algebra from rice university

RMT 2013 Algebra Test February 2, 2013

1. Nick is a runner, and his goal is to complete four laps around a circuit at an average speed of10 mph. If he completes the first three laps at a constant speed of only 9 mph, what speed doeshe need to maintain in miles per hour on the fourth lap to achieve his goal?

2. A tree has 10 pounds of apples at dawn. Every afternoon, a bird comes and eats x pounds ofapples. Overnight, the amount of food on the tree increases by 10%. What is the maximumvalue of x such that the bird can sustain itself indefinitely on the tree without the tree runningout of food?

3. Karl likes the number 17. His favorite polynomials are monic quadratics with integer coefficientssuch that 17 is a root of the quadratic and the roots differ by no more than 17. Compute thesum of the coefficients of all of Karls favorite polynomials. (A monic quadratic is a quadraticpolynomial whose x2 term has a coefficient of 1.)

4. Given that f(x) + 2f(8 x) = x2 for all real x, compute f(2).5. For exactly two real values of b, b1 and b2, the line y = bx 17 intersects the parabola y =

x2 + 2x + 3 at exactly one point. Compute b21 + b22.

6. Compute the largest root of x4 x3 5x2 + 2x + 6.7. Find all real x that satisfy 3

20x + 3

20x + 13 = 13.

8. Find the sum of all real x such that

4x2 + 15x + 17

x2 + 4x + 12=

5x2 + 16x + 18

2x2 + 5x + 13.

9. Let a = 3 +5 +7, b = 35 +7, c = 3 +57. Evaluatea4

(a b)(a c) +b4

(b c)(b a) +c4

(c a)(c b) .

10. Given a complex number z such that z13 = 1, find all possible values of z+z3+z4+z9+z10+z12.

RMT 2013 Algebra Test Solutions February 2, 2013

1. Answer: 15

Solution: Let d be the length of one lap in miles. Then he needs to complete the four laps in4d10 =

2d5 hours. He has already spent

3d9 =

d3 hours on the first three laps, so he has

2d5 d3 = d15

hours left. Therefore, he must maintain a speed of 15 mph on the final lap.

2. Answer:10

11

Solution: After removing x from 10, and then increasing that amount by 10%, we must endup with at least the amount we started with, 10 pounds. That is, the maximum value of x must

satisfy11

10(10 x) = 10. Solving for x, we get that x = 10

11.

3. Answer: 8960

Solution: All of Karls favorite quadratics take the form (x r)(x 17), where 0 r 34.The sum of the coefficients of any polynomial can be determined by evaluating the polynomial

at x = 1. This gives 16r 16.34r=0

(16r 16) = 16 34 352 16 35 = 8960 .

4. Answer:68

3

Solution: Substituting x = 2, we get that f(2) + 2f(6) = 4. Substituting x = 6, we get that

f(6) + 2f(2) = 36. Solving for f(2) and f(6) gives us that f(6) = 283

and f(2) =68

3.

5. Answer: 168

Solution: We have that b is a valid number if and only if (x2+2x+3)(bx17) = x2+(2b)x+20has exactly one real root. This means that 2 b = 220, so b = 2 220. b21 + b22 is therefore2(22) + 2(2

20)2 = 8 + 160 = 168 .

6. Answer:1 +13

2

Solution: Note that x4x35x2+2x+6 = (x45x2+6)x(x22) = (x22)(x23)x(x22) =(x2 2)(x2 x 3). The two largest candidate roots are therefore 2 and 1 +

13

2. Note that

13 > 3, so

1 +

13

2> 2 >

2, so therefore the largest root is

1 +

13

2.

7. Answer: 5465

Solution: Observe that f(a) = 3

20x + a is an increasing function in a, so the only way that

f(f(a)) = a can be true is if f(a) = a. Solving 3

20x + 13 = 13, we obtain x =546

5.

8. Answer: 11/3Solution: Let f(x) = 4x2 + 15x + 17, g(x) = x2 + 4x + 12, and h(x) = x2 + x + 1. Then, the


given equation becomes

f(x)

g(x)=f(x) + h(x)

g(x) + h(x)

= f(x)g(x) + f(x)h(x) = f(x)g(x) + g(x)h(x)= f(x)h(x) = g(x)h(x).

Since h(x) > 0 for all real x, we may divide through by h(x) to get

f(x) = g(x)

= 4x2 + 15x + 17 = x2 + 4x + 12= 3x2 + 11x + 5 = 0.

The discriminant of this quadratic is

112 4 3 5 = 61 > 0,so it has two real roots. By Vietas, the sum of these roots is 11/3 .

9. Answer: 30

Solution: Putting everything over a common denominator, we can rewrite the expression as

a4(b c) b4(a c) + c4(a b)(a b)(a c)(b c) =

a4b ab4 a4c + ac4 + b4c bc4(a b)(a c)(b c) .

Notice that if a = b, the numerator becomes a5a5a4c+ac4 +a4cac4 = 0; similarly if a = cor b = c. This means that the numerator is in fact divisible by (a b)(a c)(b c). Factoring,we find that the above expression is equal to

(a b)(b c)(a c)(a2 + b2 + c2 + ab + bc + ac)(a b)(b c)(a c) = a

2 + b2 + c2 + ab + bc + ac

as long as the original expression was well-defined. But we have

a2 + b2 + c2 + ab + bc + ac =1

2

((a + b)2 + (b + c)2 + (c + a)2

)and plugging in the given values of a, b, c gives

1

2

((2

7)2 + (2

3)2 + (2

5)2)

= 2(7 + 3 + 5) = 30 .

10. Answer: 6, 113

2

Solution: First of all, if z = 1, then the expression is simply equal to 6 . Otherwise, let = z + z3 + z4 + z9 + z10 + z12. We find that

2 = z2+z6+z8+z5+z7+z11+2(z4+z5+z10+z11+1+z7+z12+1+z2+1+z+z3+z6+z8+z9).

Applying the identity z+ z2 + z3 + + z12 = 1, we arrive at 2 = 1+ 2(3 1) = 3,

and the solutions to the quadratic are =113

2.


1. Compute the minimum possible value of

(x 1)2 + (x 2)2 + (x 3)2 + (x 4)2 + (x 5)2,

for real values of x.

2. Alice and Bob compete in Silly Math Tournament (SMT), in which a contestants score is equal to thenumber of problems he or she gets right. The product of their scores is equal to three times the sum oftheir scores. Compute the sum of all possible scores for Bob.

3. Express 231

23+1 331

33+1 431

43+1 1631

163+1 as a fraction in lowest terms.

4. The function f(x) is known to be of the formn

i=1 fi(aix), where ai is a real number and fi(x) is eithersin(x) or cos(x) for i = 1, . . . , n. Additionally, f(x) is known to have zeros at every integer between 1 and2012 (inclusive) except for one integer b. Find the sum of all possible values of b.

5. If f is a monic cubic polynomial with f(0) = 64, and all roots of f are non-negative real numbers, whatis the largest possible value of f(1)? (A polynomial is monic if it has a leading coefficient of 1.)

