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Problems

A [2] Find the product of the smallest four prime factors of: 6232 2932.

B [6] A polynomial P satisfies P(8) = 144, and the following expression is true for all x = 0, 12

:

8P(x2 + 2)

P(4x) = 1 +

x3 + 1

2x 1.

ComputeP(10).

C [4] How many integers between 1 and 1000 inclusive do not satisfy

xx

3

=

3x

4

x

12

?

D [4] There exist positive integers m and n such that

2

m+

3

n=

1

16.

What is the sum of all possible odd values ofn?

E [6] Two circles intersect at R and S. A tangent line lying closer to R touches Circle 1 at Mand Circle 2 at N.PointTlies on Circle 2 such that segment M T passes through R. Prove N Sbisects MST.

F [6] Tomorrow happens to be both Jude and Jerrys birthday! Millinova decides to invite them and 13 otherfriends to a fantabulous birthday party. At the round table which seats 16 people, Jude and Jerry sit together,and Millinova sits next to one of them. A group of three adjacent seats contains gfour84, Rolloway, andCaelestor (though not necessarily in that order). Cyleura sits across the table from Caelestor. There are atotal ofn 9! possible distinct arrangements of people sitting at the table (two arrangements are not distinct ifall the people are sitting next to the same people). Calculate n.

G [4] Compute:

n=2

2n+1 n2 + 2n 2n+1 + 2n+1 2n(n2)

22n

(n4

+ 2n3

+ n2

)

.

H [4] Positive real numbers g, f, o, u, r satisfy:

f

3+ 0 + 2f+ 6u + 7r= 20.

Maximizing the productg f o u r gives the product m/n, wherem and n are relatively prime integers.Findm + n.

I [6] Let:

a0 = cos

2010+ i sin

2010.

For all positive integers n, let an= (an1)

13n+2

n

+1 . What is the smallest n that makes an both positive and real?

J [7] The statement n8 1 is a multiple of35 is true for some integers n between 2 and 1001 inclusive. Howmany such integers are there?

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Answers and Solutions

A Answer: 1155

Solution: We can factor the expression many times via thedifference of squares:

(62 29)(62 + 29)(622 + 292)(624 + 294)(628 + 298)(6216 + 2916)(6232 + 2932).

The first two terms are 3 11 and 7 13. However, looking at the third, we find that 5 is a factor as well. Since

2 clearly does not divide the number, we have found the four smallest. Their product is 3 5 7 11 = 1155.

B Answer: 240

Solution: We combine the right side, then multiply out to getP(x2 + 2)(16x 8) = P(4x)(x3 + 2x). IfPhas degree n, then 2n + 1 = n + 3 = n= 2. Hence, Pis a quadratic. Now,

P(x2 + 2)

P(4x) =

1

8+

x3 + 1

8(2x 1)=

(2x 1) + x3 + 1

8(2x 1) =

x(x2 + 2)

4(4x 2) =

x2(x2 + 2)

4x(4x 2).

It is easy to see P as n (x)(x 2). From P(8) = 144, we have n = 3. This gives P(10) = 3 10 8 = 240.

C Answer: 251

Solution: Assume the equation can be represented by:

f(x) = x x

3

3x

4

+ x

12

.

Subistuting

f(x + 12) =x + 12

x + 12

3

3(x + 12)

4

+

x + 12

12

= x + 12

x3

4

3x

4

9 +

x12

+ 1 = f(x).

This function can be seen to have a period of 12. There are 83 groups of 12 integers in 1000, with 4 extraremaining at the end. We need to find values ofx such thatf(x) = 0. All we have to do is consider values

between 1 and 12. One finds that 1, 2, 5, 13, 14, 17, ..., 997, 998 do not work, so the answer is 83 3 + 2 = 251.

D Answer: 100

Solution: We get rid of the denominators in the equation, and then use SFTT to factor it into the form:

(m 32)(n 48) = 1536 = 29 3.

Clearly, there are only 2 possible odd factors: 1, and 3. Hence n is either 48 + 1 = 49 or 48 + 3 = 51. The sumis 49 + 51 = 100.

E Solution:

N ST = N RT= MN R + RN M since N RM+ N RT = 180 (1)N SR = MN R since MN is a tangent line to Circle 2, which contains the point S (2)

M SR = RM Nsince MN is a tangent line to Circle 1, which contains Sas well (3)

Now, sum both sides of the equations MSR + N SR = MN R + RMN= N RT [because of (1)] (4)

So, MSN= NRT, and thus, N Sbisects M ST (Q.E.D) (5)

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F Answer: 3072

Solution: There are 16 6 ways to pick the spots of gfour84, Rolloway, and Caelestor. Then Cyleurasposition is fixed, and of the 12 remaining spots, there are 8 ways to place the block consisting of Jude, Jerry,and Millinova. This block can be arranged in 4 different ways (Millinova can not be in the middle of it).Finally, there are 9! ways to distribute the rest. This gives n = 16 6 8 4 = 3072.

G Answer: 1/8

Solution: Decompose:

2n+1(n2 + 2n + 1) 2n(n2)

22n(n2)(n + 1)2 =

2n+1(n + 1)2 2n(n2)

22n(n2)(n + 1)2 =

2n+1(n + 1)2

22n(n2)(n + 1)2

2n(n2)

22n(n2)(n + 1)2 =

2n+1

22n(n2)

2n

22n(n + 1)2 =

1

22nn1(n2)

1

22nn(n + 1)2 =

1

2n1(n2)

1

2n(n + 1)2.

Now, telescope:

n=2

1

2n1(n2)

1

2n(n + 1)2 =

n=2

1

2n1(n2)

n=2

1

2n(n + 1)2 =

1

8+

1

36+

1

36+

=

1

8.

H Answer: 263

Solution: By AMGM,

4 =g3

+ 2f+ o + 6u + 7r

5 5

28gfour =

45

28=

256

7 gfour = m + n= 263

I Answer: 6

Solution: By De Moivres Formula, (cis(x))n

= cis(n x). Its easy to see that the problem will thus boildown to finding the smallestn such that

kn=1

13n + 2

n + 1 ,

is an integer multiple of 2010. We have 2010 = 2 3 5 67, and the first occurrence of a multiple of 67 in thenumerator is when n = 5, so the minimal n is 5. However, we can see that for n = 5, this product is not aninteger yet for n = 6, the product is. The product for n = 6 is indeed a multiple of 2010 and hence, the answeris 6.

J Answer: 227

Solution: n8 1 (mod 5) is true ifn 0 (mod 5), andn8 n2 1 (mod 7) is true ifn 1 (mod 7). Wesee thatn8 1 (mod 35) for n 1, 6, 8, 13, 22, 27, 29, 34 (mod 35), 8 distinct values in total. Noting that1001 = 28 35 + 21, we have 28 8 + 4 integers from 1 to 1001 satisfying the condition, and uncounting 1, wesee there are 227 such integers.

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