MathPath 2015 Qualifying Test Solutions

MathPath 2015 Qualifying Test Solutions Anonymous December 31, 2014 Abstract

If you want to get into this camp, edit the solutions solutions so it doesn’t look like you cheated. cheated. I’m publishing publishing this because I hate MathPath. MathPath. Happy Happy New Year!

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1. Let your your opponent choose choose the set S set S .. Let S Let S have a positive numbers, b negative numbers, and c zeros. Since |S Since |S | = n , we have a + b + c = n . There are n2 pairs in total. A pair has a negative product if it has one negative and one positive number in it, so there are ab pairs with negative products. Therefore, there are n2 − ab pairs that are nonnegative.





Your opponent would choose a,b, a,b, c to minimize the value of n2 − ab because he wishes to keep your score down. Therefore, the highest score you can be sure to get against a smart enemy is the minimum value of x x = n2 − ab. n





  is a fixed value, the minimum of occurs when is maximized. By AM-GM, we have  +   −   − 0    ≤ = ≤ = 2 2 2 2  0. which is attained if and only if ( )=  0 to attain this maximum of , giving a minimum of If is even, then take ( )=  ( − 1) ( − 2)

Since

2

x

a

ab

2

b

n

c

2

2

n

n2

n 2, 2,

n

2

−

2

2, 2,

n

a,b,c

n

n n

a,b,c

n n

ab

4

n 2

4

=

n n

2

−

n2

4

=

n n

4

n 1 If n n is odd, then take (a,b,c) = n+1 2 , 2 , 0 to attain a maximum of ab ≤ 2 is clearly the largest integer less than n4 ), giving a minimum of



 n

2

−



−

n2 − 1

4

=

n(n − 1)

2

−

n 2 − 1

4

=

(n − 1)2 4

(n+1)(n−1) 4

=

n2 −1

4

(which





2. (a) First, we we know that that Cecille was was 99 years years old when she died died on October 30th, 30th, 2005. Therefore Therefore,, her birthday birthday was some date between between October 31st, 1905 and October October 30th, 1906. William decided decided to marry Cecile when she was eighteen, so he decided some date between October 31st, 1923 and October 29th, 29th, 1925. They married married in 1928, which which is January January 1st, 1928 to December December 31st, 1928. The final answer is: 31-Oct-1923 – 29-Oct-1925; 1-Jan-1928 – 31-Dec-1928 (b) Write whatever whatever here, and mak makee it your your own so you don’t get caught. caught. May Maybe be something something along the lines of “bounding is important in real math.”





3. We know that p(1) is equal to the sum of the coefficients of p. Since each each coefficient coefficient is positive, positive, p(1) is greater than or equal to the largest coefficient of p . Therefore, we can ask for p (10) to get an upper bound on the coefficients of p p . Specifically, if p p (1) has m digits, then each coefficient can have at most m digits. Ask for p(10m+1 ), and the coefficien coefficients ts will be mag magica ically lly encoded encoded in the value value they give. give. Let t = 2 2m+2 m+1 n m+1 p(10 ) If p p (x) = a 0 + a1 x + a2 x + · · · + an x , then t = a 0 + a1 10 + a2 10 + · · · + an 10nm+n . m+1 m+1 m+1 Take t (mod 10 ) to get a 0 (mod 10 ). However, a 0 (mod 10 ) is just the last m digits of a a 0 . Since a0 has at most m digits, this must be all of the digits of a0 , so now we know a0 . To find find a1 , t a0 (n 1)m+(n 1) m+1 take t = 10m+1 = a 1 + a2 10 + · · · + an 10 . Using the same process, process, we have have t ≡ a 1 (mod 10m+1 ), which is the last m digits of a a 1 , which is all of the digits of a a 1 . Repeat this process until you have all of the coefficients, and then you have your enemy’s polynomial. 