6. The quartic (4th-degree) polynomial P (x) satisfies P (1) = 0 and attains its maximum value of 3 at bothx = 2 and x = 3. Compute P (5).

7. There exist two triplets of real numbers (a, b, c) such that a 1b , b 1c , and c 1a are the roots to the cubicequation x35x215x+ 3, listed in increasing order. Denote those triplets as (a1, b1, c1) and (a2, b2, c2).If a1, b1, and c1 are the roots to monic cubic polynomial f and a2, b2, and c2 are the roots to monic cubicpolynomial g, find f(0)3 + g(0)3.

8. If x, y, and z are integers satisfying xyz + 4(x+ y + z) = 2(xy + xz + yz) + 7, list all possibilities for theordered triple (x, y, z).

9. z1 and z2 are complex numbers that satisfy the equation 3z21 2z1z2 + 2z22 = 0, and z12z1+2 is a purely

imaginary number, i.e. Re(z12z1+2

)= 0. If P1, P2, and O are points in the complex plane corresponding

to z1, z2, and 0, respectively, find the area of 4P1OP2.10. Let X1, X2, . . . , X2012 be chosen independently and uniformly at random from the interval (0, 1]. In other

words, for each Xn, the probability that it is in the interval (a, b] is b a. Compute the probability thatdlog2X1e+ dlog4X2e+ + dlog4024X2012e is even. (Note: For any real number a, dae is defined as thesmallest integer not less than a.)


1. Answer: 10

Solution: We know that this expression has to be a concave-up parabola (i.e. a parabola that facesupwards), and there is symmetry across the line x = 3. Hence, we conclude that the vertex of the

parabola occurs at x = 3. Plugging in, we get 4 + 1 + 0 + 1 + 4 = 10 .

2. Answer: 22

Solution: Let a and b be the number of problems Alice and Bob solve, respectively. Then ab = 3(a+ b).Adding 9 to both sides and rearranging, (a 3)(b 3) = 9. The possible solutions are (a, b) = (0, 0),(4, 12), (6, 6), and (12, 4) which sum to 22 .

3. Answer: 91136

Solution: We note

kn=2

n3 1n3 + 1

=

kn=2

(n 1)(n2 + n+ 1)(n+ 1)(n2 n+ 1) =

(k

n=2

n 1n+ 1

)(k

n=2

n2 + n+ 1

n2 n+ 1

)

Each product telescopes, yielding 12k(k+1) k2+k+1

3 . Evaluating at k = 16 yields91

136.

4. Answer: 2047

Solution: The possible values of b are precisely the powers of two not exceeding 2012 (including 20 = 1).The following proof uses the fact that the zeroes of sine and cosine are precisely numbers of the form tpiand (t+ 1/2)pi, respectively, for t an integer.

Suppose b is not a power of 2. Then it can be written as 2m(1 + 2k) for m 0, k > 0. Since 2m < b, byassumption f must have a root at 2m. But then f must have a root at b, too: If sin(2ma) = 0, then 2ma = tpi for some integer t, so sin(ba) = sin((1 + 2k)2ma) = sin((1 + 2k)tpi) = 0and b is a root of f . If cos(2ma) = 0, then 2ma = (t+ 1/2)pi for some integer t so

cos(ba) = cos((1 + 2k)2ma) = cos((1 + 2k)(t+ 1/2)pi) = cos((t+ k + 2kt+ 1/2)pi) = 0

and thus b is a root of f .This is a contradiction, so b can only be a power of 2.

For each b of the form 2m, we can construct an f that works by using cosine terms to cover integerspreceding b and sine terms thereafter:

f(x) =

(mi=1

cos(pix/2i)

) 2012j=b+1

sin(pix/j)

has a root at every positive integer at most 2012 except b.

Hence, our final answer is 1 + 2 + 4 + ...+ 1024 = 2048 1 = 2047 .5. Answer: 125

Solution: If the three roots of f are r1, r2, r3, we have f(x) = x3(r1+r2+r3)x2+(r1r2+r1r3+r2r3)x

r1r2r3, so f(1) = 1 (r1 + r2 + r3) (r1r2 + r1r3 + r2r3) r1r2r3. Since r1r2r3 = 64, the arithmeticmean-geometric mean inequality reveals that r1 + r2 + r3 3(r1r2r3)1/3 = 12 and r1r2 + r1r3 + r2r3 3(r1r2r3)

2/3 = 48. It follows that f(1) is at most 1 12 48 64 = 125 . We have equality whenall roots are equal, i.e. f(x) = (x 4)3.

6. Answer: 24Solution: Consider the polynomial Q(x) = P (x) 3. Q has roots at x = 2 and x = 3. Moreover, sincethese roots are maxima, they both have multiplicity 2. Hence, Q is of the form a(x 2)2(x 3)2, and soP (x) = a(x2)2(x3)2+3. P (1) = 0 = a = 34 , allowing us to compute P (5) = 34 (9)(4)+3 = 24 .


7. Answer: 14Solution: By Vietas Formulas, we have that f(0) = a1b1c1 and g(0) = a2b2c2. Additionally,(a 1b )(b 1c )(c 1a ) = 3 and (a 1b ) + (b 1c ) + (c 1a ) = 5. Expanding the first expression yields3 = abc 1abc ((a+ b+ c) ( 1a + 1b + 1c )) = abc 1abc 5. This is equivalent to (abc)2 2(abc) 1 = 0,so abc = 12. It follows that f(0)3 + g(0)3 = (1 +2)3 (12)3 = 14 .

8. Answer: (1,1,1) (1,3,3) (3,1,3) (3,3,1)

Solution: Rearranging the given equality yields xyz 2(xy + xz + yz) + 4(x + y + z) 8 = 1. Butthe left side factors as (x 2)(y 2)(z 2). Since all quantities involved are integral, we must have eachfactor equal to 1. It is easy to verify that the only possibilities for (x, y, z) are those listed.

9. Answer:5

Solution: We shall use the formula S4P1OP2 =12 |z1||z2| sin , where is the angle between z1 and z2.

Solving for z2 using the quadratic equation, we obtain z2 =15i

2 z1. From this relationship we see that

tan =

5, so sin =56; also, |z2| =

62 |z1|. Thus

S4P1OP2 =1

2|z1|

(6

2|z1|)(

56

)=

5

4|z1|2.

Now, we note that |z1| must be 2. There are many ways to see this. Geometrically, z12 and z1 + 2 forma right triangle with the origin because they are pi2 apart. z1 is then the median from the origin to thehypotenuse, so its magnitude is equal to half the length of the hypotenuse, which is (z1 +2) (z12) = 4.Or we can set z12z1+2 = bi. Solving, we obtain z1 =

2(bi)b+i which satisfies |z1| = 2.

Therefore, S4P1OP2 =54 22 =

5 .

10. Answer: 20134025

Solution: To simplify notation, define Yn = dlog2nXne.We begin by computing the probability that Yn is odd. Yn = 1 if 2 < log2nXn 1, or 1(2n)2 < Xn 12n . Similarly, Yn = 3 if 1(2n)4 < Xn 1(2n)3 , and so on. Adding up the lengths of these intervals, we seethat the probability that Yn is odd is

k=1

1

(2n)2k1 1

(2n)2k=

12n (1 12n )1 1(2n)2

=12n

(1 + 12n )=

1

2n+ 1.

