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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
25、 quarter, he gets five nickels back, so the amount of money he has doesnt change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesnt change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents. Th
26、is implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is 18 Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in
27、 all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number? Solution The area she sees looks at follows: . ; The part inside the walk has area . The part outside the walk consists of
28、 four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is , which rounded to the nearest integer is . 19 Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of
29、the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is Solution . ; Let the square have area , th
30、en it follows that the altitude of one of the triangles is . The area of the other triangle is . By similar triangles, we have This is choice (Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that sca
31、ling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.) 20 Let , , and be nonnegative integers such that . What is the maximum value of ? Solution The trick is to realize that the
32、sum is similar to the product . If we multiply , we get . We know that , therefore . Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum. Suppose that some two of , , and differ by at least . Then this triple is surely not optimal. . ; Proof: WLOG let . We
33、 can then increase the value of by changing and . Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is . 21 If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?
34、 I. All alligators are creepy crawlers. II. Some ferocious creatures are creepy crawlers. III. Some alligators are not creepy crawlers. Solution We interpret the problem statement as a query about three abstract concepts denoted as “alligators“, “creepy crawlers“ and “ferocious creatures“. In answer
35、ing the question, we may NOT refer to reality - for example to the fact that alligators do exist. To make more clear that we are not using anything outside the problem statement, lets rename the three concepts as , , and . We got the following information: If is an , then is an . There is some that
36、is a and at the same time an . We CAN NOT conclude that the first statement is true. For example, the situation “Johnny and Freddy are s, but only Johnny is a “ meets both conditions, but the first statement is false. We CAN conclude that the second statement is true. We know that there is some that
37、 is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an . ; . Bobby is an , therefore Bobby is an . And this is enough to prove the second statement - Bobby is an that is also a . We CAN NOT conclude that the third statement is true. For exam
38、ple, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false. Therefore the answer is . 22 One morning each member of Angelas family drank an -ounce mixture of coffee with milk. The amounts of
39、coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family? Solution The exact value “8 ounces“ is not important. We will only use the fact that each member of the family
40、 drank the same amount. Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids. Let be the number of family members. Then each family member drank ounces of fluids. We know that Angela drank ounc
41、es of fluids. As Angela is a family member, we have . Multiply both sides by to get . If , we have . If , we have . . ; Therefore the only remaining option is . 23 When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is t
42、he sum of all possible real values of ? Solution As occurs three times and each of the three other values just once, regardless of what we choose the mode will always be . The sum of all numbers is , therefore the mean is . The six known values, in sorted order, are . From this sequence we conclude:
43、 If , the median will be . If , the median will be . Finally, if , the median will be . We will now examine each of these three cases separately. In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression. In the case we have , because . Theref
44、ore our three values in order are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore the third term must be . Solving we get the only solution for this case: . The case remains. Once again, we have , therefore the order is . The only solution
45、is when , i. e., . The sum of all solutions is therefore . . ; 24 Let be a function for which . Find the sum of all values of for which . Solution In the definition of , let . We get: . As we have , we must have , in other words . One can now either explicitly compute the roots, or use Vietas formul
46、as. According to them, the sum of the roots of is . In our case this is . (Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .) 25 In year , the day of the ye
47、ar is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur? Solution Clearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem. Let be the day of year , the day of year and the day of year . If year is not a l
48、eap year, the day will be days after . As , that would be a Monday. . ; Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a . (Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have =Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)