−

−

−



Thus, the strategy is to ask for p(1) to find m, then p(10m+1 ) to find the coefficients encoded in its digits. Note: the process is simpler simpler than it looks, it’s just hard to explain explain on paper. If p p (x) = 100x2 + x +157, then we ask for p (1) = 258, which has m = 3 digits. Then ask for p (103 ) = p (1000) to get 100(1000 2 ) + 1000 + 157 = 100001157. 100001157. Group these by sets of m = 3 digits to get the coefficients: 100 | 001 | 157. Thus, the polynomial is 100x2 + 1 · x + 157.





4. (a) The diag diagram ram is show shown n below. below.

A

E D F

B

C

(b) We have

∠BF C = =

◦

◦

180 − ∠C − ∠F BC = 180 − ∠C − (∠B − ∠EB F ) =

∠A + ∠EB F

◦

= 70 = ∠C

Thus, BC F is isosceles with BF = BC . Since Since the two triangl triangles es are equilate equilateral ral,, they they are congruent. (c) We must have have B F = B C , so ∠C = = ∠BF C = =

◦

◦

◦

180 − ∠BF A = 180 − (180 − ∠A − ∠ABF ) = ◦

◦

∠A + 60

◦

Therefore, all triangles with angles (∠A, ∠B, ∠C ) = (θ, 120 − 2θ, θ + 60 ) work. Howeve However, r, we must have 120 − 2θ, θ + 60 > 60 , so 0 < θ < 30 . ◦

◦

◦

◦





5. (a) Let Let n = a k ak

1 . . . a 1 a0 so

−

that we want 2n = a 0 ak ak

1 . . . a 2 a1 =

n − a0

−

10

+ 10k a0

or 19n = (10k+1 − 1)a0 First, we must have 19 | 10 k+1 − 1 since 0 ≤ 0 ≤ a0 ≤ 9. From Fermat’ Fermat’ss Little Little Theorem, Theorem, we know know that 19 | 10k+1 − 1 if k ≡ −1 (mod 18). 18). Therefore, Therefore, we must must have have k = 18 j − 1 (technically, we would have to check all 0 ≤ k ≤ 17, but I don’t think you’ll get points off for not stating this). 10(18j −1)+1 −1 19

Knowing that

is an an intege integer, r, all all that that remain remainss is to find find out out when 10(18j

n =

1)+1

−

19

−1

a0 =

1018j − 1 a0 19

actually has a final digit of a 0. That is, this n only works if 1018j − 1 a0 ≡ a0 19 18j

(mod 10)

We now prove that 10 19 1 ≡ 1 (mod (mod 10). To do this, this, note that 19 1 ≡ 9 ≡ −1 (mod 10) and 1018j − 1 ≡ −1 (mod 10), 10), from which the conclusion follows. follows. Finally, Finally, before concluding concluding,, note that a 0 = 0 gives n = 0, so 1 ≤ a0 ≤ 9. Thus, the solution set is all −

−

n =

1018j − 1 a0 19

where j ≥ 1, 1 ≤ a0 ≤ 9. (b) Let n = a k ak

1 . . . a 1 a0 so

−

that we want

2n = a k

1 ak−2 . . . a1 a0 ak =

−

10(n − 10k ak ) + ak

or 8n = (10k+1 − 1)ak Since k ≥ 0, 10k+1 − 1 is odd, so 8 | ak . Sinc Sincee 0 ≤ ak ≤ 9, we must have ak = 8. Ho Howe weve ver, r, k+1 that would imply n = 10 − 1, so n is a string of 9s, which contradicts that its first digit is 8. Therefore, no numbers have this property.







