2013 amc10a

AMC10 2005,GRADE 9/10 MATH,CONTEST,PRACTICE QUESTIONS. Josh and Mike live miles apart. Yesterday Josh started to ride his bicycle toward Mike's house.

Junior Balkan Math Olympiad (JBMO) 2013: Deputy Leader of the Team USA ... AMC 10A perfect score (2017); 2015 National Mathcounts qualifier; Miller MathCounts ...All AMC 10 Problems and Solutions. The problems on this page are copyrighted by the Mathematical Association of America 's American Mathematics Competitions. Category: Introductory Number Theory Problems.2002 AMC 10A. 2002 AMC 10A problems and solutions. The first link contains the full set of test problems. The second link contains the answers to each problem. The rest contain each individual problem and its solution. 2002 AMC 10A Problems. Answer Key.2014 AMC 10 A Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Created Date: 2/5/2014 12:11:46 PMTHE *Education Center 25 For some positive integers p, there is a quadrilateral ABCD with positive inte- ger side lengths, perimeter p, right angles at B and C, AB 2 ...

AMC 10 Problems and Solutions. AMC 10 problems and solutions. Year. Test A. Test B. 2022. AMC 10A. AMC 10B. 2021 Fall.2013 AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 16: Followed by Problem 18: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 …2013 AMC10A Problems 4 12. In ˜ABC, AB = AC = 28 and BC = 20. Points D, E, and F are on sides AB, BC, and AC, respectively, such that DE and EF are parallel to AC and AB, respectively. What is the perimeter of parallelogram ADEF? A D B E C F (A) 48 (B) 52 (C) 56 (D) 60 (E) 72 13. How many three-digit numbers are not divisible by 5, have digits that …Junior Balkan Math Olympiad (JBMO) 2013: Deputy Leader of the Team USA ... AMC 10A perfect score (2017); 2015 National Mathcounts qualifier; Miller MathCounts ...Resources Aops Wiki 2013 AMC 10B Page. Article Discussion View source History. Toolbox. Recent ... 2012 AMC 10A, B: Followed by 2014 AMC 10A, B: 1 ...

Resources Aops Wiki 2013 AMC 10A Problems/Problem 1 Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. 2013 AMC 10A Problems/Problem 1. Contents. 1 Problem; 2 Solution; 3 Video Solution (CREATIVE THINKING) 4 Video Solution;2021 Fall AMC 10A. 2021 Fall AMC 10A problems and solutions. The test was held on Wednesday, November , . 2021 Fall AMC 10A Problems. 2021 Fall AMC 10A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.The test was held on February 7, 2017. 2017 AMC 10A Problems. 2017 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.If we can find this N, then the next number, N+1, will make P (N)<321/400. You can do a few tries as above (N=5, 10, 15, etc.), and you will see that the ball "works" in places. from 1 to 2/5 * N + 1, and places 3/5 * N +1 to N+1. This is a total of 4/5 * N + 2 spaces, over a total of N+1 spaces: (4/5 * N + 2)/ (N + 1) Let the above = 321/400 ...

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Solution. Let the number of students on the council be . To select a two-person committee, we can select a "first person" and a "second person." There are choices to select a first person; subsequently, there are choices for the second person. This gives a preliminary count of ways to choose a two-person committee.2013 AMC 10A Problems Problem 1 A taxi ride costs $1.50 plus $0.25 per mile traveled. How much does a 5-mile taxi ride cost? Solution There are five miles which need to be …The test was held on February 15, 2017. 2017 AMC 10B Problems. 2017 AMC 10B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.2013 AMC10A Problems 3 6. Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey? (A) 3 (B) 7 (C) 9 (D) 11 (E) 13 7.[AMC 10A 2013] Square ABCD has side length 10. Point E is on BC, and the area of △ABE is. 40. What is BE? A. B.

