2006AMC10A真题与答案解析

2006AMC10A真题

答案解析请参考文末

Problem 1

Sandwiches at Joe's Fast Food cost  each and sodas cost  each. How many dollars will it cost to purchase 5 sandwiches and 8 sodas?

Problem 2

Define . What is ?

Problem 3

The ratio of Mary's age to Alice's age is . Alice is  years old. How many years old is Mary?

Problem 4

A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?

Problem 5

Doug and Dave shared a pizza with 8 equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half of the pizza. The cost of a plain pizza was 8 dollars, and there was an additional cost of 2 dollars for putting anchovies on one half. Dave ate all of the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each then paid for what he had eaten. How many more dollars did Dave pay than Doug?

Problem 6

What non-zero real value for  satisfies ?

Problem 7

The  rectangle  is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is ?

Problem 8

A parabola with equation  passes through the points  and . What is ?

Problem 9

How many sets of two or more consecutive positive integers have a sum of 15?

Problem 10

For how many real values of  is  an integer?

Problem 11

Which of the following describes the graph of the equation ?

Problem 12

Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.

Which of these arrangements give the dog the greater area to roam, and by how many square feet?

Problem 13

A player pays  to play a game. A die is rolled. If the number on the die is odd, the game is lost. If the number on the die is even, the die is rolled again. In this case the player wins if the second number matches the first and loses otherwise. How much should the player win if the game is fair? (In a fair game the probability of winning times the amount won is what the player should pay.)

Problem 14

A number of linked rings, each 1 cm thick, are hanging on a peg. The top ring has an outside diameter of 20 cm. The outside diameter of each of the other rings is 1 cm less than that of the ring above it. The bottom ring has an outside diameter of 3 cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?

Problem 15

Odell and Kershaw run for 30 minutes on a circular track. Odell runs clockwise at 250 m/min and uses the inner lane with a radius of 50 meters. Kershaw runs counterclockwise at 300 m/min and uses the outer lane with a radius of 60 meters, starting on the same radial line as Odell. How many times after the start do they pass each other?

Problem 16

A circle of radius 1 is tangent to a circle of radius 2. The sides of  are tangent to the circles as shown, and the sides  and  are congruent. What is the area of ?

Problem 17

In rectangle , points  and  trisect , and points  and  trisect . In addition, , and . What is the area of quadrilateral  shown in the figure?

Problem 18

A license plate in a certain state consists of 4 digits, not necessarily distinct, and 2 letters, also not necessarily distinct. These six characters may appear in any order, except that the two letters must appear next to each other. How many distinct license plates are possible?

Problem 19

How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?

Problem 20

Six distinct positive integers are randomly chosen between 1 and 2006, inclusive. What is the probability that some pair of these integers has a difference that is a multiple of 5?

Problem 21

How many four-digit positive integers have at least one digit that is a 2 or a 3?

Problem 22

Two farmers agree that pigs are worth $300 and that goats are worth $210. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a $390 debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way?

Problem 23

Circles with centers  and  have radii  and , respectively. A common internal tangent intersects the circles at  and , respectively. Lines  and  intersect at , and . What is ?

Problem 24

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

Problem 25

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

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2006AMC10A详细解析

1. The  sandwiches cost  dollars. The  sodas cost  dollars. In total, the purchase costs  dollars. The answer is .
2. By the definition of , we have . Then . The answer is .
3. Let  be Mary's age. Then . Solving for , we obtain . The answer is .We can see this is a combined ratio of , . We can equalize by doing , and . With the common ratio of  and difference ratio of , we see  therefore we can see our answer is correct.
4. From the greedy algorithm, we have  in the hours section and  in the minutes section. With a matrix we can see  The largest digit sum we can see is . For the minutes digits, we can combine the largest  digits, which are  which we can then do 
5. Dave and Doug paid  dollars in total. Doug paid for three slices of plain pizza, which cost . Dave paid  dollars. Dave paid  more dollars than Doug. The answer is .
6. Taking the seventh root of both sides, we get .Simplifying the LHS gives , which then simplifies to .Thus, , and the answer is .
7. Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is . This means the square will have four sides of length 12. The only way to do this is shown below.As you can see from the diagram, the line segment denoted as  is half the length of the side of the square, which leads to .

8. Solution 1

   Substitute the points  and  into the given equation for .

   Then we get a system of two equations:

   

   

   Subtracting the first equation from the second we have:

   

   

   Then using  in the first equation:

   

    is the answer.

   Solution 2

   Alternatively, notice that since the equation is that of a conic parabola, the vertex is likely . Thus, the form of the equation of the parabola is . Expanding this out, we find that .

   Solution 3

   The points given have the same -value, so the vertex lies on the line .

