A1. Find all ordered pairs (a, b) of positive integers for which 1 a + 1 b = 3

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1 A. Find all ordered pairs a, b) of positive integers for which a + b = Answer. The six ordered pairs are 009, 08), 08, 009), , 674) = 35043, 674), , 673) = 3584, 673), 674, ) = 674, 35043), and 673, ) = 673, 3584). Solution. First rewrite the equation as 009a + b) = 3ab, and note that 009 is prie, so at least one of a and b ust be divisible by 009. If both a and b are divisible by 009, say with a = 009q, b = 009r, then we have q + r) = 3qr. But qr q + r for integers q, r, so at least one of q, r is. This leads to the solutions q =, r = and r =, q =, corresponding to the ordered pairs a, b) = 009, 08) and a, b) = 08, 009). In the reaining case, just one of a and b is divisible by 009, say a = 009q. This yields q + b) = 3 009qb, which can be rewritten as 009q = 3q )b. Because the prie 009 does not divide b, it ust divide 3q ; say 3q = 009k. Then 009k + = 3 336k + k + is divisible by 3, so k od 3). For k =, we get q = 337, a = , b = q = 674. For k = 4, we get q = 346, a = , b = q/ = 673. We now show there is no solution with k > 4. Assuing there is one, the corresponding value of q is greater than 346, and so the corresponding b = q 3q 009 is less than 673. Because b is an integer, it follows that b 67, which iplies b 67 > 3 08, contradicting a + b = Finally, along with the two ordered pairs a, b) for which a is divisible by 009 and b is not, we get two ore ordered pairs by interchanging a and b. A. Let S, S,..., S n be the nonepty subsets of {,,..., n} in soe order, and let M be the n ) n ) atrix whose i, j) entry is { 0 if S i S j = ; ij = otherwise. Calculate the deterinant of M. Answer. The deterinant is if n = and if n >. Solution. Note that if we take the nonepty subsets of {,,..., n} in a different order, this will replace the atrix M by P MP for soe perutation atrix P, and the deterinant will stay unchanged. Denote the atrix M by M n to indicate its dependence on n. Then we will show that detm n+ ) = detm n )), fro which the given answer follows by induction on n because M has the single entry. Using the order S, S,..., S n for the nonepty subsets of {,,..., n}, we arrange the nonepty subsets of {,,..., n + } in the order S, S,..., S n, {n + }, S {n + }, S {n + },..., S n {n + }. Then any of the first n subsets has epty intersection with {n + }, any two of the last n subsets have nonepty intersection because they both contain n +, and for any other choice of two nonepty subsets of {,,..., n + }, whether they have

2 nonepty intersection is copletely deterined by the relevant entry in M n. Thus the atrix M n+ has the following block for: M n+ = M n 0 n M n T T 0 n n, M n n E n where 0 n, n denote colun vectors of length n, all of whose entries are 0, respectively, and E n is the n ) n ) atrix, all of whose entries are. To find detm n+ ), we subtract the iddle row fro each of the rows below it. This will not affect the lower left block, it will change the lower part of the iddle colun fro n to 0 n, and it will replace the lower right block E n by a block of zeros. Then we can expand the deterinant using the iddle colun whose only nonzero entry is now the central ) to get ) ) Mn M detm n+ ) = det n Mn O = det, O M n M n M n where the last step is carried out by switching the ith row with the n + i)th row for all i =,,..., n. Because this is an odd nuber of row swaps, the sign of the deterinant ) is reversed.) Finally, the deterinant of the block triangular atrix Mn O is equal to the product of the deterinants of the diagonal blocks, so it M n M n equals detm n )), and the result follows. A3. Deterine the greatest possible value of 0 cos3x i ) for real nubers x, x,..., x 0 satisfying 0 cosx i ) = 0. i= i= Answer. The axiu value is Solution. Let z i = cosx i ). Then the real nubers z, z,..., z 0 ust satisfy z i, and the given restriction on the x i is equivalent to the additional restriction z + z + + z 0 = 0 on the z i. Also, note that cos3x i ) = cosx i ) cosx i ) sinx i ) sinx i ) = cos x i ) ) cosx i ) sin x i ) cosx i ) = cos 3 x i ) cosx i ) cos x i )) cosx i ) = 4z 3 i 3z i. Thus we can rephrase the proble as follows: Find the axiu value of the function f given by on the set 0 0 fz,..., z 0 ) = 4zi 3 3z i ) = 4 i= i= S = {z,..., z 0 ) [, ] 0 z + + z 0 = 0}. Because f is continuous and S is closed and bounded copact), f does take on a axiu value on this set; suppose this occurs at,..., 0 ) S. Let z 3 i