Solution. We use a casework approach to solve the problem. These three digit numbers are of the form . ( denotes the number ). We see that and , as does not yield a three-digit integer and yields a number divisible by 5. The second condition is that the sum . When is , , , or , can be any digit from to , as . This yields numbers.Resources Aops Wiki 2021 Fall AMC 10A Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. 2021 Fall AMC 10A. 2021 Fall AMC 10A problems and solutions. The test was held on Wednesday, November , . 2021 Fall AMC 10A Problems; 2021 Fall AMC 10A Answer Key.Video transcript. - We've get a geometry problem here, so you know where we're gonna start, we're gonna draw the diagram. Got a triangle, couple of side lengths. Have a circle centered at one of the vertices of the triangle, and the radius is one of the side lengths of the triangle, so, it's gonna go through one of the vertices. Explanations of Awards. Average score: Average score of all participants, regardless of age, grade level, gender, and region. AIME floor: Before 2020, approximately the top 2.5% of scorers on the AMC 10 and the top 5% of scorers on the AMC 12 were invited to participate in AIME.2021 AMC 10A Problems Problem 1 What is the value of Problem 2 Portia's high school has times as many students as Lara's high school. The two high schools have a total of students. How many students does Portia's high school have? Problem 3 The sum of two natural numbers is . One of the two numbers is divisible by 10. If the2016 AMC 10A problems and solutions. The test was held on February 2, 2016. 2016 AMC 10A Problems. 2016 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.Solution We want the integers such that is a factor of . Since , it has factors. Since cannot equal or , as these cannot have the digit in their base representations, our answer is …01-Jan-2021 ... 10. 2009 AMC 12A Problem 25: · 9. 2007 AMC 12A Problem 17: · 8. 2017 AMC 10A Problem 24/12A Problem 23: · 7. 2011 AMC 12B Problem 21: · 6. 2013 AMC ...Problem. In base , the number ends in the digit .In base , on the other hand, the same number is written as and ends in the digit .For how many positive integers does the base--representation of end in the digit ?. Solution. We want the integers such that is a factor of .Since , it has factors. Since cannot equal or , as these cannot have the digit in their base representations, our answer is2013 AMC10A Problems 3 6. Joey and his five brothers are ages 3, 5, 7, 9, 11, and 13. One afternoon two of his brothers whose ages sum to 16 went to the movies, two brothers younger than 10 went to play baseball, and Joey and the 5-year-old stayed home. How old is Joey? (A) 3 (B) 7 (C) 9 (D) 11 (E) 13 7.

2014 AMC10A Problems 2 1. What is 10·(1 2 + 5 + 1 10) −? (A) 3 (B) 8 (C) 25 2 (D) 170 3 (E) 170 2. Roy’s cat eats 1 3 of a can of cat food every morning and 1 4 of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing 6 cans of cat food. On what day of the week did the cat finish eating all ...

