2024年4月1日发(作者：止博)

2010AIME试题及参考答案

2010/03/16 1：30~4：30

lists all the positive divisors of

2010

.She then randomly selects two distinct divisors

from this list. Let p be the probability that exactly one of the selected divisors is a perfect square.

The probability p can be expressed in the form

integers. Find m+n.

22222

解：因为

201023567

，所以它的所有的因数有

(12)81

个，其中是完全

4

2

m

,when m and n are relatively prime positive

n

11

C

16

C

81

26

16

平方数的有

CCCCC216

，所以

p

，所以m+n=107.



2

81

C

81

0

4

1

4

2

4

3

4

4

4

4

the remainder when

999999



99



9



9999‘s

is divided by 1000.

解：

999999



99



9999(10001)



(10

9999

1)



9999‘s

999(1000M1)1000N891

所以余数是109.

e the

y

3

x

and

x

y

y

x

.The quantity x+y can be expressed as a rational number

4

r

,where r and s are relatively prime positive integers. Find r+s.

s

3

yx

解：由

xy

知x>0,y>0,且

ylgxxlgy

，以

yx

带入得：

4

333344

xlgxxlgx,

于是

lgxlglgx

，所以

x()

4

，所以

y()

3

，所以

444433

448

x+y=，所以r+s=529.

81

and Phil have two fair coins and a third coin that comes up heads with probability

4m

.Jackie flips the three coins,and then Phill flips the three be the probablity that

7n

Jackie gets the same number of heads as Phill,where m and n are relatively prime positive integers.

Find m+n.

解：投掷三枚硬币的概率分布列如下：

头像朝上的次数

0

P

1 2 3

3

28

5

14

11

28

1

7

所以两人投掷硬币出现头像朝上的次数相同的概率为：

(

3

2

5111123

)()

2

()

2

()

2



，所以m+n=515.

2814287392

ve integers a,b,c, and d satisfy a>b>c>d，a+b+c+d=2010,and

a

2

b

2

c

2

d

2

2010

. Find the number of possible values of a.

解：易知

abcdabcd

，于是

2222

(ab)(ab1)(cd)(cd1)0

，又因为a>b>c>d，且a,b,c,d均是正整数，所以

a-b=1,c-d=1,即b=a-1,d=c-1，带入a+b+c+d=2010中得，a+c=1006,但a>c，所以a可以取值

504，505，…1004，所以a可能的取值为501个.

P(x) be a quadratic polynomial with real coefficients satisfying

x

2

2x2P(x)2x

2

4x3

for all real numbers x, and suppose P(11)= P(16).

解：

x2x2P(x)2x4x3

可化为

(x1)1P(x)2(x1)1

，所以可

令

P(x)a(x1)b

，由此

P(x)(x1)1(a1)(x1)b10

，所以

222

2222

a1,b1

,而

P(x)2(x1)

2

1(a2)(x1)

2

b10

,所以

a2,b1

，所以

b=1，但P(11)=181,所以100a+1=181，所以a=1.8,所以P(16)=406.

解法二：直接令

P(x)a(x1)1

，易得a=1.8.

an ordered triple (A,B,C) of sets to be minimally intersecting if

2

|AB||BC||CA|1

,

ABC

. For example, ({1,2},{2,3},{1,3,4}) is a

minimally intersecting N be the number of minimally intersecting ordered triples of sets

for which each set is a subset of {1,2,3,4,5,6,7}.Find the remainder when N is dividedi by 1000.

3

解：易知若

AB{a},BC{b},CA{c}

，共有

P

7

种办法，且

a,cA,a,bB,b,cC

，不妨设a=1,b=2,c=3,把剩下的4个数分成4份(有一份是不安插

进来的)，安插到A,B,C中，共有

4

种办法，故N=53760.

a real number a,let [a] denote the greatest integer less than or equal to R denote the

region in the coordinate plane consisting of points (x,y) such that

4

[x]

2

[y]

2

25

.