Armed with this fact, we are now ready to solve the problem. One way to continue would be to note thatthe probability that Y1 is even is 2/3, the probability that Y1 + Y2 is even is 3/5, the probability thatY1 + Y2 + Y3 is even is 4/7 and to show by induction that the probability that Y1 + Yn is even is n+12n+1 .Below, we present an alternate approach.

Note that Y1 + Y2 + + Y2012 is even if and only if (1)Y1+Y2++Y2012 = 1. Rewrite (1)Y1+Y2++Y2012as (1)Y1(1)Y2 (1)Y2012 , and note that because the Yn are independent,

E[(1)Y1(1)Y2 (1)Y2012] = E[(1)Y1 ]E[(1)Y2 ] E[(1)Y2012 ], (1)

where E denotes the expected value of the quantity. But E[Yn] = (+1)P (Yn is even)+(1)P (Yn is odd).We computed earlier that the probability that Yn is odd is

12n+1 , so E[Yn] =

2n12n+1 and product in (1) is

13 35 40234025 , which telescopes to yield 14025 . Let p be the probability that Y1 + Y2 + Y2012 is even. Wejust found that (+1)(p) + (1)(1 p) = 14025 , which we can solve to obtain p =

2013

4025.


1. Sammy is a child math prodigy born on Jan 1. At age 0.1, he already excels at multiplication. Atage 8, he decides to start saving money to attend Rice University, his top choice. He estimates thatRices tuition plus room and board will cost $45000 per year for each of the four years he will be incollege. Being a clever investor of money, he can invest it in such a way that the annual interest rate is100% compounded yearly ($100 will earn $100 in interest at the end of the first year, which will earn100% (100 + 100) = $200 in interest at the end of the next year). If Sammy starts to invest some ofthe cash he receives as birthday gift for his college fund every year at each of his birthdays, startingwith his eighth and ending with his seventeenth (inclusive), what is the minimum integer amount ofmoney he needs to deposit into the bank each year so that he will have enough money for Rice on hiseighteenth birthday? Assume he invests the same amount of money every year (he receives sufficientcash as a birthday boy), and the interest applies at the end of each year. Also assume no inflation andconstant interest rate.

2. Consider the curves x2 + y2 = 1 and 2x2 + 2xy + y2 2x 2y = 0. These curves intersect at twopoints, one of which is (1, 0). Find the other one.

3. Find all rational roots of |x 1| |x2 2| 2 = 0.4. If r, s, t, and u denote the roots of the polynomial f(x) = x4 + 3x3 + 3x + 2, find

1

r2+

1

s2+

1

t2+

1

u2

5. Computen=1

(7n + 32) 3nn (n + 2) 4n .

6. Find the remainder when (x + 2)2011 (x + 1)2011 is divided by x2 + x + 1.7. There are 2011 positive numbers with both their sum and their sum of reciprocals equal to 2012. Let

x be one of these numbers. Find the maximum of x + x1.

8. Let P (x) be a polynomial of degree 2011 such that P (1) = 0, P (2) = 1, P (4) = 2, ... , and P (22011) =2011. Compute the coefficient of the x1 term in P (x).

9. It is a well-known fact that the sum of the first n k-th powers can be represented as a polynomial inn. Let Pk(n) be such a polynomial for integers k and n. For example,

ni=1

i2 =n(n + 1)(2n + 1)

6,

so one has

P2(x) =x(x + 1)(2x + 1)

6=

1

3x3 +

1

2x2 +

1

6x.

Evaluate P7(3) + P6(4).10. How many polynomials P of degree 4 satisfy P (x2) = P (x)P (x)?

RMT 2011 Algebra Solutions February 19, 2011

1. Answer: $88

Let x be the amount of money he invests each year. We make the following table about the amount ofmoney he has:

Year Money on Jan 1 Money on Dec 318 x 2x9 2x+ x 22x+ 2x10 22x+ 2x+ x 23x+ 22x+ 2x. . . . . . . . .17 29x+ 28x+ . . .+ 2x+ x 210x+ 29x+ . . .+ 2x

Sammy needs 4$45000 = $180000. Then 180000 = 210x+29x+ . . .+2x = 2(29 +28 + . . .+2+1)x =2(210 1)x = 2046x x = 180000/2046 ' 87.97 so the least integer amount of money he needs toinvest is $88.

2. Answer:(3

5, 45

)From the first equation, we get that y2 = 1 x2. Plugging this into the second one, we are left with

2x2 2x

1 x2 + 1 x2 2x 2

1 x2 = 0 (x 1)2 = 2

1 x2(x 1) x 1 = 2

1 x2 assuming x 6= 1

x2 2x+ 1 = 4 4x2 5x2 2x 3 = 0.

The quadratic formula yields that x = 2810 = 1, 35 (we said that x 6= 1 above but we see that it is stillvalid). If x = 1, the first equation forces y = 0 and we easily see that this solves the second equation.If x = 35 , then clearly y must be positive or else the second equation will sum five positive terms.Therefore y =

1 925 =

1625 =

45 . Hence the other point is

( 35 , 45).3. Answer: x = 1, 0, 2

There are four intervals to consider, each with their own restrictions. Consider the case in whichx >

2. Then the equation becomes (x 1)(x2 2) 2 = x(x 2)(x + 1) = 0. Thus, x = 2 is

the only rational root for x >

2. Consider the case in which 2 < x < 1. Then the equationbecomes (x 1)(x2 2) 2 = x(x 2)(x + 1) = 0. Thus, x = 0 and x = 1 are the rational rootsfor 2 < x < 1. Consider the case in which x < 2 or the case in which 1 < x < 2. In thesecases, the equation becomes (1x)(x22)2 = x3 +x2 + 2x4. By the rational root theorem, therational roots of this polynomial can only be 4,2,1 and a quick check shows that none of theseare roots, so this polynomial has no rational roots.

4. Answer: 94

First notice that the polynomial

g(x) = x4(

1

x4+

3

x3+

3

x+ 2

)= 2x4 + 3x3 + 3x+ 1

is a polynomial with roots 1r ,1s ,

1t ,

1u . Therefore, it is sufficient to find the sum of the squares of the

roots of g(x), which we will denote as r1 through r4. Now, note that

r21 + r22 + r

23 + r

24 = (r1 + r2 + r3 + r4)

2 (r1r2 + r1r3 + r1r4 + r2r3 + r2r4 + r3r4) = (a3a4

)2 a2a4

by Vietas Theorem, where an denotes the coefficient of xn in g(x). Plugging in values, we get that

our answer is ( 32 )2 0 = 94 .


5. Answer: 332

Note that 7n+32n(n+2) =16n 9n+2 so that

n=1

(7n+ 32)

n(n+ 2)

3n

4n=

n=1

16

n

3n

4nn=1

9

n+ 2

3n

4n

=

n=1

16

n

3n

4nn=1

16

n+ 2

3n+2

4n+2

=

n=1

16

n

3n

4nn=3

16

n

3n

4n

=16

1

3

4+

16

2

9

16=

33

2.