The straight lines will be joined together to form a single line on the surface of the cone, so 10 will be the slant height of the cone. The curve line will form the circumference of the base. We can compute its length and use it to determine the radius. The length of the curve line is 252/360 * 2 * pi *10 = 14 * pi.Solution 2. First, realize that is the midpoint of and is the midpoint of . Connect to to form . Let the midpoint of be . Connect to . is a median of . Because is isosceles, is also an altitude of . We know the length of and from the given coordinates. The area of is . Let the intesection of , and be . is the centroid of .Junior Balkan Math Olympiad (JBMO) 2013: Deputy Leader of the Team USA ... AMC 10A perfect score (2017); 2015 National Mathcounts qualifier; Miller MathCounts ...2013 AMC10A Solutions 6 O E A0 B F A B0 21. Answer (D): For 1 • k • 11, the number of coins remaining in the chest before the kth pirate takes a share is 12 12¡k times the number remaining afterward. Thus if there are n coins left for the 12th pirate to take, the number of coins originally in the chest is 1211 ¢n 11! = 222 ¢311 ¢n 28 ¢34 ¢52 ¢7¢11 214 ¢37 ¢n …2013 AMC 10A (Problems • Answer Key • Resources) Preceded by Problem 16: Followed by Problem 18: 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 …First pirate's gonna come along and take 1/12 of the gold that's in the chest. Second pirate's gonna come along, take 2/12 of the whatever's left after the first pirate is finished. Third pirate's gonna take 3/12 of whatever's left after the second pirate finished, and on, and on, and on.These mock contests are similar in difficulty to the real contests, and include randomly selected problems from the real contests. You may practice more than once, and each attempt features new problems. Archive of AMC-Series Contests for the AMC 8, AMC 10, AMC 12, and AIME. This achive allows you to review the previous AMC-series contests.Mock (Practice) AMC 10 Problems and Solutions (Please note: Mock Contests are significantly harder than actual contests) Problems Answer Key SolutionsAMC 10A American Mathematics Competition 10A Wednesday, February 7, 2018. INSTRUCTIONS 1. DO NOT OPEN THIS BOOKLET UNTIL YOUR COMPETITION MANAGER TELLS YOU. 2. This is a 25-question multiple-choice exam. ... Document of doument 2013. Thyeadi Tungson. 2007PascalContest (1) 2007PascalContest (1) Daniel …Circle & Triangle segment lengths (AMC 10A 2013 #23) In ABC A B C, AB = 86 A B = 86, and AC = 97 A C = 97. A circle with center A A and radius AB A B intersects BC¯ ¯¯¯¯¯¯¯ B C ¯ at points B B and X X. Moreover BX¯ ¯¯¯¯¯¯¯ B X ¯ and CX¯ ¯¯¯¯¯¯¯ C X ¯ have integer lengths. What is BC B C?

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2018 AMC 10A Solutions 2 1. Answer (B): Computing inside to outside yields: (2 + 1) 1 + 1 41 + 1 1 + 1 = 3 1 + 1! 1 + 1 = 7 4 1 + 1 = 11 7: Note: The successive denominators and numerators of numbers ob-tained from this pattern are the Lucas numbers. 2. Answer (A): Let L, J, and A be the amounts of soda that Liliane, Jacqueline, and Alice have ...AMC 10 Problems and Solutions. AMC 10 problems and solutions. Year. Test A. Test B. 2022. AMC 10A. AMC 10B. 2021 Fall.Solution 3. The meaning of sharing costs equally is meaning that, after the vacation, they are equally dividing the money in a way such that, each person would have the same amount left. As each person spends an amount of money, greater than 100, let it be that they all had dollars to spend. This means that after the vacation we want the amount ...2012-Problems-AMC10A.indd 4 11/11/2011 9:47:03 AM. 2012 AMC10A Problems 4 14. Chubby makes nonstandard checkerboards that have 31 squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are thereThe first link contains the full set of test problems. The rest contain each individual problem and its solution. 2006 AMC 10B Problems. Answer Key. 2006 AMC 10B Problems/Problem 1. 2006 AMC 10B Problems/Problem 2. 2006 AMC 10B Problems/Problem 3. 2006 AMC 10B Problems/Problem 4. 2006 AMC 10B Problems/Problem 5.11-Jun-2014 ... 2014 AMC 10A 25 / AMC 12A 22. The number $latex 5^{867}$ is between $latex 2^{2013}$ and $latex 2^{2014}$. How many pairs of integers $latex ...2016 AMC 10A. 2016 AMC 10A problems and solutions. The test was held on February 2, 2016. 2016 AMC 10A Problems. 2016 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3. Problem 4.To learn more about the AMC 10 exam, please contact Think Academy at [email protected] or +1 (844) 844-6587. Subscribe to our newsletter for more K-12 educational information! As one of the most challenging high school-level math competitions in the US, the AMC 10 will take place in November 2023, following its annual tradition.AMC 10 Problems and Solutions. AMC 10 problems and solutions. Year. Test A. Test B. 2022. AMC 10A. AMC 10B. 2021 Fall.HOMEAMC10AMC10B 2014AMC10A 2014AMC10B 2015AMC10A 2015AMC10A 2013AMC10B 2013AMC10A 2012AMC10B 2012AMC10A 2011AMC10B 2011AMC10A 2010AMC10B 2010AMC10A 2009 ...If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? Solution. The nth item for the sequence is: An=An-1+4n. We add increasing multiples of 4 each time we go up a figure. So, to go from Figure 0 to 100, we add. 4 *1+4*2+...+4*99+4*100=4*5050=20200. ….