The region R is completely contained in a disk of radius r (a disk is the union of a circle and its

interior). The minimum value of r can be written as

divisible by the square of any prime. Find m+n.

解：易知

[x]0,[y]5

或

[x]5,[y]0

，或

[x]3,[y]4

，或

[x]4,[y]3

，

m

,where m and n are integers and m is not

n

于是



0x1



0x1

或





5y6



5y4

，或



5x6



5x4

或





0y1



0y1

，或



3x4



3x4



3x2



3x2

，或

或或





4y5



4y3



4y5



4y3



4x5



4x5



4x3



4x3

或



，或



或



或



3y43y23y43y2



在平面直角坐标系中画出区域来如下

图：容易得到此图的中心是

(,)

，

4

11

22

6

边界点到其距离的最大值为

130

，所

2

2

以m+n=132.

(a,b,c) be a real solution of the

system of equations

-5510

x

3

xyz2

y

3

xyz6

z

3

xyz20

The

3

-2

-4

greatest

33

possible value of

-6

m

abc

can be written in the form ,where m and n are relatively prime positive

n

m+n.

-8

x

3

2xyz

3

解：易知

y6xyz

,所以

(xyz)(2xyz)(6xyz)(20xyz)

，于是

xyz4

,或

3

z

3

20xyz

xyz

1515

333

，取

xyz

，得

xyz

的最大值为158.

77

N be the number of ways to write 2010 in the form

2010a

3

10

3

a

2

10

2

a

1

10a

0

,where the

a

i

's

are integers, and

0a

i

99

.An

example of such a represention is

110310671040

.Find N.

解：易知

a

3

只能取0，1，2三个值，因此根据

a

3

讨论如下：

（1）当

a

3

=0时，

a

2

10,11,12,13,14,15,16,17,18,19,20

，当

a

2

10

时，

a

1

有8种取法，

当

a

2

20

时，

a

1

有2种取法，其余各有10种取法，共100种取法

32

，2，3，4，5，6，7，8，9，10

，当

a

2

0

时，

a

1

有8种取法，（2）当

a

3

=1时，

a

2

0,1

当

a

2

10

时，

a

1

有2种取法，其余各有10种取法，共100种取法

（3）当

a

3

=2时，

a

2

0

，当

a

2

10

时，

a

1

有2种取法.

故N=202.

R be the region consisting of the set of points in the coordinate plane that satify both

|8x|y10

and

3yx15

,when R is revolved around the line whose equation is

3yx15

,the volume of the resulting solid is

m



np

,where m,n,and p are positive integers,m

and n are relatively prime,and p is not divisible by the square of any m+n+p.

解：画出区域如右图：

9133933

易得

A(,),B(,),C(8,10)

2244

且C到AB的距离为

C

B

d

|310815|

31

2



7

10

A

所以旋转体的体积为

1343



V



d

2

|AB|

3

1210

m+n+p=365.

，所以

m3

be an integer an let

S{3,4,5,,m}

.Find the smallest value of m such that for

every partition of S into two subsets,at least one of the subsets contains integers a,b,and c(not

necessarily distinct) such that ab=c.

Note:a partition of S is a pair of sets A,B such that

AB,ABS

.

解：m=120.

},B{6,8,9,10}

，所以

m12

.下证m=12符合条件. 设

A{3,4,5,7,11

证明：设

A

1

{3,6}，A

2

{3,12}，A

3

{4,8}，A

4

{4,12}，A

5

{5,10}，A

6

{6,12}，

B

1

{7}，B

2

{11}

设S集合的一个划分为A,B,若不存在

A

i

A，A

i

B

，则各

A

i

必有一个元素属于A,也必

有一个元素属于B.考察

A

1

，不妨设

3A

，则对于

A

6

，

6A,12A

，于是

A

6

B

矛盾.

故m=12符合条件.

