2006年2007 美国AMC10

amc10评分标准AMC10是美国数学竞赛的一部分，旨在评估初中学生在数学领域的能力。

在这项竞赛中，学生将面临一系列挑战性的数学问题。

为了公正地评估学生的表现，AMC10有一套评分标准，本文将介绍这些标准。

一、选择题评分标准AMC10中的多项选择题占据了竞赛的大部分内容。

这些题目提供了若干个选项供学生选择，需要学生选出正确答案。

选择题的评分标准如下：1.正确答案：如果选手选择了正确的答案，将得到4分。

2.错误答案：如果选手选择了错误的答案，不得分也不扣分。

3.未解答：如果选手未选择任何答案，将不得分也不扣分。

需要注意的是，选择题没有部分得分。

学生的分数完全取决于他们的选择是否正确。

二、填空题评分标准AMC10中的填空题考察学生解答数学问题的能力，学生需要计算出准确的答案并填入相应的空白中。

填空题的评分标准如下：1.正确答案：如果选手填入了正确的答案，将得到6分。

2.错误答案：如果选手填入了错误的答案，不得分也不扣分。

3.未解答：如果选手未填入任何答案，将不得分也不扣分。

填空题同样没有部分得分，只有完全正确才能得到分数。

三、证明题评分标准AMC10中的证明题考察学生证明数学命题的能力，学生需要利用所学的数学知识和推理能力，给出合理的证明过程。

证明题的评分标准如下：1.正确证明：如果学生给出了完全正确的证明过程，将得到7分。

2.部分正确：如果学生给出了部分正确的证明过程，将根据证明的准确性和完整性给予部分分数。

3.未解答：如果选手未给出任何证明过程，将不得分也不扣分。

需要注意的是，证明题可以有部分得分，学生的分数取决于他们的证明过程的准确性和完整性。

四、解答题评分标准AMC10中的解答题考察学生解决复杂数学问题的能力，学生需要给出详细的解题步骤和答案。

解答题的评分标准如下：1.完整解答：如果学生给出了完整的解题过程和正确的答案，将得到10分。

2.部分解答：如果学生给出了部分解题过程或者答案，将根据解题的准确性和完整性给予部分分数。

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6@2 中文：定义 a @ b=ab-b2 ，a # b=a + b - ab2。求 的值。 6#2 3. An aquarium has a rectangular base that measures 100cm by 40cm and has a height of 50cm. It is filled with water to a height of 40cm. A brick with a rectangular base that measures 40cm by 20cm and a height of 10cm is placed in the aquarium. By how many centimeters does that water rise? (A) 0.5 (B) 1 (C) 1.5 (D) 2 (E)2.5 中文：一个养鱼缸有 100cm×40cm 的底，高为 50cm。它装满水到 40cm 的高度。把一 个底为 40cm×20cm，高为 10cm 的砖块放在这个养鱼缸里。鱼缸里的水上升了多少厘 米？ 4. The larger of two consecutive odd integers is three times the smaller. What is their sum? (A) 4 (B) 8 (C) 12 (D) 16 (E) 20 中文：2 个连续的奇整数中较大的数是较小的数的 3 倍。它们的和是多少？ 5. A school store sells 7 pencils and 8 notebooks for $4.15. It also sells 5 pencils and 3 notebooks for $1.77. How much do 16 pencils and 10 notebooks cost? (A) $4.76 (B) $5.84 (C) $6.00 (D) $6.16 (E) $6.32 中文：在一个学校商店花 4.15 美元可以买到 7 支铅笔和 8 个笔记本，花 1.77 美元可以 买到 5 个铅笔和 3 个笔记本。问 16 支铅笔和 10 个笔记本需要多少钱？ 6. At Euclid High School, the number of students taking the AMC10 was 60 in 2002, 66 in 2003, 70 in 2004, 76 in 2005, and 78 in 2006, and is 85 in 2007. Between what two consecutive years was there the largest percentage increase? (A) 2002 and 2003 (B) 2003 and 2004 (C) 2004 and 2005