YouTube 频道 Kevin's Math Class，相关视频：AMC 10 几何专题 Geometry 2009-2000，2022 AMC 10A 难题讲解 18-23，2019 AMC 12A 真题讲解 1-15，2014 AMC 10B 真题讲解 1-20，2022 AMC 10A 真题讲解 1-17，2015 AMC 10A 难题讲解 #19-25，2014 AMC 10B 难题讲解 #21-25，2013 AMC 10B 难题讲解 #21-25，新鲜出炉！Resources Aops Wiki 2020 AMC 10A Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. PREPARE FOR THE AMC 10 Join outstanding instructors and top-scoring students in our online AMC 10 Problem Series course. ...2016 AMC 10A 7 The mean, median, and mode of the 7 data values 60,100,x,40,50,200,90 are all equal to x. What is the value of x? (A)50 (B)60 (C)75 (D)90 (E)100 8 Trickster Rabbit agrees with Foolish Fox to double Fox’s money every time Fox crosses the bridge by Rabbit’s house, as long as Fox pays 40 coins in toll to Rabbit after each crossing.Resources Aops Wiki 2014 AMC 10A Problems Page. Article Discussion View source History. Toolbox. Recent changes Random page Help What links here Special pages. Search. GET READY FOR THE AMC 10 WITH AoPS ... 2013 AMC 10B Problems: Followed by2013 AMC10A Problems 4 12. In ˜ABC, AB = AC = 28 and BC = 20. Points D, E, and F are on sides AB, BC, and AC, respectively, such that DE and EF are parallel to AC and AB, respectively. What is the perimeter of parallelogram ADEF? A D B E C F (A) 48 (B) 52 (C) 56 (D) 60 (E) 72 13. How many three-digit numbers are not divisible by 5, have digits that sum to2013 AMC10A Solutions 4 14. Answer (D): The large cube has 12 edges, and a portion of each edge remains after the 8 small cubes are removed. All of the 12 edges of each small cube are also edges of the new solid, except for the 3 edges that meet at a vertex of the large cube. Thus the new solid has a total of 12+8(12−3) = 84 edges. 15. 2013 AMC10A Solutions 4 14. Answer (D): The large cube has 12 edges, and a portion of each edge remains after the 8 small cubes are removed. All of the 12 edges of each small cube are also edges of the new solid, except for the 3 edges that meet at a vertex of the large cube. Thus the new solid has a total of 12+8(12−3) = 84 edges. 15. 2010 AMC 10B problems and solutions. The test was held on February 24 th, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2010 AMC 10B Problems. 2010 AMC 10B Answer Key.The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2006 AMC 10B Problems. Answer Key. 2006 AMC 10B Problems/Problem 1. 2006 AMC 10B Problems/Problem 2. 2006 AMC 10B Problems/Problem 3. 2006 AMC 10B Problems/Problem 4. 2006 AMC 10B Problems/Problem 5. 2013 amc10a, Jan 1, 2021 · 5. 2013 AMC 10A Problem 19: In base 10, the number 2013 ends in the digit 3. In base 9, on the other hand, the same number is written as (2676)_9 and ends in the digit 6. For how many positive integers b does the base-b-representation of 2013 end in the digit 3? , View Triangle_Geometry_-_November_25_2014.pdf from MATH GEOMETRY at Rosemont High. Triangle Geometry November 25, 2014 Level I 1. (2012 AMC10A #4) Let ∠ABC = 24 and ∠ABD = 20 . What is the smallest, Solution 1. We can use Euler's polyhedron formula that says that . We know that there are originally faces on the cube, and each corner cube creates more. . In addition, each cube creates new vertices while taking away the original , yielding vertices. Thus , so., 2013 and 22014 How many pairs of integers (m, n) are there such that 1 < m < 2012 and (A) 278 (B) 279 (C) 280 (D) 281 (E) 282 AMC 10 2014 product .. . 