattern using mods. But I thought this was easier. -TheMagician
Solution 4 (Lazy solution)：Assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} mod 6$ (which happens to be 4). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 mod 6$ or just $4$ . -Patrick4Presiden

Problem 17

答案：B
Solution 1：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 $2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}$ . Our answer is $8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{\text{(B)}~7}$ . (Mudkipswims42)
Solution 2 ：Denote the length of the equilateral octagon as . The length of can be expressed as . By Pythagoras, we find that:
```
 
```
Since $\overline{CQ}=\overline{DR}$ , we can say that $x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}$ . We can discard the negative solution, so $k+m+n=-7+3+11=\boxed{\textbf{(B) }7}$ ~ blitzkrieg21

Problem 18：

答案：D
Solution 1 (Casework)：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 aCase 2: Second Row: a c b Third Row: c b aCase 3: Second Row: a b c Third Row: c a bCase 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 $32 \cdot 3 = 96$ . So, the correct answer is $\boxed{D}$ .

Written By: Archimedes15
Solution 2：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 $2018 AMC 10B Answer Key答案解析$ takes it (a $\circ$ is an empty seat): $A_1 \circ \circ \\ \circ \ \circ \ \circ$ Then there are 4 choices for the second seat ( $B_1$ , $B_2$ , $C_1$ , or $C_2$ ). Again, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$ : $A_1 B_1 \circ \\ \circ \ \circ \ \circ$
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 $C_2$ . Suppose WLOG that it is $C_1$ . There are two ways to create a second row:

$A_1 B_1 C_1 \\ B_2 C_2 A_2$

$A_1 B_1 C_1 \\ C_2 A_2 B_2$

Therefore, there are $6 \cdot 4 \cdot 2 \cdot 2= \boxed{96}$ possible seating arrangements.

Written by: R1ceming
Solution 3 (Using the Answers)：Notice how given an arrangement of the children that works (the answers tell us there is at least one) we can swap each pair of the siblings in one of 2 ways for = 8 arrangements, and each of the 3 pairs can take each others' spaces in 3! = 6 ways. This means that the answer must be divisible by 48.
```
Written by: Kevin Schmidt
```

Problem 19

答案：E
Solution 1： 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 $k\in\mathbb{Z}$ 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 $\boxed{\text{(E) }11}$
Solution 2： Here's a different way of saying your 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 = $\boxed{\text{(E) }11}$
Solution 3：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 if $Z$ 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, $C-Z=36$ $J-Z=37$ 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 $J+n=74$ . $7+4=\boxed{(\textbf{E}) 11}$

Problem 20

答案：B
Solution 1： $f\left(n\right) = f\left(n - 1\right) - f\left(n - 2\right) + n$ $= \left(f\left(n - 2\right) - f\left(n - 3\right) + n - 1\right) - f\left(n - 2\right) + n$ $= 2n - 1 - f\left(n - 3\right)$ $= 2n - 1 - \left(2\left(n - 3\right) - 1 - f\left(n - 6\right)\right)$ $= f\left(n - 6\right) + 6$
Thus, $f\left(2018\right) = 2016 + f\left(2\right) = 2017$ . $\boxed{B}$
Solution 2：Start out by listing some terms of the sequence. $f(1)=1$ $f(2)=1$ $f(3)=3$ $f(4)=6$
$f(5)=8$

$f(6)=8$

$f(7)=7$

$f(8)=7$

$f(9)=9$

$f(10)=12$

$f(11)=14$

$f(12)=14$

$f(13)=13$

$f(14)=13$

$f(15)=15$ $.....$ Notice that $f(n)=n$ whenever $n$ is an odd multiple of $3$ , and the pattern of numbers that follow will always be +3, +2, +0, -1, +0. The closest odd multiple of $3$ to $2018$ is $2013$ , so we have

$f(2013)=2013$

$f(2014)=2016$

$f(2015)=2018$

$f(2016)=2018$

$f(2017)=2017$

$f(2018)=\boxed{(B) 2017}.$
Solution 3 (Bashy Pattern Finding)：Writing out the first few values, we get: $1,1,3,6,8,8,7,7,9,12,14,14,13,13,15,18,20,20,19,19...$ . Examining, we see that every number $x$ where $x \equiv 1 \pmod 6$ 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 (mod 6) and less $2018$ is $2017$ , so we have $f(2017)=f(2018)=2017.$ $\boxed B$

Problem 21

答案：C
Solution 1：Prime factorizing $323$ gives you $17 \cdot 19$ . Looking at the answer choices, $\fbox{(C) 340}$ is the smallest number divisible by $17$ or $19$ .
Solution 2：Let the next largest divisor be $k$ . Suppose $\gcd(k,323)=1$ . Then, as $323|n, k|n$ , therefore, $323\cdot k|n.$ 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 $323+17=\boxed{\text{(C) }340}$ .

Problem 22

答案：C
Solution：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 1st quadrant.

By observing that $x^2+y^2<1$ is the equation for a circle, the amount that is in the 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 $\frac{\pi}{4} - \frac{1}{2} \approx \boxed{0.29}$ .