2007 AMC 10A ProblemsProblem 1One ticket to a show costs $20 at full price. Susan buys 4 tickets using a coupon that gives her a 25% discount. Pam buys 5 tickets using a coupon that gives her a 30% discount. How many more dollars does Pam pay than Susan?（A ）2 （B ）5 （C ）10 （D ）15 （E ）20Problem 2Defineand . What is ? （A ）21- （B ）41- （C ）81 （D ）41 （E ）21 Problem 3An aquarium has a rectangular base that measures 100 cm by 40 cm and has a height of 50 cm. It is filled with water to a height of 40 cm. A brick with a rectangular base that measures 40 cm by 20 cm and a height of 10 cm is placed in the aquarium. By how many centimeters does the water rise?（A ）0.5 （B ）1 （C ）1.5 （D ）2 （E ）2.5Problem 4The larger of two consecutive odd integers is three times the smaller. What is their sum?（A ）4 （B ）8 （C ）12 （D ）16 （E ）20Problem 5A school store sells 7 pencils and 8 notebooks for $4.15. It also sells 5 pencils and 3 notebooks for $1.77. How much do 16 pencils and 10 notebooks cost?（A ）$1.76 （B ）$5.84 （C ）$6.00 （D ）$6.16 （E ）$6.32Problem 6At Euclid High School, the number of students taking the AMC 10 was 60 in 2002, 66 in 2003, 70 in 2004, 76 in 2005, 78and 2006, and is 85 in 2007. Between what two consecutive years was there the largest percentage increase?（A ）2002 and 2003 （B ）2003 and 2004 （C ）2004 and 2005（D ）2005 and 2006 （E ）2006 and 2007Problem 7Last year Mr. Jon Q. Public received an inheritance. He paid 20% in federal taxes on the inheritance, and paid 10% of what he had left in state taxes. He paid a total of 10500 for both taxes. How many dollars was his inheritance?（A ）30000 （B ）32500 （C ）35000 （D ）37500 （E ）40000 Problem 8Triangles ABC and ADC are isosceles with AB=BC and AD=DC . Point D is inside triangle ABC , angle ABC measures 40 degrees, and angle ADC measures 140 degrees. What is the degree measure of angle BAD ?（A ）20 （B ）30 （C ）40 （D ）50 （E ）60Real numbers and satisfy the equations 2813+=b a and 35125-=a b . What is ?（A ）-60 （B ）-17 （C ）9 （D ）12 （E ）60 Problem 10The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is 20, the father is 48 years old, and the average age of the mother and children is 16. How many children are in the family?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6 Problem 11The numbers from 1 to 8 are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?（A ）14 （B ）16 （C ）18 （D ）20 （E ）24 Problem 12Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?（A ）56 （B ）58 （C ）60 （D ）62 （E ）64 Problem 13Yan is somewhere between his home and the stadium. To get to the stadium he can walk directly to the stadium, or else he can walk home and then ride his bicycle to the stadium. He rides 7 times as fast as he walks, and both choices require the same amount of time. What is the ratio of Yan's distance from his home to his distance from the stadium?（A ）32 （B ）43 （C ）54 （D ）65 （E ）87 Problem 14A triangle with side lengths in the ratio 3:4:5 is inscribed in a circle with radius 3. What is the area of the triangle?（A ）8.64 （B ）12 （C ）π5 （D ）17.28 （E ）18 Problem 15Four circles of radius 1 are each tangent to two sides of a square and externally tangent to a circle of radius 2, as shown. What is the area of the square?（A ）32 （B ）21222+ （C ）31616+ （D ）48 （E ）21636+ Problem 16Integers c b a ,, and d , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that bc ad - is even?（A ）83 （B ）167 （C ）21 （D ）169 （E ）85Suppose that m and are positive integers such that 375n m = . What is the minimumpossible value of n m +?（A ）15 （B ）30 （C ）50 （D ）60 （E ）5700 Problem 18Consider the 12-sided polygon ABCDEFGHIJKL , as shown. Each of its sides has length 4, and each two consecutive sides form a right angle. Suppose that AG and CH meet at M . What is the area of quadrilateral ABCM ?（A ）344 （B ）16 （C ）588 （D ）20 （E ）362 Problem 19A paint brush is swept along both diagonals of a square to produce the symmetric painted area, as shown. Half the area of the square is painted. What is the ratio of the side length of the square to the brush width?（A ）122+ （B ）23 （C ）222+ （D ）123+ （E ）223+ Problem 20Suppose that the number satisfies the equation 14-+=a a . What is the value of44-+a a ? （A ）164 （B ）172 （C ）192 （D ）194 （E ）212 Problem 21A sphere is inscribed in a cube that has a surface area of 24 square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?