1、- 1 -2003 AMC 10A1、What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers? Solution The first even counting numbers are . The first odd counting numbers are . Thus, the problem is asking for the value of . . Alternatively, using the su
2、m of an arithmetic progression formula, we can write . 2、Members of the Rockham Soccer League buy socks and T-shirts. Socks cost per pair and each T-shirt costs more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away ga
3、mes. If the total cost is , how many members are in the League? Solution Since T-shirts cost dollars more than a pair of socks, T-shirts cost dollars. Since each member needs pairs of socks and T-shirts, the total cost for member is dollars. Since dollars was the cost for the club, and was the cost
4、per member, the number of members in the League is . - 2 -3、A solid box is cm by cm by cm. A new solid is formed by removing a cube cm on a side from each corner of this box. What percent of the original volume is removed? Solution The volume of the original box is The volume of each cube that is re
5、moved is Since there are corners on the box, cubes are removed. So the total volume removed is . Therefore, the desired percentage is . 4、It takes Mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school to home along the same route. What is her avera
6、ge speed, in km/hr, for the round trip? Solution Since she walked km to school and km back home, her total distance is km. Since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours. Therefore her average speed in km/hr is 5、Let and denote the solution
7、s of . What is the value of ? Solution Using factoring: - 3 -or So and are and . Therefore the answer is OR we can use sum and product. 6、Define to be for all real numbers and . Which of the following statements is not true? Solution Examining statement C: when , but statement C says that it does fo
8、r all . Therefore the statement that is not true is “ for all “ Alternatively, consider that the given “heart function“ is actually the definition of the distance between two points. Examining all of the statements, only C is not necessarily true; if c is negative, the distance between and is the ab
9、solute value of , not itself, because distance is always nonnegative. 7、How many non-congruent triangles with perimeter have integer side lengths? Solution - 4 -By the triangle inequality, no one side may have a length greater than half the perimeter, which is Since all sides must be integers, the l
10、argest possible length of a side is Therefore, all such triangles must have all sides of length , , or . Since , at least one side must have a length of Thus, the remaining two sides have a combined length of . So, the remaining sides must be either and or and . Therefore, the number of triangles is
11、 . 8、What is the probability that a randomly drawn positive factor of is less than ? Solution For a positive number which is not a perfect square, exactly half of the positive factors will be less than . Since is not a perfect square, half of the positive factors of will be less than . Clearly, ther
12、e are no positive factors of between and . Therefore half of the positive factors will be less than . So the answer is . 9、Simplify Solution . Therefore: - 5 -10、The polygon enclosed by the solid lines in the gure consists of congruent squares joined edge-to-edge. One more congruent square is attach
13、ed to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? Solution Let the squares be labeled , , , and . When the polygon is folded, the “right“ edge of square becomes adjacent to the “bottom edge“ of square , a
14、nd the “bottom“ edge of square becomes adjacent to the “bottom“ edge of square . So, any “new“ square that is attached to those edges will prevent the polygon from becoming a cube with one face missing. Therefore, squares , , and will prevent the polygon from becoming a cube with one face missing. S
15、quares , , , , , and will allow the polygon to become a cube with one face missing when folded. - 6 -Thus the answer is . Another way to think of it is that a cube missing one face has of its faces. Since the shape has faces already, we need another face. The only way to add anopther face is if the
16、added square does not overlap any of the others. , , and overlap, while 9 do not. The answer is 11、The sum of the two -digit numbers and is . What is ? Solution Since , , and are digits, , , . Therefore, . 12、A point is randomly picked from inside the rectangle with vertices , , , and . What is the
17、probability that ? Solution The rectangle has a width of and a height of . The area of this rectangle is . The line intersects the rectangle at and . - 7 -The area which is the right isosceles triangle with side length that has vertices at , , and . The area of this triangle is Therefore, the probab
18、ility that is 13、The sum of three numbers is . The rst is times the sum of the other two. The second is seven times the third. What is the product of all three? Solution Solution 1 Let the numbers be , , and in that order. The given tells us that Therefore, the product of all three numbers is . Solu
19、tion 2 Alternatively, we can set up the system in matrix form: - 8 -Or, in matrix form To solve this matrix equation, we can rearrange it thus: Solving this matrix equation by using inverse matrices and matrix multiplication yields Which means that , , and . Therefore, 14、Let be the largest integer
20、that is the product of exactly distinct prime numbers, , , and , where and are single digits. What is the sum of the digits of ? Solution Since is a single digit prime number, the set of possible values of is . Since is a single digit prime number and is the units digit of the prime number , the set
21、 of possible values of is . Using these values for and , the set of possible values of is Out of this set, the prime values are Therefore the possible values of are: - 9 -The largest possible value of is . So, the sum of the digits of is 15、What is the probability that an integer in the set is divis
22、ible by and not divisible by ? Solution There are integers in the set. Since every 2nd integer is divisible by , there are integers divisible by in the set. To be divisible by both and , a number must be divisible by . Since every 6th integer is divisible by , there are integers divisible by both an