6. Answer: (31005 1)x+ (2 31005 1)The standard method is to use the third root of unity , 2 ++1 = 0. Let (x+2)2011 (x+1)2011 =(x2+x+1)Q(x)+ax+b and substitute x = . Then a+b = (+2)2011(+1)2011. Note that +2has size

3 and argument pi/6, so ( + 2)6 = 33. Also + 1 has magnitude 1 and argument pi/3, so

( + 1)6 = 1. Using this and 2011 = 6 335 + 1, we get that a + b = (31005 1) + (2 31005 1).Another solution is to note that (x + 2)2 x2 + 4x + 4 3x2 (mod x2 + x + 1) and (x + 1)2 x2 + 2x + 1 x (mod x2 + x + 1). Then we have x3 1 (mod x2 + x + 1) and we can proceed byusing periodicity.

7. Answer: 80452012

Let y1, y2, , y2010 be the 2010 numbers distinct from x. Then y1 + y2 + + y2010 = 2012 x and1y1

+ 1y2 + + 1y2010 = 2012 1x . Applying the Cauchy-Schwarz inequality gives(2010i=1

yi

)(2010i=1

1

yi

)= (2012 x)(2012 1

x) 20102

so 20122 2012(x+ x1) + 1 20102 0, x+ x1 8045/2012.8. Answer: 2 1

22010

We analyze Q(x) = P (2x) P (x). One can observe that Q(x) 1 has the powers of 2 starting from1, 2, 4, , up to 22010 as roots. Since Q has degree 2011, Q(x) 1 = A(x 1)(x 2) (x 22010)for some A. Meanwhile Q(0) = P (0) P (0) = 0, so

Q(0) 1 = 1 = A(1)(2) (22010) = 2(20102011)/2A.

Therefore A = 2(10052011). Finally, note that the coefficient of x is same for P and Q 1, so it equalsA(20)(21) (22010)((20) + (21) + + (22010)) = A210052011(220111)22010 = 2 122010 .

9. Answer: 665Since the equation

Pk(x) = Pk(x 1) + xk

has all integers 2 as roots, the equation is an identity, so it holds for all x. Now we can substitutex = 1,2,3,4, to prove

Pk(n) = n1i=1

(i)k

so P7(3) + P6(4) = (1)6 (2)6 (3)6 (1)7 (2)7 = 665.


10. Answer: 10

Note that if r is a root of P then r2 is also a root. Therefore r, r2, r22

, r23

, , are all roots of P . SinceP has a finite number of roots, two of these roots should be equal. Therefore, either r = 0 or rN = 1for some N > 0.

If all roots are equal to 0 or 1, then P is of the form axb(x 1)(4b) for b = 0, ..., 4.Now suppose this is not the case. For such a polynomial, let q denote the largest integer such thatr = e2piip/q is a root for some integer p coprime to q. We claim that the only suitable q > 1 are q = 3and q = 5.

First note that if r is a root then one ofr or r is also a root. So if q is even, then one of

e2piip/2q or e2piip+q/2q should also be root of p, and both p/q and (p+ q)/2q are irreducible fractions.This contradicts the assumption that q is maximal. Therefore q must be odd. Now, if q > 6, thenr2, r1, r, r2, r4 should be all distinct, so q 6. Therefore q = 5 or 3.If q = 5, then the value of p is not important as P has the complex fifth roots of unity as its roots,so P = a(x4 + x3 + x2 + x + 1). If q = 3, then P is divisible by x2 + x + 1. In this case we letP (x) = a(x2 + x + 1)Q(x) and repeating the same reasoning we can show that Q(x) = x2 + x + 1 orQ(x) is of form xb(x 1)2b.Finally, we can show that exactly one member of all 10 resulting families of polynomials fits the desiredcriteria. Let P (x) = a(x r)(x s)(x t)(x u). Then, P (x)P (x) = a2(x2 r2)(x2 s2)(x2 t2)(x2 u2). We now claim that r2, s2, t2, and u2 equal r, s, t, and u in some order. We can provethis noting that the mapping f(x) = x2 maps 0 and 1 to themselves and maps the third and fifth rootsof unity to another distinct third or fifth root of unity, respectively. Hence, for these polynomials,P (x)P (x) = a2(x2 r)(x2 s)(x2 t)(x2 u) = aP (x2), so there exist exactly 10 polynomials thatfit the desired criteria, namely the ones from the above 10 families with a = 1.


1. No math tournament exam is complete without a self referencing question. What is the product ofthe smallest prime factor of the number of words in this problem times the largest prime factor of thenumber of words in this problem?

2. King Midas spent 100x % of his gold deposit yesterday. He is set to earn gold today. What percentageof the amount of gold King Midas currently has would he need to earn today to end up with as muchgold as he started?

3. Find all integer pairs (a, b) such that ab+ a 3b = 5.4. Find all the solutions for which f(x) + xf

(1x

)= x.

5. Find the minimum possible value of 2x2 + 2xy + 4y + 5y2 x for real numbers x and y.6. The dollar is now worth 1980 ounce of gold. After the n

th $7001 billion No Bank Left Behind bailoutpackage passed by congress, the dollar gains 1

22n1of its (n 1)th value in gold. After four bank

bailouts, the dollar is worth 1b(1 12c

)in gold, where b,c are positive integers. Find b+ c.

7. Evaluate2009k=1

b k60c

8. Balanced tertiary is a positional notation system in which numbers are written in terms of the digits1 (negative one), 0, and 1 with the base 3. For instance, 1011 = (1)30 + (1)31 + (0)32 + 1(3)3 = 2510.Calculate (1100)(11) + (111) and express your answer in balanced tertiary.

9. All the roots of x3 + ax2 + bx + c are positive integers greater than 2, and the coefficients satisfya+ b+ c+ 1 = 2009. Find a.

10. Let (n) be the number of 1s in the binary expansion of n (e.g. (1) = 1, (2) = 1, (3) = 2, (4) = 1).Evaluate:

10

( n=1

(n)n2

n=0(1)n1(n)

n2

).


1. Answer: 1681

There are 41 words in the problem statement. Since 41 is itself a prime, the answer is 412 = 1681.

2. Answer: 100x1%

After yesterday, the fraction of the initial gold remaining is 1 1x = x1x . Therefore, in order to to reachthe original amount of gold, we must multiply by xx1 = 1 +

1x1 . Thus, the gold must be increased

by 100x1 percent.