（A ）3 （B ）6 （C ）8 （D ）9 （E ）12 Problem 22A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might begin with the terms 247, 475, and 756 andend with the term 824. Let be the sum of all the terms in the sequence. What is thelargest prime factor that always divides ?（A ）3 （B ）7 （C ）13 （D ）37 （E ）43How many ordered pairs ),(n m of positive integers, with n m ≥, have the property that their squares differ by 96?（A ）3 （B ）4 （C ）6 （D ）9 （E ）12 Problem 24 Circles centered at A and B each have radius 2, as shown. Point O is the midpoint of AB , and 22=OA . Segments OC and OD are tangent to the circles centered at A and B, respectively, and EF is a common tangent. What is the area of the shaded region ECODF ?（A ）328 （B ）π--428 （C ）24 （D ）824π+ （E ）2228π-- Problem 25For each positive integer , let )(n S denote the sum of the digits of For how manyvalues of is 2007))(()(=++n S S n S n ?（A ）1 （B ）2 （C ）3 （D ）4 （E ）52007 AMC 12A SolutionsProblem 1P = the amount Pam spent S = the amount Susan spent70)75.020(5=∙∙=P 60)75.020(4=∙=SProblem 221168)2()6()2()6()2()2()6(2#62@622-=-=∙-+-⨯= Problem 3The volume of the brick is 40*20*10=8000. Thus the water volume rose 8000=100*40*h ,so h =2 .Problem 4Let the two consecutive odd integers be 2,+a a . Then a a 32=+, so 1=a and theirsum is 4Problem 5We let =p cost of pencils in cents, =n number of notebooks in cents. Then2075403541587=+⇒=+n p n p 1239213517735=+⇒=+n p n pSubtracting these equations yields 4483619=⇒=n n . Backwards solving gives 9=p .Thus the answer is .5841016=+n p .Problem 6We compute the percentage increases: 1.%10606066=- 2. %6666670≈- 3. %6.8707076≈- 4. %6.2767678=- 5. %9787885≈-In fact, the answer follows directly from examining the differences between each year. The largest differences are 6,6,7, and it is easy to see that due to the decreased starting number of students in 2002 that that will be our answer.Problem 7After paying his taxes, he has 72.09.0*8.0= of the inheritance left. Since 10500 is0.28 of the inheritance, the whole inheritance is 3750028.010500=. Problem 8Solution 1We angle chase, and find out that:202140180=-=DAC 70240180=-=BAC BAD=BAC-DAC=50 Solution 2Since triangle ABC is isosceles we know that angle ∠BAC =∠BCA .Also since triangle ADC is isosceles we know that ∠DAC =∠DCA .This implies that ∠BAD =∠BCD .Then the sum of the angles in quadrilateral ABCD is 40+220+2∠BAD =360.Solving the equation we get ∠BAD=50. Therefore the answer is (D).Problem 9843381)2(42+=⇒==++b a a b b And b a a b b 335512533=-⇒==-Substitution gives 53384-=⇒=-+b b b , and solving for yields -12. So 60=ab Problem 10 Let be the number of children. Then the total ages of the family is )1(1648++n , and the total number of people in the family is 2+n . So6641640202)1(164820=⇒+=+⇒+++=n n n n n Problem 11The sum of the numbers on the top face of a cube is equal to the sum of the numbers on the bottom face of the cube; these 8 numbers represent all of the vertices of the cube.Thus the answer is 1828...21=+++ . Problem 12 Each tourist has to pick in between the 2 guides, so for 6 tourists there are 62 possiblegroupings. However, since each guide must take at least one tourist, we subtractthe 2 cases where a guide has no tourist. Thus the answer is 62226=-. Problem 13Let the distance from Yan's initial position to the stadium be and the distance from Yan's initial position to home be . We are trying to find a b /, and we have the following identity given by the problem:b a b a b a 867=⇒++= Thus 4/38/6/==a b and the answer is 43Problem 14Since 3-4-5 is a Pythagorean triple, the triangle is a right triangle. Since the hypotenuse is a diameter of the circumcircle, the hypotenuse is 62=r . Then the other legs are24/5=4.8 and 18/5=3.6. The area is 64.826.38.4=∙ Problem 15 Solution 1The diagonal has length 226212212+=+++++. Therefore the sides havelength 232+, and the area is 21222)232(2+=+=A Solution 2Extend two radii from the larger circle to the centers of the two smaller circles above. This forms a right triangle of sides 23,3,3. The length of the hypotenuse of the right triangle plus twice the radius of the smaller circle is equal to the side of the square. It follows, then 21222)232(2+=+=AProblem 16The only