6. (a) Using Using the given given iden identit tity y ϕ 2 = ϕ + 1, we have (ϕ + 2)2 = ϕ 2 + 4ϕ + 4 = ( ϕ + 1) + 4ϕ + 4 = 5 ϕ + 5 = 5(ϕ + 1) = 5ϕ2 as desired. (b) Using the given identity identity ϕ 2 = ϕ + 1, we have ϕ5 = (ϕ2 )2 ϕ = (ϕ + 1)2 ϕ = (ϕ2 + 2ϕ + 1)ϕ = (3ϕ + 2)ϕ = 3ϕ2 + 2ϕ = 5ϕ + 3

Rewriting the given identity as ϕ 5

−

ϕ

1

−

= ϕ − 1, we have

= (ϕ − 1)5 = (( ϕ − 1)2 )2 (ϕ − 1) = (ϕ2 − 2ϕ + 1)2 (ϕ − 1) = (− (−ϕ + 2)2 (ϕ − 1) = (ϕ2 − 4ϕ + 4)(ϕ − 1) = (− (−3ϕ + 5)(ϕ − 1) = −3ϕ2 + 8ϕ − 5 = 5ϕ − 8

(c) Rewrite as ϕ + 2 = ( pϕ + q )(3 )(3ϕ + 4) = 3 pϕ2 + (4 p + 3 q )ϕ + 4q = = 3 p(ϕ + 1) + (4 p + 3 q )ϕ + 4 q

= (7 p + 3 q )ϕ + (3 p + 4q ) 2 Solving the system 7 p + 3q = = 1, 3 p + 4q = = 2, we get p = − 19 and q =

11 19 .

(d) Notice that ϕ2 − ϕ − 1 = (1 − ϕ)2 − (1 − ϕ) − 1 = (1 − 2ϕ + ϕ2 ) + ϕ − 2 = ϕ 2 − ϕ − 1 = 0

so ϕ also satisfies the identity ϕ2 = ϕ + 1. 1. Since the above above solutions solutions manipulated manipulated this identit identity y only, it doesn’t matter which root of 0 = x 2 − x − 1 we use.





7. (a) Independen Independentt of the the actual actual terms terms of the the sequence sequence f n , as long as f n satisfies f n = f n we have

1 + f n−2 ,

−

f n = f n

1 + f n−2 =

−

(f n

2 + f n−3 ) + f n−2 =

−

2f n

2 + f n−3 =

−

2f n

2 + (f n−2 − f n−4 ) =

−

(b) Independent of the actual actual terms of the sequence t n , as long as t n satisfies t n = t n then we have

3f n

then

2 − f n−4

−

1 + tn−2 + tn−3 ,

−

tn = t n

1 + tn−2 + tn−3 =

−

= 7 tn = 7 tn

− t n 5 − 3t n 3 − 5tn 6 + tn 3

−

−

−

2tn 2 + 2 tn 3 + tn 4 = 4tn 3 + 3 tn 6 = 7tn 3 − 4tn 6 − tn 7 − tn 8

−

−

−

−

−

−

−

−

−

4 + 2 tn−5

−

−

9

−

(c) Let sn = sn 1 + s n 2 + · · · + s n k . This linear recurrenc recurrencee has a charact characteristi eristicc polynomial of k k 1 k 2 χk (x) = x − x −x − · · · − 1, so let the roots of χk be r1 , r2 , . . . , rk . By the princi principle pless of linear recurrence and characteristic polynomials, we know that −

−

−

−

−

sn = c 1 r1n + c2 r2n + · · · + ck rkn

for some coefficients c1 , c2 , . . . , ck . Therefore, a recurrence for sn in terms of s sn k , sn 2k , . . . , sn k2 k k k would have a characteristic polynomial with roots r 1 , r2 , . . . , rk . For example, f n = f n 1 + f n 2 has a characteristic polynomial of x2 − x − 1 with roots r1 , r2 . From Vieta’s formulas, a polynomial with roots r12 , r22 would be x2 − (r12 + r22 )x + r 12 r22 = x2 − (12 − 2(− 2(−1))x + (− (−1)2 = x 2 − 3x + 1. Therefore, Therefore, f n = 3f n 2 − f n 4 . −

−

−

−

−

−

−

MathPath 2015 Qualifying Test Solutions

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