8), where the second factor has k digits, is an integer whose digits have a sum of 1000. What is k? (A) 901 (B) 911 (C) 919 (D) 991 (E) 999 Positive integers a and b are such that the graphs of y, 2017 AMC 10A 真题讲解 1-19. 美国数学竞赛AMC10，历年真题，视频完整讲解。. 真题解析，视频讲解，不断更新中. 你的数学竞赛辅导老师。. YouTube 频道 Kevin's Math Class. 十年老玩家都哭了！. 刀刀暴击，满地神装. 新鲜出炉！. 2021 AMC 10A 难题讲解 20-25., 2012 AMC 10A. 2012 AMC 10A problems and solutions. The test was held on February 7, 2012. 2012 AMC 10A Problems. 2012 AMC 10A Answer Key. Problem 1. Problem 2. Problem 3., 2021 AMC 10A problems and solutions. The test will be held on Thursday, February , . Please do not post the problems or the solutions until the contest is released. 2021 AMC 10A Problems. 2021 AMC 10A Answer Key. , The rest contain each individual problem and its solution. 2000 AMC 10 Problems. 2000 AMC 10 Answer Key. 2000 AMC 10 Problems/Problem 1. 2000 AMC 10 Problems/Problem 2. 2000 AMC 10 Problems/Problem 3. 2000 AMC 10 Problems/Problem 4. 2000 AMC 10 Problems/Problem 5. 2000 AMC 10 Problems/Problem 6., 2011 AMC 10A. 2011 AMC 10A problems and solutions. The test was held on February 8, 2011. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2011 AMC 10A Problems. , Direct link to Daniel Chaviers's post “The AMC 10 is more about ...”. The AMC 10 is more about analysis and "abuse" of the various laws and properties of any number of things, which is seemingly unrelated. The AMC 10 has a bit more algebra than the AMC 8, would, but it's otherwise pretty similar: lot of analysis., YouTube 频道 Kevin's Math Class，相关视频：AMC 10 几何专题 Geometry 2009-2000，2022 AMC 10A 难题讲解 18-23，2019 AMC 12A 真题讲解 1-15，2014 AMC 10B 真题讲解 1-20，2022 AMC 10A 真题讲解 1-17，2015 AMC 10A 难题讲解 #19-25，2014 AMC 10B 难题讲解 #21-25，2013 AMC 10B 难题讲解 #21-25，新鲜出炉！, Solving problem #15 from the 2013 AMC 10A test., Explanations of Awards. Average score: Average score of all participants, regardless of age, grade level, gender, and region. AIME floor: Before 2020, approximately the top 2.5% of scorers on the AMC 10 and the top 5% of scorers on the AMC 12 were invited to participate in AIME., Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚. Books for Grades 5-12 Online Courses, 2010. 188.5. 188.5. 208.5 (204.5 for non juniors and seniors) 208.5 (204.5 for non juniors and seniors) Historical AMC USAJMO USAMO AIME Qualification Scores., The test was held on February 20, 2013. 2013 AMC 10B Problems. 2013 AMC 10B Answer Key. Problem 1. Problem 2. Problem 3. Problem 4., 2013 and 22014 How many pairs of integers (m, n) are there such that 1 < m < 2012 and (A) 278 (B) 279 (C) 280 (D) 281 (E) 282 AMC 10 2014 product .. . 8), where the second factor has k digits, is an integer whose digits have a sum of 1000. What is k? (A) 901 (B) 911 (C) 919 (D) 991 (E) 999 Positive integers a and b are such that the graphs of y, If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? Solution. The nth item for the sequence is: An=An-1+4n. We add increasing multiples of 4 each time we go up a figure. So, to go from Figure 0 to 100, we add. 4 *1+4*2+...