times when bc ad - is even is when ad and bc are of the same parity. Thechance of ad being odd is 412121=∙, so it has a 43 probability of being even. Therefore,the probability that will be even is 85)43()41(22=+ . Problem 17m 253∙must be a perfect cube, so each power of a prime in the factorization for m 253∙ must be divisible by 3. Thus the minimum value of is 45532=∙, whichmakes 1553333=∙=n . These sum to 60Problem 18Solution 1We can obtain the solution by calculating the area of rectangle ABGH minus the combined area of triangles △AHG and △CGM .We know that triangles △AMH and △GMC are similar because CG AH ||. Also, since AH/CG =3/2, the ratio of the distance from M to AH to the distance from M to CG is also 3/2. Solving with the fact that the distance from AH to CG is 4, we see that the distance from M to CG is 8/5.The area of △AHG is simply 2412421=∙∙, the area of △CGM . Is 53285821=∙∙, and the area of rectangle ABGH is 4*12=48.Taking the area of rectangle ABGH and subtracting the combined area of △AHG and△CGM yields 588)53224(48=+- . Solution 2Extend AB and CH and call their intersection N .The triangles CBN and CGH are clearly similar with ratio 1:2, hence BN =2 and thus AN =6 . The area of the triangle BCN is (2·4)/2=4 .The triangles MAN and MGH are similar as well, and we now know that the ratio of their dimensions is AN:GH =6:4=3:2 .Draw altitudes from M onto AN and GH , let their feet be M 1 and M 2 . We get that 2:3:21=MM MM . Hence 53612531=∙=MM . Then the area of AMN is 5108211=∙∙MM AN , and the area of ABCM can be obtained by subtracting the area of BCN , which is 4. Hence the answer is 58845108=-. Problem 19Solution 1Without loss of generality , let the side length of the square be 1 unit. The area of the painted area is 1/2 of the area of the larger square, so the total unpainted area is also 1/2. Each of the 4 unpainted triangle has area 1/8. It is easy to tell that these triangles are isosceles right triangles, so let be the side length of one of the smaller triangles:218122=⇒=a a The diagonal of the triangle is 22. The corners of the painted areas are also isosceles right triangles with side length 42212122-=-. Its diagonal is equal to the width of the paint, and is 2122-. The answer we are looking for is thus 21221-. Multiply the numerator and the denominator by 2122+ to simplify, and you get 41422122-+ whichis .Solution 2Again, have the length of the square equal to 1 and let the width of each individual stripe be . Note that you can split each stripe into two rectangles and two isosceles righttriangles at the corners. Then the area of each stripe is 22)22(2n n n n -=-. The area covered by the two total stripes is twice the area of one stripe, minus the area in the intersection of the stripes, which is a square with side length . This area is equalto 1/2 So: 0212222122222=-+-⇒=-n n n n . By the quadratic formula , 212)2(2))(2(4)22(22212±=⇒----±-=n n It's easy to tell that 212+ is too large, so 212-=n . We want to find n 1, and 1221-=n .Multiply the numerator and the denominator by 12+, 222)12)(12()12(21+=+-+=n Problem 20Solution 1Notice that 2)(222++=+--k k k k a a a a .Thus 1942]2)[(2)(22122244=--+=-+=+---a a a a a a Solution 2 142+=a a . We apply the quadratic formula to get 322124±=±=a . Thus 444444)32()32()32(1)32(-++=+++=+-a a(so it doesn't matter which root of we use). Using thebinomial theorem we can expand this out and collect terms to get 194.Solution 3We know that 41=+a a . We can square both sides to get 162122=++a a , So 14122=+a a . Squaring both sides again gives 1961421244==++a a , So 194144=+aa . Q. E. D. Solution 4We let a and a /1 be roots of a certain quadratic. Specifically 0142=+-x x . We use Newton's Sums given the coefficients to find S 4. S 4=194.Problem 21Solution 1We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.Let S be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length 3s where is a side of the cube, the ratio of a side of the inner square to that of theouter square (and the side of the outer square = the diagonal of the inner square), wehave 2)31(24=S . Thus S =8 . Solution 2 (computation)The area of each face of the outer cube is 24/6=4, and the edge length of the outer cube is 2. This is also the diameter of the sphere, and