AMC美国数学竞赛2000AMC10试题及答案解析(0619101406).pdf
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1、. ; USA AMC 10 2000 1 In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . Whats the largest possible value of the sum ? Solution The sum is the highest if two factors are the lowest. So, and . 2 Solution
2、 . 3 Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally? Solution . ; 4 Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her
3、December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee? Solution Let be the fixed fee, and be the amount she pays for the minutes she used in the first month. We want the fixed fee, which is 5 Points and are the midp
4、oints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change? (a) the length of the segment (b) the perimeter of (c) the area of (d) the area of trapezoid . ; Solution (a) Clearly does not change, and , so doesnt change either. (b) Obvi
5、ously, the perimeter changes. (c) The area clearly doesnt change, as both the base and its corresponding height remain the same. (d) The bases and do not change, and neither does the height, so the trapezoid remains the same. Only quantity changes, so the correct answer is . 6 The Fibonacci Sequence
6、 starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence? Solution The pattern of the units digits are In order of appearance: . is the last. 7 In rectangle , , is on ,
7、and and trisect . What is the perimeter of ? . ; Solution . Since is trisected, . Thus, . Adding, . 8 At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true? There ar
8、e five times as many sophomores as freshmen. There are twice as many sophomores as freshmen. There are as many freshmen as sophomores. There are twice as many freshmen as sophomores. There are five times as many freshmen as sophomores. Solution . ; Let be the number of freshman and be the number of
9、sophomores. There are twice as many freshmen as sophomores. 9 If , where , then Solution , so . . . 10 The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possibl
10、e value of ? Solution From the triangle inequality, and . The smallest positive number not possible is , which is . 11 Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? Solution . ; Two prime numbers
11、 between and are both odd. Thus, we can discard the even choices. Both and are even, so one more than is a multiple of four. is the only possible choice. satisfy this, . 12 Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonov
12、erlapping unit squares would there be in figure 100? Solution Solution 1 We have a recursion: . I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add . . ; Solution 2 We can divide up figure to get the sum of the sum of the first odd numbers and the
13、sum of the first odd numbers. If you do not see this, here is the example for : The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice 13 There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a tria
14、ngular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color? . ; Solution In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go
15、 there. In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column. By similar logic, we can fill in the yellow pegs as shown: After this we can proceed to fill in
16、the whole pegboard, so there is only arrangement of the pegs. The answer is 14 Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that afte
17、r each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered? Solution . ; The sum of the first scores must be even, so we must choose evens or the odds to be the first two scores. Let us look at th
18、e numbers in mod . If we choose the two odds, the next number must be a multiple of , of which there is none. Similarly, if we choose or , the next number must be a multiple of , of which there is none. So we choose first. The next number must be 1 in mod 3, of which only remains. The sum of the fir
19、st three scores is . This is equivalent to in mod . Thus, we need to choose one number that is in mod . is the only one that works. Thus, is the last score entered. 15 Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ? Solution Substituting , we get . ; 16 The
20、 diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment . Solution Solution 1 Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and . The line is given by the equa
21、tion . The -intercept is , so . We are given two points on , hence we can compute the slope, to be , so is the line Similarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line . . ; At , the intersection point, both of the equations must be true, so We h
22、ave the coordinates of and , so we can use the distance formula here: which is answer choice Solution 2 Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which . , and , so by AA similarity, By the Pythagorean Theorem, we have , , and . Let , so , then Th
23、is is answer choice Also, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B. . ; 17 Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels;
24、 when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly? Solution Consider what happens each time he puts a coin in. If he puts in a
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