   The -coordinate of the vertex is also equal to , so set this equal to  and solve for , given that :

   

   

   

   

   Now the equation is of the form . Now plug in the point  and solve for :

   

   

   

   

   

   Solution 4

   Substituting y into the two equations, we get:

   

   Which can be written as:

   

   4, 2, are the solutions to the quadratic. Thus:

   

   

   

9. Notice that if the consecutive positive integers have a sum of 15, then their average (which could be a fraction) must be a divisor of 15. If the number of integers in the list is odd, then the average must be either 1, 3, or 5, and 1 is clearly not possible. The other two possibilities both work:
   • 
   • 

   If the number of integers in the list is even, then the average will have a . The only possibility is , from which we get:

   • 

   Thus, the correct answer is 3, answer choice .

10. For  to be an integer,  must be a perfect square.Since  can't be negative, .The perfect squares that are less than or equal to  are , so there are  values for .Since every value of  gives one and only one possible value for , the number of values of  is .
11. Expanding the left side, we haveThus there are two lines described in this graph, the horizontal line  and the vertical line . Thus, our answer is .
12. Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement I allows the dog  square feet of area. Arrangement II allows  square feet plus a little more on the top part of the fence. So we already know that Arrangement II allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is . Thus the answer is .
13. The probability of rolling an even number on the first turn is  and the probability of rolling the same number on the next turn is . The probability of winning is . If the game is to be fair, the amount paid,  dollars, must be  the amount of prize money, so the answer is 

14. Solution 1

    The inside diameters of the rings are the positive integers from 1 to 18. The total distance needed is the sum of these values plus 2 for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series, the answer is .

    Solution 2

    Alternatively, the sum of the consecutive integers from 3 to 20 is . However, the 17 intersections between the rings must be subtracted, and we also get .

    • For , x must be in the form of , where  denotes any integer.
    • For , .
15. The smallest possible value of  will be that of .
16.  and  are vertical angles so they are congruent, as are angles  and  (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, . By the Pythagorean Theorem, line segment . The sides are proportional, so . This makes  and .
17. It is not difficult to see by symmetry that  is a square.Draw . Clearly . Then  is isosceles, and is a . Hence , and .There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.Draw the lines as shown above, and count the squares. There are 12, so we have .We see that if we draw a line to  it is half the width of the rectangle so that length would be , and the resulting triangle is a  so using the Pythagorean Theorem we can get that each side is  so the area of the middle square would be  which is our answer.Since  and  are trisection points and , we see that . Also, , so triangle  is a right isosceles triangle, i.e. . By symmetry, triangles , , and  are also right isosceles triangles. Therefore, , which means triangle  is also a right isosceles triangle. Also, triangle  is a right isosceles triangle.

    Then , and . Hence, .

    By symmetry, quadrilateral  is a square, so its area is

18. There are  ways to choose 4 digits.There are  ways to choose the 2 letters.For the letters to be next to each other, they can be the 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or the 5th and 6th characters. So, there are  choices for the position of the letters.Therefore, the number of distinct license plates is .
19. The sum of the angles of a triangle is  degrees. For an arithmetic progression with an odd number of terms, the middle term is equal to the average of the sum of all of the terms, making it  degrees. The minimum possible value for the smallest angle is  and the highest possible is  (since the numbers are distinct), so there are  possibilities .
20. For two numbers to have a difference that is a multiple of 5, the numbers must be congruent  (their remainders after division by 5 must be the same). are the possible values of numbers in . Since there are only 5 possible values in  and we are picking  numbers, by the Pigeonhole Principle, two of the numbers must be congruent .Therefore the probability that some pair of the 6 integers has a difference that is a multiple of 5 is .
21. Since we are asked for the number of positive 4-digit integers with at least 2 or 3 in it, we can find this by finding the total number of 4-digit integers and subtracting off those which do not have any 2s or 3s as digits.The total number of 4-digit integers is , since we have 10 choices for each digit except the first (which can't be 0).Similarly, the total number of 4-digit integers without any 2 or 3 is .Therefore, the total number of positive 4-digit integers that have at least one 2 or 3 in their decimal representation is 

22. Solution 1

    The problem can be restated as an equation of the form , where  is the number of pigs,  is the number of goats, and  is the positive debt. The problem asks us to find the lowest x possible.  and  must be integers, which makes the equation a Diophantine equation. Bezout’s Identity tells us that the smallest  for the Diophantine equation  to have solutions is when  is the greatest common divisor of  and . Therefore, the answer is , which is , 

    Solution 2

    Alternatively, note that  is divisible by 30 no matter what  and  are, so our answer must be divisible by 30. In addition, three goats minus two pigs gives us  exactly. Since our theoretical best can be achieved, it must really be the best, and the answer is . debt that can be resolved.