thus the length of a long diagonal of the inner cube.A long diagonal of a cube is the hypotenuse of a right triangle with a side of the cube and a face diagonal of the cube as legs. If a side of the cube is , we see that32)2(222=⇒+=x x x . Thus the surface area of the inner cube is 8)32(6622==x . Problem 22A given digit appears as the hundreds digit, the tens digit, and the units digit of a term the same number of times. Let k be the sum of the units digits in all the terms. Then S =111k =3*37k , so S must be divisible by 37 (D ). To see that it need not be divisible by any larger prime, the sequence 123, 231, 312 gives S=666=2·32·37.Problem 233296))((522∙==-+=-n m n m n mFor every two factors 96=xy , we have 2,2,y x n y x m y n m x n m -=+=⇒=-=+ Since 0,0>-≥+>≥y x y x n m , from which it follows that the number of ordered pairs ),(n m is given by the number of ordered pairs 0,96:),(>>=y x xy y x . There are 12)11)(15(=++ factors of 96, which give us six pairs ),(y x . However, since n m , are positive integers, we also need that 2/)(y x +, 2/)(y x -are positive integers, so x and y must have the same parity . Thus we exclude the factors )32,3)(96,1(),(=y x ,and we are left with four pairs Problem 24The area we are trying to find is simply ??Undefined control sequence.). Obviously, AB EF ||. Thus, ABFE is a rectangle, and so its area is28)22(22)(2=⨯=+⨯=⨯OB AO h bSince OC is tangent to circle A , △ACO is a right △. We know 22=AO and AC =2 , so △ACO is isosceles, a 45-45 right △, and has CO with length 2. The area of△ACO=221=bh . By symmetry, , and so the area of △BDO is also 2.?? (Error compiling ?? Undefined control sequence.) (or ??) Undefined control sequence.), for that matter) is 1/8 the area of its circle. Thus ?? Undefined control sequence.) and ?? (Error compiling ?? Undefined control sequence.) both have an area of2/π. Plugging all of these areas back into the original equation yields)4(28)2222(28πππ+-=+++- Problem 25Solution 1For the sake of notation let ))(()()(n S S n S n n T ++=. Obviously 2007<n . Then the maximum value of ))(()(n S S n S + is when 1999=n , and the sum becomes 28+10=38. So the minimum bound is 1969. We do casework upon the tens digit:Case 1: 9196=⇒u u . Easy to directly disprove.Case 2: u 197. u u n S +=+++=17791)(, And u n S S +=8))((, if 2≤uAnd 1)3(2))((-=-+=u u n S S otherwise.Subcase a:42007319958171970)(=⇒=+=+++++=u u u u u n T . This exceeds our bounds, so no solution here.Subcase b: 72007319861171970)(=⇒=+=-++++=u u u u u n T . First solution. Case 3:u 198. u n S +=18)(, and u n S S +=9))(( if 1≤u and u u =-+)2(2 otherwise.Subcase a: 02007320079181980)(=⇒=+=+++++=u u u u u n TSecond solution.Subcase b: 3200731998181980)(=⇒=+=++++=u u u u u n T .Third solution.Case 4: u 199. But 19)(>n S , and )(n S n + clearly sum to >2007.Case 5: u 200. So u n S +=2)( and u n S S +=2))(( (recall that 2007<n ), and 1200732004222000=⇒=+=+++++u u u u u . Fourth solution.In total we have 4 (D) solutions, which are 1977, 1980, 1983, and 2001.Solution 2Clearly, 1950>n . We can break this into three cases:Case 1: 2000>nInspection gives 2001=n .Case 2: 2000<n , xy n 19=(not to be confused with 10),**19(<+y x y xIf you set up an equation, it reduces to 324=+y xwhich has as its only solution satisfying the constraints 0,8==y x .Case 3: 2000<n , xy n 19=, 10≥+y xThis reduces to 354=+y x . The only two solutions satisfying the constraints for this equation are 7,7==y x and 3,8==y x .The solutions are thus 1997, 1980, 1983, 2001 4.Solution 3As in Solution 1, we note that 28)(≤n S and 10))((≤n S S . Obviously,)9))(mod (()(n S S n S n ≡≡As 2007≡0 (mod9), this means that n mod }6,3,0{9∈, or equivalently that)3(mod 0))(()(≡≡≡n S S n S n .Thus }9,6,3{))((∈n S S . For each possible ))((n S S we get three possible )(n S .(E. g., if ))((n S S =6, then x n S =)( is a number such that 28≤x and 6)(=x S , therefore }24,15,6{)(∈n SFor each of these nine possibilities we computeas ))(()(2007n S S n S -- and checkwhether )(?)(n S n S =.We'll find out that out of the 9 cases, in 4 the valuehas the correct sum of digits.This happens for )2001,1983,1980,1977{?∈n .2007 AMC 10A Answer Key 1. C 2. A3.D4. A5. B6.A7.D8.D 8. E 10.E 11.C 12. D13.B 14.A 15. B 16.E 17. D 18.C 19.C 20.D 21.C 22.D23.B 24.B 25.D。