3. Answer: (5, 0), (4, 1), (1,2), and (2,3)We factor the expression as follows:

ab+ a 3b 3 = 5 3(a 3)(b+ 1) = 2

We can use a table to find appropriate values for a and b.Thus, (5, 0), (4, 1), (1,2), and (2,3) are the desired solutions.

a 3 b+ 1 a b2 1 5 0-2 -1 1 -21 2 4 1-1 -2 2 3

4. Answer: x = 1

f(x) + xf(

1x

)= x

f

(1x

)+

1xf(x) =

1x

f

(1x

)=

1x 1xf(x)

f(x) + x(

1x 1xf(x)

)= x

x = 1

5. Answer: 54

We complete the square:

2x2 + 2xy + 4y + 5y2 x = (x2 + 2xy + y2) + (x2 x+ 14

) + (4y2 + 4y + 1) (14

+ 1)

= (x+ y)2 + (x 12

)2 + (2y + 1)2 54

Notice that x = 12 and y = 12 would yield the minimum, which is 54 .6. Answer: 506


Let Pn be the value of the dollar in gold after the nth bailout. Let s = 12 . Then after the nth bailout,

the dollar is a factor of (1 + s2n1

) of its (n 1)th value. Thus,

P4 =1

980(1 + s)(1 + s2)(1 + s4)(1 + s8)

=1

980(1 + s+ s2 + s3)(1 + s4)(1 + s8)

=1

980(1 + s+ s2 + s3 + s4 + s5 + s6 + s7)(1 + s8)

=1

980(1 + s+ s2 + s3 + s4 + s5 + s6 + s7 + s8 + s9 + s10 + s11 + s12 + s13 + s14 + s15)

=1

980

(1 s161 s

).

Plug in s = 12 , and we find that P4 =1490

(1 1216

). So b+ c = 490 + 16 = 506.

7. Answer: 32670

Largest multiple of 60 below 2009 is 1980, so find the sum for k = 1 to 1979, so that we have eachvalue of bk/60c exactly 60 times. This sum is therefore 60(1 + 2 + . . . + 32) = 60(1 + 32) 3222 = 31680.The remaining terms are all 33, and there are 2009 1980 + 1 = 30 of them, giving an answer of31680 + 30 33 = 32670.

8. Answer: 1011

(1100)(11) = 11000 + 1100 = 11001100 + 111 = 1011

9. Answer: 58Let the roots be r, s, and t. Then they satisfy r + s+ t = a, rs+ st+ rt = b, and rst = c. So wehave (a+ b+ c+ 1) = r + s+ t rs rt st+ rst 1 = (r 1)(s 1)(t 1) = 2009 = 7 7 41.Thus the roots are 8, 8, and 42, and a = (r + s+ t) = 58.

10. Answer: 20

For convenience, set x =n=1

(n)n2 and y =

n=0

(1)n1(n)n2 .

The crucial observation is that 12 (x+ y) and12 (x y) give the same summation as x, restricted to the

terms with odd n and even n respectively. The latter summation is easily related to x using the factthat (2n) = (n) (since multiplying by 2 is simply appending a 0 in the binary expansion), as follows.

12

(x y) =

even n2

(n)n2

=n=1

(2n)(2n)2

=n=1

(n)4n2

=14x.

Thus we have 12 (x y) = 14x. It follows that x = 2y, so x/y = 2. Thus, the desired answer is 20.


1. Reid is twice as old as Gabe. Four years ago, Gabe was twice as old as Dani. In 10 years, Reid will betwice as old as Dani. How many years old is Reid now?

2. Let P (x) = x6 + ax5 + bx4 +x3 + bx2 + ax+ 1. Given that 1 is a root of P (x) = 0 and 1 is not, whatis the maximum number of distinct real roots that P could have?

3. If a, b, c C and a + b + c = ab + bc + ac = abc = 1, find a, b, c. (The order in which you write youranswers does not matter.)

4. Find x4 + y4 + z4, given that {0 = x + y + z1 = x2 + y2 + z2

5. The product of a 13x5 matrix and a 5x13 matrix contains the entry x in exactly two places. If D(x) isthe determinant of the matrix product, D(x = 0) = 2008, D(x = 1) = 1950, and D(x = 2) = 2142.Find D(x).

6. For how many integers k, with 0 k 2008, does x2 x k = 0 have integer solutions for x?7. Find all ordered pairs of positive integers (p, q) such that 2p2 + q2 = 4608.

8. How many monic polynomials P (x) are there with P (x)Q(x) = x41 for some other polynomial Q(x),where the coefficients of P and Q are in C?

9. Find the number of distinct ordered integer pairs (x,y) with x + y xy = 43.10. Evaluate

k=1

k

5k.


1. Answer: 28 r = 2gg 4 = 2(d 4)r + 10 = 2(d + 10)

So r = 2g = 2(2(d 4) + 4) = 4d 8 = 4(r/2 5) 8 = 2r 28 r = 28.2. Answer: 5

Divide the equation P (x) = 0 by x3 to get x3 + ax2 + bx + 1 + b 1x + a1x2 +

1x3 = 0. In this equation,

replacing x by 1x doesnt change anything, so anytime x is a root of P (x) = 0,1x is also a root. Any

root other than 1 (since 0 isnt a root) must be paired with another, namely its reciprocal. And 1is a root, while 1 is not. So the total number of real roots must be odd. Note that having an oddnumber of distinct real roots requires that 1 be a double root. This makes the maximum number ofreal roots 5.

3. Answer: {i,i, 1}First, notice that the three solutions are symmetric. We write our conditions as a system of equations: a + b + c = 1 (1)ab + bc + ca = 1 (2)

abc = 1 (3)

(3) can be rewritten c = 1/ab. Substituting that in (2), we get

ab + bab +aab = 1

ab + 1/a + 1/b = 1a2b + 1 + a/b = 1a(ab + 1/b) = 0

Because we know from (3) that a = 0 cannot be a solution, we throw it out:

ab + 1/b = 0a = 1/b2

Substituting this as well as our expression for c in (1), we get:1b2 + b +

11/b2(b) = 1

1b2 + b b = 11

b2 = 1b = i

Letting any two variables be i and i, we easily find using any of our three equations that the thirdmust equal 1.

4. Answer: 12

0 = (x + y + z)2 = x2 + y2 + z2 + 2(xy + xz + yz)12

= xy + yz + xz

14

= (xy + yz + xz)2 = x2y2 + x2z2 + y2z2 + 2(x2yz + xy2z + xyz2)

= x2y2 + x2z2 + y2z2 + 2xyz(x + y + z) = x2y2 + x2z2 + y2z2

1 = (x2 + y2 + z2)2 = x4 + y4 + z4 + 2(x2y+x2z2 + y2z2)

1 = x4 + y4 + z4 + 2 14


5. Answer: 3x2 + 61x+ 2008

The highest power of x that can occur in the determinant is x2, so D(x) must be quadratic; let it beax2 + bx + c. The constant term is c = D(0) = 2008, so we have D(1) 2008 = 58 = a b andD(2) 2008 = 134 = 4a + 2b. Solving the pair of linear equations gives a = 3 and b = 61.

6. Answer: 45

By the quadratic formula, the solutions to x2 x k = 0 are precisely

11 + 4k2

.

These solutions are integers precisely when 11 + 4k is an even integer, i.e. when 1 + 4k is an oddinteger. Since 1 + 4k is itself odd,

1 + 4k is an odd integer precisely when 1 + 4k is a perfect square.