2024年4月1日发(作者：止博)

2010AIME试题及参考答案

2010/03/16 1：30~4：30

lists all the positive divisors of

2010

.She then randomly selects two distinct divisors

from this list. Let p be the probability that exactly one of the selected divisors is a perfect square.

The probability p can be expressed in the form

integers. Find m+n.

22222

解：因为

201023567

，所以它的所有的因数有

(12)81

个，其中是完全

4

2

m

,when m and n are relatively prime positive

n

11

C

16

C

81

26

16

平方数的有

CCCCC216

，所以

p

，所以m+n=107.



2

81

C

81

0

4

1

4

2

4

3

4

4

4

4

the remainder when

999999



99



9



9999‘s

is divided by 1000.

解：

999999



99



9999(10001)



(10

9999

1)



9999‘s

999(1000M1)1000N891

所以余数是109.

e the

y

3

x

and

x

y

y

x

.The quantity x+y can be expressed as a rational number

4

r

,where r and s are relatively prime positive integers. Find r+s.

s

3

yx

解：由

xy

知x>0,y>0,且

ylgxxlgy

，以

yx

带入得：

4

333344

xlgxxlgx,

于是

lgxlglgx

，所以

x()

4

，所以

y()

3

，所以

444433

448

x+y=，所以r+s=529.

81

and Phil have two fair coins and a third coin that comes up heads with probability

4m

.Jackie flips the three coins,and then Phill flips the three be the probablity that

7n

Jackie gets the same number of heads as Phill,where m and n are relatively prime positive integers.

Find m+n.

解：投掷三枚硬币的概率分布列如下：

头像朝上的次数

0

P

1 2 3

3

28

5

14

11

28

1

7

所以两人投掷硬币出现头像朝上的次数相同的概率为：

(

3

2

5111123

)()

2

()

2

()

2



，所以m+n=515.

2814287392

ve integers a,b,c, and d satisfy a>b>c>d，a+b+c+d=2010,and

a

2

b

2

c

2

d

2

2010

. Find the number of possible values of a.

解：易知

abcdabcd

，于是

2222

(ab)(ab1)(cd)(cd1)0

，又因为a>b>c>d，且a,b,c,d均是正整数，所以

a-b=1,c-d=1,即b=a-1,d=c-1，带入a+b+c+d=2010中得，a+c=1006,但a>c，所以a可以取值

504，505，…1004，所以a可能的取值为501个.

P(x) be a quadratic polynomial with real coefficients satisfying

x

2

2x2P(x)2x

2

4x3

for all real numbers x, and suppose P(11)= P(16).

解：

x2x2P(x)2x4x3

可化为

(x1)1P(x)2(x1)1

，所以可

令

P(x)a(x1)b

，由此

P(x)(x1)1(a1)(x1)b10

，所以

222

2222

a1,b1

,而

P(x)2(x1)

2

1(a2)(x1)

2

b10

,所以

a2,b1

，所以

b=1，但P(11)=181,所以100a+1=181，所以a=1.8,所以P(16)=406.

解法二：直接令

P(x)a(x1)1

，易得a=1.8.

an ordered triple (A,B,C) of sets to be minimally intersecting if

2

|AB||BC||CA|1

,

ABC

. For example, ({1,2},{2,3},{1,3,4}) is a

minimally intersecting N be the number of minimally intersecting ordered triples of sets

for which each set is a subset of {1,2,3,4,5,6,7}.Find the remainder when N is dividedi by 1000.

3

解：易知若

AB{a},BC{b},CA{c}

，共有

P

7

种办法，且

a,cA,a,bB,b,cC

，不妨设a=1,b=2,c=3,把剩下的4个数分成4份(有一份是不安插

进来的)，安插到A,B,C中，共有

4

种办法，故N=53760.

a real number a,let [a] denote the greatest integer less than or equal to R denote the

region in the coordinate plane consisting of points (x,y) such that

4

[x]

2

[y]

2

25

.