AMC10美国数学竞赛讲义全AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let is the prime number.if21p - is also the prime number. then1(21)2p p --is the perfect number. For example:6,28,496.3: Let ,0n abc a =≠ is three digital integer .if 333n a b c =++Then the number n is called Daffodils number . There are only four numbers: 153 370 371 407 Let ,0n abcd a =≠ is four digital integer .if 4444d n a b c +=++Then the number n is called Roses number . There are only three numbers: 1634 8208 94744：The Fundamental Theorem of ArithmeticEvery natural number can be written as a product of primes uniquely up to order.5：Suppose that a and b are integers with b =0. Then there exists unique integers q a d such that 0 ≤ < |b| a d a = bq + .6：(1)G eatest Commo D v so : Let gcd (a, b) = max {d ∈ Z: d | a a d d | b}. For any integers a and b, we havegcd(a, b) = gcd(b, a) = gcd(±a, ±b) = gcd(a, b ? a) = gcd(a, b + a). For example: gcd(150, 60) = gcd(60, 30) = gcd(30, 0) = 30 (2)Least commo mult le:Let lcm(a,b)=m {d∈Z: a | d a d b | d }. (3)We have that: ab= gcd(a, b) lcm(a,b) 7：Congruence modulo nIf ,0a b mq m -=≠,then we call a congruence b modulo m and we rewritemod a b m ≡.(1)Assume a b c d m Z 0 m 0 .If a b mod m c d mod m then we havemod a c b d m ±≡± , mod ac bd m ≡ , mod k k a b m ≡(2) The equat o ax ≡ b (mod m) has a solut o f a d o ly f gcd(a, m) d v des b.8：How to find the unit digit of some special integers (1)How many zero at the end of !nFor example, when 100n =, Let N be the number zero at the end of 100!then10010010020424525125N=++=+=(2) ,,a n Z ∈Find the unit digit n a . For example, when 100,3n a ==9：Palindrome, such as 83438, is a number that remains the same when its digits are reversed.There are some number not only palindrome but 2 2 ，222 ， (1)Some special palindrome n that 2n is also palindrome. For example :222221111121111123211111123432111111111112345678987654321=====(2)How to create a palindrome? Almost integer plus the number of its reverseddigits and repeat it again and again. Then we get a palindrome. For example: 87781651655617267266271353135335314884+=+=+=+=But whether any integer has this Property has yet to prove(3) The palindrome equation means that equation from left to right and right to left it all set up. For example ：1242242112231132211121241388888831421211====Let ab and cde are two digital and three digital integers. If the digits satisfy the,,9a c b e d c e d ?=?=+≤, then ab cde edc ba ?=?.10: Features of an integer divisible by some prime number If n is even ，then 2|n⼀个整数n 的所有位数上的数字之和是3（或者9）的倍数，则n 被3（或者9）整除⼀个整数n 的尾数是零，则n 被5整除⼀个整数n 的后三位与截取后三位的数值的差被7、11、13整除，则n 被7、11、13整除⼀个整数n 的最后两位数被4整除，则n 被4整除⼀个整数n 的最后三位数被8整除，则n 被8整除⼀个整数n 的奇数位之和与偶数位之和的差被11整除，则n 被11整除 11. The number Theoretic functions If 312123t rr rrt n p p p p =(1) {}12()#0:|(1)(1)(1)t n a a n r r r χ=>=+++(2) 12222111222|()(1)(1)(1)t r r r t t t a nn a p pp p p p p p p δ==+++++++++∑(3) {}11221111122()#:,gcd(,)1()()()t t r r rr rr t t n a N a n a n p p p p p p φ---=∈≤==---For example: 2(12)(23)(21)(11)6χχ=?=++= 22(12)(23)(122)(13)28δδ=?=+++= 22(12)(23)(22)(31)4φφ=?=--= Exercise1. The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?(A) 4 (B) 5 (C) 6 (D) 7(E) 83. For the positive integer n, let denote the sum of all the positive divisors of n with the exception of n itself. For example,<4>=1+2=3 and <12>=1+2+3+4+6=16. What is <<<6>>>?