Thus, we are interested in how many (nonnegative, to avoid double counting) integers a give an integersolution for k with 0 k 2008 in 1 + 4k = a2, or equivalently to 4k = a2 1. Notice that a2 1 isdivisible by 4 precisely when a is odd. The only other restriction on a is that 4 2008 a2 1. Since89 0 and x2 = 4x1 are solutions to ax2 + bx+ c and that 3a = 2(c b), what is x1?3. Let a, b, c be the roots of x3 7x2 6x+ 5 = 0. Compute (a+ b)(a+ c)(b+ c).4. How many positive integers n, with n 2007, yield a solution for x (where x is real) in the equation

bxc+ b2xc+ b3xc = n?5. The polynomial 400x5 + 2660x4 3602x3 + 1510x2 + 18x 90 has five rational roots. Suppose you

guess a rational number which could possibly be a root (according to the rational root theorem). Whatis the probability that it actually is a root?

6. What is the largest prime factor of 49 + 94?

7. Find the minimum value of xy + x+ y + 1xy +1x +

1y for x, y > 0 real.

8. If r + s+ t = 3, r2 + s2 + t2 = 1, and r3 + s3 + t3 = 3, compute rst.

9. Find a2 + b2 given that a, b are real and satisfy

a = b+1

a+ 1b+ 1a+

; b = a 1b+ 1

a 1b+

10. Evaluate2007k=1

(1)kk2

Algebra Solutions2007 Rice Math Tournament

February 24, 2007

1. Answer: 41/9,1f(x1/9) = (x 4)(x+ 1) so f = 0 means x = 4 or x = 1, so f(41/9) = f(1) = 0.

2. Answer: 14

Since (x x1)(x 4x1) = x2 5x1x + 4x21, we know 3 = 2(4x21 + 5x1) so x1 = 14 . (The other root isnegative.)

3. Answer: 37Simply note that (a+ b)(a+ c)(b+ c) = (ab+ ac+ bc)(a+ b+ c) abc = 6 7 + 5 = 37.

4. Answer: 1339

Let k be a nonnegative integer. Let f(x) = bxc + b2xc + b3xc. If k x < k + 13 , then f(x) = 6k. Ifk+ 13 x < k+ 12 , then f(x) = 6k+1. If k+ 12 x < k+ 23 , then f(x) = 6k+2. If k+ 23 x < k+1,then f(x) = 6k+ 3. There is therefore only a solution if n is 0, 1, 2, or 3 mod 6; there are 2004 46 + 3of these.

5. Answer: 5144

There are clearly five correct guesses; counting the number of possible guesses is the difficult part. Apossible guess q is 1 times a divisor of 90 divided by a divisor of 400. We count these by extendingthe idea of prime factorization: from the factorizations of 90 and 400: we have q = 2i3j5k where4 i 1, 0 j 2, and 2 k 1. There are thus 6 3 4 = 72 possible fractions making 144possible guesses.

6. Answer: 881

We can factor 4x4+ y4 = (4x4+4x2y2+ y4) 4x2y2 = (2x2+ y2)2 (2xy)2 = (2x2+2xy+ y2)(2x22xy + y2). Since 49 + 94 = 4(16)4 + 94, we plug in to obtain the factoring 881 305. Quick checking(up to 29) shows 881 to be prime.

7. Answer: 6

The expression is 6 times the arithmetic mean of the terms, which is is always greater than or equalto the geometric mean, which is xy x y 1x 1y 1xy = 1. The minimum is achieved when all terms areequal, i.e. x = y = 1.

8. Answer: 4

(r + s+ t)3 3(r + s+ t)(r2 + s2 + t2) + 2(r3 + s3 + t3) = 6rst - just plug in!9. Answer:

5

Note that the equations reduce by substitution to a = b+ 1a+1/a and b = a 1b+1/b . Solving the secondfor a, substituting into the first, and reducing yields b4 + b2 1 = 0; solving this as a quadratic in b2yields only one positive value for b2 =

512 . Plugging back in and solving for a gives a

2 =5+12 .

10. Answer: 2015028Note that (x+ 1)2 x2 = 2x+ 1 so:

2007k=1

(1)kk2 = 20072 +1003k=1

(2(2k 1) + 1)

= 20072 + 41003 10032

+ 1003

= 20072 + 1003 2007 = 2007(1003 2007)

Algebra Test2006 Rice Math Tournament

February 25, 2006

1. A finite sequence of positive integers mi for i = 1, 2, . . . , 2006 are defined so that m1 = 1 and mi =10mi1 + 1 for i > 1. How many of these integers are divisible by 37?

2. Find the minimum value of 2x2 + 2y2 + 5z2 2xy 4yz 4x 2z + 15 for real numbers x, y, z.3. A Gaussian prime is a Gaussian integer z = a + bi (where a and b are integers) with no Gaussian

integer factors of smaller absolute value. Factor 4+ 7i into Gaussian primes with positive real parts.i is a symbol with the property that i2 = 1.

4. Simplify: a3

(ab)(ac) +b3

(ba)(bc) +c3

(ca)(cb)

5. Jerry is bored one day, so he makes an array of Cocoa pebbles. He makes 8 equal rows with the pebblesremaining in a box. When Kramer drops by and eats one, Jerry yells at him until Kramer realizes hecan make 9 equal rows with the remaining pebbles. After Kramer eats another, he finds he can make10 equal rows with the remaining pebbles. Find the smallest number of pebbles that were in the boxin the beginning.

6. Let a, b, c be real numbers satisfying:

ab a = b+ 119bc b = c+ 59ca c = a+ 71

Determine all possible values of a+ b+ c.

7. Find all (a, b) so that aabb = n4 6n3 for some integer n, where aabb is a four digit number with aand b non-zero digits. (The answer form was added after the contest)

8. Evaluate:10x=2

2x(x2 1)

.

9. Principal Skinner is thinking of two integers m and n and bets Superintendent Chalmers that he willnot be able to determine these integers with a single piece of information. Chalmers asks Skinner thenumerical value of mn+13m+13nm2n2. From the value of this expression alone, he miraculouslydetermines both m and n. What is the value of the above expression?

10. Evaluate:k=1

kak1 for |a| > 1. (Due to a typographical error, this was printed as |a| < 1 on contest

day; this is the intended question)

1


February 25, 2006

1. Answer: 668Note that 111 = 3 37. It follows that mi is divisible by 37 for all i = 3, 6, 9, . . . , 2004. The others willclearly leave remainders of 1 or 11.

2. Answer: 10The expression can be written as (x 2)2 + (x y)2 + (y 2z)2 + (z 1)2 + 10. This clearly must beat least 10. Indeed, if x = 2, y = 2, z = 1, this value is achieved.

3. Answer: (1 + 2i)(2 + 3i)We write 4 + 7i = (a+ bi)(c+ di). The solution can be intuitive after the first line of expansion, inthe same way as factoring of polynomials. However, we can assume a = 1 and then move factors from(c+ di) back to (a+ bi) if we dont end up with integers (fortunately, in this case were lucky).

4 + 7i = (a+ bi)(c+ di)= ac bd+ (ad+ bc)i= c bd+ (d+ bc)i

We then know c should be positive (and not too large), so we can try c = 1, giving 1 bd = 4 andb+d=7, which clearly has no rational solution. We then try c = 2, giving 6 = bd and 2b+d = 7, whichis easily solved giving the final solution.