The region R is completely contained in a disk of radius r (a disk is the union of a circle and its

interior). The minimum value of r can be written as

divisible by the square of any prime. Find m+n.

解：易知

[x]0,[y]5

或

[x]5,[y]0

，或

[x]3,[y]4

，或

[x]4,[y]3

，

m

,where m and n are integers and m is not

n

于是



0x1



0x1

或





5y6



5y4

，或



5x6



5x4

或





0y1



0y1

，或



3x4



3x4



3x2



3x2

，或

或或





4y5



4y3



4y5



4y3



4x5



4x5



4x3



4x3

或



，或



或



或



3y43y23y43y2



在平面直角坐标系中画出区域来如下

图：容易得到此图的中心是

(,)

，

4

11

22

6

边界点到其距离的最大值为

130

，所

2

2

以m+n=132.

(a,b,c) be a real solution of the

system of equations

-5510

x

3

xyz2

y

3

xyz6

z

3

xyz20

The

3

-2

-4

greatest

33

possible value of

-6

m

abc

can be written in the form ,where m and n are relatively prime positive

n

m+n.

-8

x

3

2xyz

3

解：易知

y6xyz

,所以

(xyz)(2xyz)(6xyz)(20xyz)

，于是

xyz4

,或

3

z

3

20xyz

xyz

1515

333

，取

xyz

，得

xyz

的最大值为158.

77

N be the number of ways to write 2010 in the form

2010a

3

10

3

a

2

10

2

a

1

10a

0

,where the

a

i

's

are integers, and

0a

i

99

.An

example of such a represention is

110310671040

.Find N.

解：易知

a

3

只能取0，1，2三个值，因此根据

a

3

讨论如下：

（1）当

a

3

=0时，

a

2

10,11,12,13,14,15,16,17,18,19,20

，当

a

2

10

时，

a

1

有8种取法，

当

a

2

20

时，

a

1

有2种取法，其余各有10种取法，共100种取法

32

，2，3，4，5，6，7，8，9，10

，当

a

2

0

时，

a

1

有8种取法，（2）当

a

3

=1时，

a

2

0,1

当

a

2

10

时，

a

1

有2种取法，其余各有10种取法，共100种取法

（3）当

a

3

=2时，

a

2

0

，当

a

2

10

时，

a

1

有2种取法.

故N=202.

R be the region consisting of the set of points in the coordinate plane that satify both

|8x|y10

and

3yx15

,when R is revolved around the line whose equation is

3yx15

,the volume of the resulting solid is

m



np

,where m,n,and p are positive integers,m

and n are relatively prime,and p is not divisible by the square of any m+n+p.

解：画出区域如右图：

9133933

易得

A(,),B(,),C(8,10)

2244

且C到AB的距离为

C

B

d

|310815|

31

2



7

10

A

所以旋转体的体积为

1343



V



d

2

|AB|

3

1210

m+n+p=365.

，所以

m3

be an integer an let

S{3,4,5,,m}

.Find the smallest value of m such that for

every partition of S into two subsets,at least one of the subsets contains integers a,b,and c(not

necessarily distinct) such that ab=c.

Note:a partition of S is a pair of sets A,B such that

AB,ABS

.

解：m=120.

},B{6,8,9,10}

，所以

m12

.下证m=12符合条件. 设

A{3,4,5,7,11

证明：设

A

1

{3,6}，A

2

{3,12}，A

3

{4,8}，A

4

{4,12}，A

5

{5,10}，A

6

{6,12}，

B

1

{7}，B

2

{11}

设S集合的一个划分为A,B,若不存在

A

i

A，A

i

B

，则各

A

i

必有一个元素属于A,也必

有一个元素属于B.考察

A

1

，不妨设

3A

，则对于

A

6

，

6A,12A

，于是

A

6

B

矛盾.

故m=12符合条件.