(A) 6 (B) 12 (C) 24 (D) 32 (E) 36 8. What is the sum of all integer solutions to 21<(x-2)<25? (A) 10 (B) 12(C) 15(D) 19(E) 5(A) 6 (B) 7 (C) 8 (D) 9 (E) 10(A) 1(B) 2(C) 3(D) 4(E) 515．The figures 123,,F F F and 4F shown are the first in a sequence of figures. For3n ≥, n F is constructed from -1n F by surrounding it with a square and placing one more diamond on each side of the new square than -1n F had on each side of its outside square. For example, figure 3F has 13 diamonds. How many diamonds are there in figure 20F ?18. Positive integers a, b, and c are randomly and independently selected with replacement from the set {1, 2, 3,…, 2010}. What is the probability that abc ab a ++ is divisible by 3? (A)13(B)2981(C)3181(D)1127(E)132724. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250 (C) 251 (D) 252 (E)2535. In multiplying two positive integers a and b, Ron reversed the digits of the two-digit number a. His erroneous product was 161. What is the correct value of the product of a and b?(A) 116 (B) 161 (C) 204 (D) 214 (E) 224 23. What is the hundreds digit of 20112011?(A) 1 (B) 4 (C) 5 (D) 6 (E) 99. A palindrome, such as 83438, is a number that remains the same when its digits are reversed. The numbers x and x+32 are three-digit and four-digit palindromes, respectively. What is the sum of the digits of x?(A) 20 (B) 21 (C) 22 (D) 23 (E) 2421. The polynomial 322010x ax bx -+- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 11824. The number obtained from the last two nonzero digits of 90! Is equal to n. What is n?(A) 12 (B) 32 (C) 48 (D) 52 (E) 6825. Jim starts with a positive integer n and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with n=55, then his sequence contain 5 numbers:55 55-72= 6 6-22= 2 2-12= 1 1-12= 0Let N be the smallest umbe fo wh ch J m’s seque ce has umbe s. What is the units digit of N?(A) 1 (B) 3 (C) 5 (D) 7 (E) 9 21．What is the remainder when 01220093+3+3++3is divided by 8?(A) 0 (B) 1 (C) 2 (D) 4 (E) 65．What is the sum of the digits of the square of 111,111,111? (A) 18 (B) 27(C) 45(D) 63(E) 81ABC 100643a S 2(A) 6 (B) 7(C) 8(D) 9(E) 1024. Let 2200820082k =+. What is the units digit of 222k +? (A) 0 (B) 1(C) 4(D) 6(E) 8AMC about algebraic problems⼀、Linear relations(1) Slope y-intercept form: y kx b =+ (k is the slope, b is the y-intercept) (2)Standard form: 0Ax By C ++= (3)Slope and one point 0000(,),()()P x y k slope y y k x x -=- (4) Two points 1122(,),(,)P x y P x y12121212y y y y y y x x x x x x ---==--- (5)x,y-intercept form: (,0),(0,),(0,0)1x yP a Q b a b a b≠≠+= ⼆、the relations of the two lines 111222:0,:0l A x B y C l A x B y C ++=++=(1) 1l ∥2l 122112210,0A B A B C B C B ?-=-≠ (1) 1l ⊥2l 12120A A B B ?-=三、Special multiplication rules:222223322332212211222112222()()()2()()()()()()(2)()((1)(1))(1)()n n n n n n n n n n n n n n a b a b a b a b a ab b a b a b a ab b a b a b a ab b a b a b a a b ab b n a b a b a a b ab b n is odd n a b c ab bc ac a b -----------=-+±=±+-=-+++=+-+-=-++++≥+=+-++-+->++=++?-22()()0b c c a a b c+-+-=?