4. Answer: a+ b+ c

a3

(a b)(a c) +b3

(b a)(b c) +c3

(c a)(c b) =a3(c b) + b3(a c) + c3(b a)

(a b)(b c)(c a)=

a3(c b) + a(b3 c3) + bc3 cb3(a b)(b c)(c a)

=(c b)(a3 a(b2 + bc+ c2) + b2c+ c2b

(a b)(b c)(c a)= b

2(c a) + b(c2 ac) + a3 ac2(a b)(c a)

= (c a)(b2 + bc ac a2)

(a b)(c a)=

(a+ b)(a b) + c(a b)a b

= a+ b+ c

5. Answer: 352

Let N represent the number of remaining pebbles after Kramer eats the second. Then N is divisibleby 10, and N + 1, which must end in 1, is divisible by 9. Put N + 1 = 100a+ 10b+ 1, where a and bare digits summing to 8 or 17 (so the sum of the digits will be divisible by 9 - hence the number willbe divisible by 9). Now we need N + 2 to be divisible by 8. Try 82, 172, 262, and 352 to get 352 asthe answer.

1

6. Answer: 31,25From the first equation:

ab a = b+ 119a(b 1) = (b 1) + 120

(a 1)(b 1) = 120

Similarly, (b 1)(c 1) = 60 and (a 1)(c 1) = 72. Therefore a1c1 = 2, and so 2(c 1)2 = 72. Thisgives c = 7, and then it is easy to find a = 13 and b = 11. The other solution is c = 5, so a = 11,and b = 9. The sums are 31 and -25.

7. Answer: (6, 5)Since 11|aabb, aabb = 11 a0b. Factor n4 6n3 = (n 6)n3, so clearly n > 6, as aabb > 0. Also,a0b < 1000, so unless n = 11, n < 10. Trying n = 7, 8, 9 yields no solutions, so n = 11 must be theonly solution, if it exists. Indeed we get 6655 = (11 6) 113.

8. Answer: 2755

2x(x21) =

1x

(1

x1 1x+1)= 1x(x1) 1x(x+1)

Let f(x) = 1x(x1) . Then:10x=2

2x(x21) =

10x=2(f(x) f(x + 1)) =

10x=2 f(x)

11x=3 f(x) = f(2) f(11) = 121 11110 =

12 1110 = 2755

9. Answer: 169Let A be the value of the expression. We have: m2 + n2 13m 13nmn+ A = 0. Multiplying by2 yields:

m2 2mn+ n2 +m2 26m+ n2 26n+ 2A = 0(m n)2 + (m 13)2 + (n 13)2 = 2 13 2 2A

In order for there to be a single solution, the sum of the squares must equal zero, yielding A = 169.If instead the sum is a positive integer with a solution (m,n), then (n,m) will provide an additionalsolution unlessm = n. In that case, (26m, 26n) is an additional solution. Hence, it is both sufficientand necessary that the sum of the squares equal zero in order that the solution be unambiguous.

10. Answer:(

aa1

)2k=1

k

ak1=

11+2a+

3a2

+4a3

+

=(11+1a+

1a2

+1a3

+ )+(1a+

1a2

+1a3

+ )+(1a2

+1a3

+ )+

=(11+1a+

1a2

+1a3

+ )+1a

(11+1a+

1a2

+ )+

1a2

(11+1a+

)+

=(11+1a+

1a2

+1a3

+ )(

11+1a+

1a2

+1a3

+ )

=(

11 1/a

)2=(

a

a 1)2

2


February 26, 2005

1. On a test, the average score for the girls in the class is 91, and the average score for the boys in theclass is 85. If the average score for the class is 89, what fraction of the class are boys?

2. How many distinct real roots does the following equation have?

x4 + 8x2 + 16 = 4x2 12x+ 9

3. At William Rices Marsh, there are an infinite number of magic lillypads numbered 1, 2, 3, and soon. A magic lillypad lights up if a frog jumps on it while it is not lit, and turns off if a frog jumpson it while it is lit. Suppose all lillipads are initially turned off. Conor the frog begins by hopping onthe first lillypad and then continues hopping on every lillypad thereafter. Conors friend Bob startshopping after Conor and begins by hopping on the second pad and continues by hopping on the fourth,sixth, eighth, and so on. Shortly after Bob, Dan hops on the third, sixth, ninth, and so on lillypads. Ifthere is a frog for each positive number n that hops on every nth pad, what is the number on the mth

lillypad that remains lit in the end?

4. Ashley Ann Allen, a hapless algebra student, sees the expression logAlogB . She mistakenly cancels thelogs, to get the expression AB . Miraculously, when she plugs in values for A and B, she gets thecorrect answer. Assuming A 6= B, find all possible ordered pairs (A,B)

5. Let f(x) and g(x) be functions which take integers as arguments. Let

f(x+ y) = f(x) + g(y) + 8

for all integers x and y. Let f(x) = x for all negative numbers x, and let g(8) = 17. What is f(0)?

6. Letx = b2007 2006 2004 20031

3 20054c,

where bxc is the greatest integer less than or equal to x. Find (x2 + 1)(((x2 + 1) x2)2 + 1).

7. If the roots of x3+ax2+bx+c are three consecutive positive integers, then what are all possible valuesof a

2

b+1?

8. Find all ordered pairs of digits (a, b) such that the 6 digit number 24ab32 is divisible by 99.

9. If a, b, c are real numbers such thata+ b+ c = 1a2 + b2 + c2 = 17

a3 + b3 + c3 = 11

Find abc.

10. A monic polynomial is one in which the coefficient of the highest order term is 1. Find the monicpolynomial p(x) (with integer coefficients) of least degree that satisfies p(

2 +

5) = 0.

1


February 26, 2005

1. Answer: 13

If g is the fraction of the class that are girls and b is the fraction of the class that are boys, then91g + 85b = 89 and g + b = 1, so 91(1 b) + 85b = 89, which simplifies to b = 13 .

2. Answer: 1

(x2 + 4)2 = (2x 3)2

(x2 + 4)2 (2x 3)2 = 0(x2 + 4 + 2x 3)(x2 + 4 2x+ 3) = 0

(x2 + 2x+ 1)(x2 2x+ 7) = 0(x+ 1)2(x2 2x+ 7) = 0

x2 2x+ 7 has no real roots since 22 4 7 < 0.3. Answer: m2

A lillypad will be off if an even number of frogs jump on it. Hence, the mth lillypad will be on if mhas an odd number of factors. m has an odd number of factors m is a perfect square.

4. Answer: {(2, 4), (4, 2)}

logAlogB

=A

B

B logA = A logB

logAB = logBA

AB = BA

(A,B) {(2, 4), (4, 2)}

5. Answer: 17

f(0) = f(-8+8)= f(-8) + g(8) + 8= -8 + 17 + 8= 17

6. Answer: 2005

x = 2 so (x2 + 1)(((x2 + 1) x2)2 + 1) = 2005.

7. Answer: 3

The roots are n, n + 1, and n + 2. Then a2 = ((n) + (n + 1) + (n + 2))2 = (3n + 3)2 = 9(n + 1)2.b+ 1 = n(n+ 1) + n(n+ 2) + (n+ 1)(n+ 2) + 1 = 3n2 + 6n+ 3 = 3(n+ 1)2. So a

2

b+1 = 3.