23、d in the set. So there are integers in this set that are divisible by and not divisible by . Therefore, the desired probability is 16、What is the units digit of ? Solution Since : Therefore, the units digit is - 10 -17、The number of inches in the perimeter of an equilateral triangle equals the numbe
24、r of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Solution Let be the length of a side of the equilateral triangle and let be the radius of the circle. In a circle with a radius the side of an inscribed equilateral triangle is . So . The perime
25、ter of the triangle is The area of the circle is So: 18、What is the sum of the reciprocals of the roots of the equation Solution Multiplying both sides by : Let the roots be and . The problem is asking for By Vietas formulas: - 11 -So the answer is . 19、A semicircle of diameter sits at the top of a
26、semicircle of diameter , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune. Solution The shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area
27、of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle. The area of the smaller semicircle is . Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures
28、 . The area of the sector of the larger semicircle is . - 12 -The area of the triangle is So the shaded area is 20、A base- three-digit number is selected at random. Which of the following is closest to the probability that the base- representation and the base- representation of are both three-digit
29、 numerals? Solution To be a three digit number in base-10: Thus there are three-digit numbers in base-10 To be a three-digit number in base-9: To be a three-digit number in base-11: So, Thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11. Therefore the desi
30、red probability is . 21、Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected? - 13 -Solution Solution 1 Le
31、t the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies. Using casework: Pat selects chocolate chip cookies: Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate
32、 chip cookie: Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate chip cookies: Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate chip cooki
33、es: Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate chip cookies: - 14 -Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate chip cookies:
34、Pat needs to select more cookies that are either oatmeal or peanut butter. The assortments are: assortments. Pat selects chocolate chip cookies: Pat needs to select more cookies that are either oatmeal or peanut butter. The only assortment is: assortment. The total number of assortments of cookies t
35、hat can be collected is Solution 2 It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number
36、of ways to do this is 22、In rectangle , we have , , is on with , is on with , line intersects line at , and is on line with . Find the length . - 15 -Solution S olution 1 (Opposite angles are equal). (Both are 90 degrees). (Alt. Interior Angles are congruent). Therefore and are similar. and are also
37、 similar. is 9, therefore must equal 5. Similarly, must equal 3. Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, . . So . . - 16 -Therefore . Solution 2 Since is a rectangle, . Since is a rectangle and , . Since is a rectangle, . So, is
38、a transversal, and . This is sufficient to prove that and . Using ratios: Since cant have 2 different lengths, both expressions for must be equal. - 17 -Solution 3 Since is a rectangle, , , and . From the Pythagorean Theorem, . Lemma Statement: Proof: , obviously. Since two angles of the triangles a
39、re equal, the third angles must equal each other. Therefore, the triangles are similar. Let . Also, , therefore We can multiply both sides by to get that is twice of 10, or Solution 4 We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let
40、GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH - 18 -and GEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B. 23、A large equilateral triangle is constructed by using toothpick
41、s to create rows of small equilateral triangles. For example, in the gure we have rows of small congruent equilateral triangles, with small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of small equil
42、ateral triangles? Solution Solution 1There are small equilateral triangles. Each small equilateral triangle needs toothpicks to make it. But, each toothpick that isnt one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles. So, the number of t
43、oothpicks on the inside of the large equilateral triangle is Therefore the total number of toothpicks is Solution 2We see that the bottom row of small triangles is formed from downward-facing triangles and upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, a
44、nd since the total number of downward-facing triangles is - 19 -, we have that the total number of toothpicks is 24、Sally has ve red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly
45、into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards? Solution Let and designate the red card numbered and the blue card numbered , respectively. is the only blue card that evenly divides, so must be at one end of the stack and must be the card next
46、 to it. is the only other red card that evenly divides , so must be the other card next to . is the only blue card that evenly divides, so must be at the other end of the stack and must be the card next to it. is the only other red card that evenly divides , so must be the other card next to . doesn
47、t evenly divide , so must be next to , must be next to , and must be in the middle. This yields the following arrangement from top to bottom: Therefore, the sum of the numbers on the middle three cards is . 25、Let be a -digit number, and let and be the quotient and remainder, respectively, when is d
48、ivided by . For how many values of is divisible by ? Solution - 20 -Solution 1 When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, . Therefore, can be any integer from to inclusive, and can be any integer from to inclusive. For each
49、of the possible values of , there are at least possible values of such that . Since there is “extra“ possible value of that is congruent to , each of the values of that are congruent to have more possible value of such that . Therefore, the number of possible values of such that is . Solution 2 Let equal , where through are digits. Therefore, We now take : The divisor trick for 11 is as follows: “Let be an digit integer. If is divisible by , then is also divisib