==四、quadratic equations and PolynomialThe quadratic equations 2(0)y ax bx c a =++≠ has two roots 12,x x then we has1212b c x x x x a a+=-=More generally, if the polynomial 121210nn n n n x a x a x a x a ---+++++= has nroots 123,,,,n x x x x ,then we have:1231122312123(1)n n n n n nx x x x a x x x x x x a x x x x a -++++=-++==-开⽅的开⽅、估计开⽅数的⼤⼩绝对值⽅程Arithmetic Sequence123(1)(2)(3)()n m a a n d a n d a n d a n m d =+-=+-=+-==+-121321()()()()2222n n n m n m n n a a n a a n a a n a a s ---+++++=====1(1)2n n n ds na -=+If n=2k, then we have 1()n k k s k a a +=+ If n=2k+1, then we have 1n k s na += Geometric sequence123123n n n n m n m a a q a q a q a q ----=====1(1)1,1n n a q q s q-≠=-Some special sequence , , 2, 3, 5, ,… 9,99,999,9999，… 1，11，111，1111，… Exercise4 .When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be left over?7 For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?21. Four distinct points are arranged on a plane so that the segments connecting them have lengths ,,,,, and . What is the ratio of to ?13. An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, and then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?16. Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?24. Let ,a b and c be positive integers with >>a b c such that 222-b -c +=2011a ab and222+3b +3c -3-2-2=-1997a ab ac bc . What is a ?(A) 249 (B) 250(C) 251(D) 252(E) 2531. What is246135135246++++-++++? (A) -1 (B) 536(C) 712(D)14760(E)43310. Consider the set of numbers {1, 10, 102, 103(010)}. The ratio of the largest element of the set to the sum of the other ten elements of the set is closest to which integer?(A) 1 (B) 9 (C) 10 (D) 11 (E) 101=(A) -64 (B) -24 (C) -9 (D) 24 (E) 5764. Let X and Y be the following sums of arithmetic sequences: X= 10 + 12 + + …+ 00. Y= 2 + + + …+ 02. What is the value of Y X -?(A) 92 (B) 98 (C) 100 (D) 102 (E) 1127. Which of the following equations does NOT have a solution?(A) 2(7)0x +=(B) -350x += 20=80= (E) -340x -=(A)(B)(C)2(D) (E) 613. What is the sum of all the solutions of 2602x x x =--?(A) 32 (B) 60 (C) 92 (D) 120 (E) 12414. The average of the numbers 1, 2, 3… 9 , 99, and x is 100x. What is x?(A)49101(B)50101(C)12 (D)51101(E)509911. The length of the interval of solutions of the inequality 23a x b ≤+≤ is 10.What is b-a?(A) 6 (B) 10 (C) 15 (D) 20 (E) 3013. Angelina drove at an average rate of 80 kph and then stopped 20 minutes for gas. After the stop, she drove at an average rate of 100 kph. Altogether she drove 250 km in a total trip time of 3 hours including the stop. Which equation could be used to solve for the time t in hours that she drove before her stop? (A) 880100()2503t t +-=(B) 80250t = (C) 100250t =(D) 90250t =(E) 880()1002503t t -+=21. The polynomial 32-2010x ax bx +- has three positive integer zeros. What is the smallest possible value of a?(A) 78 (B) 88 (C) 98 (D) 108 (E) 118 15．When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight in kilograms when the bucket is full of water?13．Suppose thatand. Which of the following is equal tofor every pair of integers16．Let ,,, and be real numbers with,, and. What is the sum of all possible values of5. Which of the following is equal to the product?81216442008............481242004n n +? (A) 251(B) 502(C) 1004(D) 2008 (E) 40167. The fraction 20082200622007220052(3)(3)(3)(3)-- simplifies to which of the following? (A) 1 (B) 9/4 (C) 3 (D) 9/2 (E) 913. Doug can paint a room in 5 hours. Dave can paint the same room in 7 hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let t be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by t ?(A) 11()(1)157t ++=(B) 11()1157t ++= (C) 11()157t +=(D) 11()(1)157t +-=(E) (57)1t +=15. Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?(A) 120 (B) 130 (C) 140 (D) 150 (E) 160AMC 中的⼏何问题⼀、三⾓形有关知识点1.三⾓形的简单性质与⼏个⾯积公式①三⾓形任何两边之和⼤于第三边；②三⾓形任何两边之差⼩于第三边；③三⾓形三个内⾓的和等于 0°；④三⾓形三个外⾓的和等于3 0°；⑤三⾓形⼀个外⾓等于和它不相邻的两个内⾓的和；⑥三⾓形⼀个外⾓⼤于任何⼀个和它不相邻的内⾓。