    Solution 3

    Let us simplify this problem. Dividing by , we get a pig to be: , and a goat to be . It becomes evident that if you exchange  pigs for  goats, we get the smallest positive difference - , since we can't made a non-integer with a linear combination of integers. Since we originally divided by , we need to multiply again, thus getting the answer: 

23.  and  are vertical angles so they are congruent, as are angles  and  (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, . By the Pythagorean Theorem, line segment . The sides are proportional, so . This makes  and .
24. By the Multinomial Theorem, the summands can be written asandrespectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:

    terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of  and  must be opposite. Now we find a pattern:

    if the exponent of  is 1, the exponent of  can be all even integers up to 2004, so there are 1003 terms.

    if the exponent of  is 3, the exponent of  can go up to 2002, so there are 1002 terms.

    

    if the exponent of  is 2005, then  can only be 0, so there is 1 term.

    If we add them up, we get  terms. However, we can switch the exponents of  and  and these terms will still have a negative sign. So there are a total of  negative terms.

    By subtracting this number from 2015028, we obtain  or  as our answer.

    Solution 2

    Alternatively, we can use a generating function to solve this problem. The goal is to find the generating function for the number of unique terms in the simplified expression (in terms of ). In other words, we want to find  where the coefficient of  equals the number of unique terms in .

    First, we note that all unique terms in the expression have the form, , where  and  is some constant. Therefore, the generating function for the MAXIMUM number of unique terms possible in the simplified expression of  is

    Secondly, we note that a certain number of terms of the form, , do not appear in the simplified version of our expression because those terms cancel. Specifically, we observe that terms cancel when  because every unique term is of the form:for all possible .

    Since the generating function for the maximum number of unique terms is already known, it is logical that we want to find the generating function for the number of terms that cancel, also in terms of . With some thought, we see that this desired generating function is the following:

    Now, we want to subtract the latter from the former in order to get the generating function for the number of unique terms in , our initial goal:which equals

    The coefficient of  of the above expression equals

    Evaluating the expression, we get , as expected.

    Solution 3

    Define  such that . Then the expression in the problem becomes: .

    Expanding this using binomial theorem gives , where  (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).

    Simplifying gives: . Note that two terms that come out of different powers of  cannot combine and simplify, as their exponent of  will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem,  will have  terms, so the answer is .

    Solution 4

    Using stars and bars we know that  has  or  different terms. Now we need to find out how many of these terms are canceled out by . We know that for any term(let's say ) where  of the expansion of  is only going to cancel out with the corresponding term  if only  is odd and  is even or  is even and  is odd. Now let's do some casework to see how many terms fit this criteria:

    Case 1:  is even

    Now we know that  is even and . Thus  is also even or both  and  are odd or both  and  are even. This case clearly fails the above criteria and there are 0 possible solutions.

    Case 2:  is odd

    Now we know that  is odd and . Thus  is odd and  is odd and  is even or  is even and  is odd. All terms that have  being odd work.

    We now need to figure out how many of the terms have  as a odd number. We know that  can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus  of the  terms will cancel out and  of the terms will work. Thus there are  terms. This number comes out to be   (Author: David Camacho)

25. Solution 1

    This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:

    How many ways are there to choose  elements from an ordered  element set without choosing two consecutive members?

    You want to choose  numbers out of  with no consecutive numbers. For each configuration, we can subtract  from the -th element in your subset. This converts your configuration into a configuration with  elements where the largest possible element is , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection. Without consideration of the second condition, we have: 

    Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than , which translates to subtracting  from the "top" of each binomial coefficient. Now we have, after we cancel all the terms  where , 

    Solution 2

    Another way of visualizing the solution above would be to use 's and 's. Denote  as the numbers we have chosen and  as other numbers. Taking an example, assuming we are picking two numbers, we imagine the shape . This notation forces a number between the two chosen numbers, which blocks the two numbers we picked from being consecutive. Now we consider the orientations with this shape. We have  remaining numbers.

    We need to find the number of ways to place the remaining 's. We can find this by utilizing stars and bars, with the following marker being placed to represent groups: *| - *|*. Now, we have to place  numbers within  groups, which is . The same concept can be used for the remaining numbers. The rest of the solution continues as above.

    Solution by: Everyoneintexas

    Solution 3

    We have the same setup as in the previous solution.

    Note that if , the answer will be 0. Otherwise, the  elements we choose define  boxes (which divide the nonconsecutive numbers) into which we can drop the  remaining elements, with the caveat that each of the middle  boxes must have at least one element (since the numbers are nonconsecutive). This is equivalent to dropping  elements into  boxes, where each box is allowed to be empty. And this is equivalent to arranging  objects,  of which are dividers, which we can do in  ways.

    Now, looking at our original question, we see that the thing we want to calculate is just 

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