-allenle873

Problem 23

答案：B
Solution ：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 $x\cdot y + 63 = 20x + 12y$ $x\cdot y - 20x - 12y + 63 = 0$ Using Simon's Favorite Factoring Trick, we rewrite this equation as $(x - 12)(y - 20) - 240 + 63 = 0$ $(x - 12)(y - 20) = 177$ 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 $(189, 21)$ . Obviously, the first pair does not work. The second pair can be ordered in two ways. Thus, the answer is $\boxed{2}$ . (awesomeag)

Problem 24

答案：C
Solution 1：

$[asy] pair A,B,C,D,E,F,W,X,Y,Z,M,N,O,P,Q,R; A=(0,sqrt(3)); B=(1,sqrt(3)); C=(3/2,sqrt(3)/2); D=(1,0); E=(0,0); F=(-1/2,sqrt(3)/2); X=(1/2, sqrt(3)); Y=(5/4, sqrt(3)/4); Z=(-1/4, sqrt(3)/4); M=(0,sqrt(3)/2); N=(3/4,3sqrt(3)/4); O=(3/4,sqrt(3)/4); P=(3/8,7sqrt(3)/8); Q=(9/8, 3sqrt(3)/8); R=(0,sqrt(3)/4); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,ESE); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,WSW); label("$X$", X, N); label("$Y$", Y, ESE); label("$Z$", Z, WSW); label("$M$", M, NW); label("$N$", N, NE); label("$O$", O, SE); label("$P$", P, NNW); label("$Q$", Q, ESE); label("$R$", R, SW); fill((0,sqrt(3)/2)--(3/8,7sqrt(3)/8)--(3/4,3sqrt(3)/4)--(9/8, 3sqrt(3)/8)--(3/4,sqrt(3)/4)--(0,sqrt(3)/4)--cycle,gray); draw(A--B--C--D--E--F--cycle); draw(A--C--E--cycle); draw(X--Y--Z--cycle); draw(M--N--O--cycle); [/asy]$

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 $\frac{[MNO]}{[ACE]} = \frac{1}{4}$ and $\frac{[MPN]}{[ACE]} = \frac{1}{8}$ . Our desired area becomes

$(\frac{1}{4}+3 \cdot \frac{1}{8}) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \frac {15}{32}\sqrt{3} = \boxed {C}$
Solution 2：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 vertice on each of X, Y, and Z). We know that one base of each trapezoid is just the side length of the hexagon which is 1, and the other base is 3/2 (It is halfway in between the side and the longest diagonal) with a height of $\sqrt{3}/4$ (by using the Pythagorean theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of $5\sqrt{3}/16$ for a total area of $15\sqrt{3}/16$ (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 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 vertice on X, is similar to the triangle with a base of YC = 1/2. Using similar triangles we calculate the base to be 1/4 and the height to be $\sqrt{3}/4$ giving us an area of $\sqrt{3}/32$ per triangle, and a total area of $3\sqrt{3}/32$ . Adding the two areas together, we get $15\sqrt{3}/16+3\sqrt{3}/32=33*sqrt{3}/32$ . Finding the total area, we get $6*1^2*\sqrt{3}/4=3\sqrt{3}/2$ . Taking the complement, we get $3\sqrt{3}/2-33\sqrt{3}/32=15\sqrt{3}/32 = \boxed {C}$
Solution 3 (Trig) ：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 $120^{\textrm{o}}$ and the trapezoid is isosceles, we know that the angle opposite is $60^{\textrm{o}}$ , and thus the side length of this triangle is $1+2(\frac{1}{2}\cos(60^{\textrm{o}})=1+\frac{1}{2}=\frac{3}{2}$ . 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 $\frac{1}{2}(\frac{1}{2})\cos(60^{\textrm{o}}))(\frac{1}{2}\sin(60^{\textrm{o}}))=\frac{\sqrt{3}}{32}$ 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 $\frac{9\sqrt{3}}{16}-\frac{3\sqrt{3}}{32}=\frac{18\sqrt{3}}{32}-\frac{3\sqrt{3}}{32}=\boxed{\frac{15\sqrt{3}}{32}}\implies \boxed{C}$

Problem 25

答案：C
Solution 1：This rewrites itself to $x^2=10,000\{x\}$ .Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with 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 $y=16\{x\}$ for visualization. $[asy] import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); [/asy]$ Now notice that when $x=\pm 100$ then graph has a hole at $(\pm 100,10,000)$ 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 $\boxed{\text{(C)}~199}$ .
Solution 2 （Alternative, Bashy Solution）：Same as the first solution, $x^2=10,000\{x\}$ .We can write $x$ as $\lfloor x \rfloor+\{x\}$ . Expanding everything, we get a quadratic in ${x}$ in terms of $\lfloor x \rfloor$ : $\{x\}^2+ (2\lfloor x \rfloor -10,000)\{x\} + \lfloor x \rfloor ^2 = 0$ We use the quadratic formula to solve for {x}: $\{x\} = \frac {-2\lfloor x \rfloor + 10,000 \pm \sqrt{( -2 \lfloor x \rfloor + 10,000^2- 4\lfloor x \rfloor^2 ) }}{2}$ Since $0 \leq \{x\} < 1$ , we get an inequality which we can then solve. After simplifying a lot, we get that $\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 9999 < 0$ .Solving over the integers, $-101 < \lfloor x \rfloor < 99$ , and since $\lfloor x \rfloor$ is an integer, there are $\boxed{\text{(C)}~199}$ solutions. Each value of $\lfloor x \rfloor$ should correspond to one value of $x$ , so we are done.
Solution 3：Let $x = a+k$ where $a$ is the integer portion of $x$ and $k$ is the decimal portion. 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 $a+k$ ...
$(a+k)^2 = 10000k$

$a+k = \pm100\sqrt{k}$

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

$-99 \leq a+k = x \leq 99$

There are 199 elements in this range, so the answer is $\fbox{C 199}$
Solution 4：As in the first solution, we can write $x^2 = 10,000\{x\}$ .Now obviously, $0<\{x\}<1$ .So, x can get be $x\to{[-99,99]}$ .Hence, the answer is $\boxed{199}$ .-Pi_3.14_Squared

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