1

8. Answer: a = 4, b = 3

Since 24ab32 is divisible by 9,2 + 4 + a+ b+ 3 + 2 0(mod9)

a+ b 2(mod9)So we have either a+ b = 7 or a+ b = 16Since 24ab32 is divisible by 11,

2 + 4 a+ b 3 + 2 0(mod11)

a b 1(mod11)So a b = 1. Hence, the only solution is a = 4, b = 3.

9. Answer: 12

Let a, b, c be the roots of x3 +Ax2 +Bx+ C = 0.

A = (a+ b+ c) = 1.

B = ab+ bc+ ac =12[(a+ b+ c)2 (a2 + b2 + c2)] = 1

2[1 17] = 8

Therefore:a3 +Aa2 +Ba+ C = 0

b3 +Ab2 +Bb+ C = 0

c3 +Ac2 +Bc+ C = 0

Add them up to get:11 +A(17) +B(1) + C = 0

11 + 17 + 8 + 3C = 0

C =363

= 12C = abc.

Thus,abc = 12.

Alternatively, solve to geta = b = 2, c = 3.

10. Answer: x4 14x2 + 9Let x =

2 +

5.

Then x2 = 210 + 7

x4 = 89 + 2810

x4 14x2 = 9so if p(x) = x4 14x2 + 9, then p(2 +5) = 0.p(x) must be of at least degree 4 since

2 +

5 is not the root of any quadratic or cubic polynomials

with integer coefficients.

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February 28, 2004

1. How many ordered pairs of integers (a,b) satisfy all of the following inequalities?

a2 + b2 < 16a2 + b2 < 8aa2 + b2 < 8b

2. Find the largest number n such that (2004!)! is divisible by ((n!)!)!.

3. Compute:

b 20053

2003 2004 20033

2004 2005c.Where bxc denotes the greatest integer less than x.

4. Given that:

x > 0, y > 0, z > 0,

xy 3x 7y + 15 = 0,xz 2x 7z + 8 = 0,yz 2y 3z + 2 = 0.

Solve for x, y, and z.

5. There exists a positive real number x such that cos(tan1(x)) = x. Find the value of x2.

6. Dan, Shravan, and Jake attempt to clean their room for the first time in six months. It would takeDan 3 hours to clean it by himself. It would take Shravan 10 hours and Jake 15 hours to clean theroom. How long does it take the three of them together to clean the room?

7. Find all real x such thatx(x 2)(x 4)

(x 1)(x 3)(x 5) < 0.

8. Let x be a real number such that x3 + 4x = 8. Determine the value of x7 + 64x2.

9. A sequence of positive integers is defined by a0 = 1 and an+1 = a2n + 1 for each n 0. Findgcd(a999, a2004).

10. There exists a polynomial P of degree 5 with the following property: if z is a complex number suchthat z5 + 2004z = 1, then P (z2) = 0. Calculate the quotient P (1)P (1) .

1

Algebra Test Solutions

2004 Rice Math Tournament

February 28, 2004

1. Answer: 6

This is easiest to see by simply graphing the inequalities. They correspond to the (strict) interiors ofcircles of radius 4 and centers at (0, 0), (4, 0), (0, 4), respectively. So we can see that there are 6 latticepoints in their intersection (circled in the figure).

2. Answer: 6

For positive integers a, b, we havea! | b! a! b! a b.

Thus,(n!)!)! | (2004!)! (n!)! 2004! n! 2004 n 6.

3. Answer: 8

Let x = 2004. Then the expression inside the floor brackets is

(x + 1)3

(x 1)x (x 1)3x(x + 1)

=(x + 1)4 (x 1)4

(x 1)x(x + 1) =8x3 + 8x

x3 x = 8 +16x

x3 x .

Since x is certainly large enough that 0 < 16x/(x3 x) < 1, the answer is 8.4. Answer: x = 10, y = 5, z = 4

Factoring, (x 7)(y 3) = 6, (x 7)(z 2) = 6, (y 3)(z 2) = 4. This implies that

x 7 = 3y 3 = 2z 2 = 2

Thus x = 10, y = 5, z = 4.

5. Answer: 1+

5

2

Draw a right triangle with legs 1, x; then the angle opposite x is tan1 x, and we can computecos() = 1

x2+1. Thus, we only need to solve x = 1

x2+1. This is equivalent to x

x2 + 1 = 1. Square

both sides to get x4 + x2 = 1 x4 + x2 1 = 0. Use the quadratic formula to get the solutionx2 = 1+

5

2 (unique since x2 must be positive).

6. Answer: 2hours.

Adding the individual rates, we get 13 +110 +

115 =

12 of the room is cleaned per hour so the whole room

takes two hours.

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7. Answer: 0 < x < 1 or 2 < x < 3 or 4 < x < 5

The sign of one of the terms switches every time x moves from the range (I, I + 1) to (I + 1, I + 2).When x is less than zero, all terms are negative so the LHS is negative. Also note that x is undefinedat 1, 3, 5.

8. Answer: 128

For any integer n 0, the given implies xn+3 = 4xn+1 + 8xn, so we can rewrite any such power of xin terms of lower powers. Carrying out this process iteratively gives

x7 = 4x5 + 8x4= 8x4 + 16x3 32x2= 16x3 64x2 + 64x= 64x2 + 128.

Thus, our answer is 128.

9. Answer: 677

If d is the relevant greatest common divisor, then a1000 = a2999 + 1 1 = a0 (mod d), which implies

(by induction) that the sequence is periodic modulo d, with period 1000. In particular, a4 a2004 0.So d must divide a4. Conversely, we can see that a5 = a

24 + 1 1 = a0 modulo a4, so (again by

induction) the sequence is periodic modulo a4 with period 5, and hence a999, a2004 are indeed bothdivisible by a4. So the answer is a4, which we can compute directly; it is 677.

10. Answer: 20100122010013

Let z1, . . . , z5 be the roots of Q(z) = z5 + 2004z 1. We can check these are distinct (by using the

fact that theres one in a small neighborhood of each root of z5 + 2004z, or by noting that Q(z) isrelatively prime to its derivative). And certainly none of the roots of Q is the negative of another, sincez5 + 2004z = 1 implies (z)5 + 2004(z) = 1, so their squares are distinct as well. Then, z21 , . . . , z25are the roots of P , so if we write C for the leading coefficient of P , we have

P (1)P (1) =

C(1z21)...(1z2

5)

C(1z21)...(1z2

5)

= [(1z1)...(1z5)][(1+z1)...(1+z5)][(iz1)...(iz5)][(i+z1)...(i+z5)]= [(1z1)...(1z5)][(1z1)...(1z5)][(iz1)...(iz5)][(iz1)...(iz5)]

= (15+200411)(15+2004(1)1)

(i5+2004i1)(i5+2004(i)1)

= (2004)(2006)(1+2005i)(12005i)= 20052120052+1= 40200244020026 = 20100122010013 .

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algebra from rice university

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