+4*99+4*100=4*5050=20200., 2016 AMC 10A 7 The mean, median, and mode of the 7 data values 60,100,x,40,50,200,90 are all equal to x. What is the value of x? (A)50 (B)60 (C)75 (D)90 (E)100 8 Trickster Rabbit agrees with Foolish Fox to double Fox’s money every time Fox crosses the bridge by Rabbit’s house, as long as Fox pays 40 coins in toll to Rabbit after each crossing., Solution. We can assume there are 10 people in the class. Then there will be 1 junior and 9 seniors. The sum of everyone's scores is 10*84 = 840. Since the average score of the seniors was 83, the sum of all the senior's scores is 9 * 83 = 747. The only score that has not been added to that is the junior's score, which is 840 - 747 = 93., Back then, there was no redemption for a poor 11th grade USAMO performance, so that single score not only lost me a chance to attend IMO 2013 ... AMC 10A 2012: ..., 2011 AMC 10A. 2011 AMC 10A problems and solutions. The test was held on February 8, 2011. The first link contains the full set of test problems. The rest contain each individual problem and its solution. 2011 AMC 10A Problems. , Direct link to Daniel Chaviers's post “The AMC 10 is more about ...”. The AMC 10 is more about analysis and "abuse" of the various laws and properties of any number of things, which is seemingly unrelated. The AMC 10 has a bit more algebra than the AMC 8, would, but it's otherwise pretty similar: lot of analysis., 2021 AMC 10A problems and solutions. The test will be held on Thursday, February , . Please do not post the problems or the solutions until the contest is released. 2021 AMC 10A Problems. 2021 AMC 10A Answer Key. , If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100? Solution. The nth item for the sequence is: An=An-1+4n. We add increasing multiples of 4 each time we go up a figure. So, to go from Figure 0 to 100, we add. 4 *1+4*2+...+4*99+4*100=4*5050=20200., 2013 AMC 10A2013 AMC 10A Test with detailed step-by-step solutions for questions 1 to 10. AMC 10 [American Mathematics Competitions] was the test conducted b..., The AMC 10 is a 25 question, 75 minute multiple choice examination in secondary school mathematics containing problems which can be understood and solved with pre-calculus concepts. Calculators are not allowed starting in 2008. For the school year there will be two dates on which the contest may be taken: AMC 10A on , , , and AMC 10B on , , ., 2014 AMC 10 A Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. Created Date: 2/5/2014 12:11:46 PM, Solution 1. Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for …, Direct link to Daniel Chaviers's post "The AMC 10 is more about ...". The AMC 10 is more about analysis and "abuse" of the various laws and properties of any number of things, which is seemingly unrelated. The AMC 10 has a bit more algebra than the AMC 8, would, but it's otherwise pretty similar: lot of analysis., 2013 AMC10A Problems 4 12. In ˜ABC, AB = AC = 28 and BC = 20. Points D, E, and F are on sides AB, BC, and AC, respectively, such that DE and EF are parallel to AC and AB, respectively. What is the perimeter of parallelogram ADEF? A D B E C F (A) 48 (B) 52 (C) 56 (D) 60 (E) 72 13. How many three-digit numbers are not divisible by 5, have digits ... , 2013 AMC 10A. 2013 AMC 10A problems and solutions. The test was held on February 5, 2013. 2013 AMC 10A Problems · 2013 AMC 10A Answer Key., In base 10, the number 2013 ends in the digit 3. In base 9, on the other hand, the same number is written as (2676)9 and ends in the digit 6. For how many positive integers b does the base-b representation of 2013 end in the digit 3? (C) 13 (D) 16 (E) 18 A unit square is rotated 450 about its center. What is the area of the region swept out by