3 a = +. If we fix z 3 = 3,..., z 0 = 0 and z + z = a, then the function gz) = 4 fz, a z, 3,..., 0 ) ) = z 3 + a z) 3 = a 3 3a z + 3az has a global axiu at z = on the interval for z that corresponds to z, a z, 3,..., 0 ) S, which is the interval [a, ] if a 0 and the interval [, a + ] if a < 0. If a 0 that axiu occurs at both endpoints of the interval that is, when either z = or a z = ), while if a < 0 the axiu occurs when z = a z = a/. We can conclude that either in the first case) one of, equals or in the second case) =. Repeating this arguent for other pairs of subscripts besides, ), we see that whenever two of the i are distinct, one of the equals. So there are at ost two distinct values aong the i, naely and one other value v. Suppose that occurs d ties aong the i, so that v occurs 0 d ties. Then because = 0, we have d + 0 d)v = 0, so v = d/d 0) and the axiu value of f on S is f,..., 0 ) = 4 d 3 d ) 3 4d d) = 4d = hd), say. d 0 0 d) In order for v = d/d 0) to be in the interval [, ], we ust have 0 d 5. So to finish, we can ake a table of values d hd) /8 5/ 3 480/ /9 5 0 and read off that the axiu value is 480 = 9 39, for d = 3. This value occurs for x = x = x 3 = 0, x 4 = x 5 = = x 0 = arccos 3/7), which corresponds to z = z = z 3 =, z 4 = z 5 = = z 0 = 3/7.) A4. Let and n be positive integers with gcd, n) =, and define k k ) a k = n n for k =,,..., n. Suppose that g and h are eleents in a group G and that gh a gh a gh an = e, where e is the identity eleent. Show that gh = hg. As usual, x denotes the greatest integer less than or equal to x.) Solution. The proof is by induction on n. If n =, we have a = and gh = e, so g = h and g, h coute. For the induction step, first consider the case that < n. Then each a k is 0 or, and a + + a n =. The values k,... k of k for which a k = are the sallest values for which k n k, n,..., k n,

4 so they are k = n, k = n,..., k = n = n. Thus the given relation gh a gh a gh an = e can be rewritten, by oitting all factors h 0, as g k hg k k h g k k h = e. Taking the inverse of both sides, we get h g ) k k h g ) k k h g ) k = e. We clai that this new relation is one of the original for, but with g and h replaced by h and g respectively, and with the roles of and n interchanged. Because < n, it will follow by the induction hypothesis that h and g coute, and thus g and h coute. To prove the clai, note that the exponents of g in the new relation are b = k k,..., b = k k, b = k, and so, with the convention k 0 = 0, we have i + )n i)n b i = k i+ k i = i + )n i)n = + in i )n = n n in i )n =, as desired. Now consider the reaining case, in which > n. Write = qn+r with 0 r < n. We have a k = k k ) = qk + rk n n n rk ) qk ) = q + a n k where a k = rk rk ). If we set g = gh q then we have n n g h a g h a g h a n = gh a gh a gh an = e. This is again a relation of the original for, but with replaced by r and g replaced by g ). So by the case considered above, g and h coute. As a result, gh)h q = g h = hg = hg)h q and it follows that gh = hg, copleting the proof. A5. Let f : R R be an infinitely differentiable function satisfying f0) = 0, f) =, and fx) 0 for all x R. Show that there exist a positive integer n and a real nuber x such that f n) x) < 0. Solution. Suppose, to the contrary, that f n) x) 0 for all integers n 0 and all x R. To begin, note that f has a iniu at x = 0, so f 0) = 0. Now we show by induction on n that for every function f satisfying the given conditions and such that f n) x) 0 for all n 0 and all x, we have fx) n fx) for all n 0 and all x 0. Since f 0, we have fx) fx) for x 0, showing the base case n = 0. Suppose

5 we have shown for soe particular n that fx) n fx) for all relevant functions f and all x 0. Note that because f0) = 0 and f) =, f x) > 0 for soe x [0, ]; because f is nondecreasing, it follows that f ) > 0. Therefore, we can define a function g by gx) = f x)/f ), and for this infinitely differentiable function we have that g0) = 0, g) =, and all derivatives of g are nonnegative. By the induction hypothesis, we then have gx) n gx), and therefore f x) n f x), for all x 0. Integrating with respect to x fro 0 to y gives fy) n fy) for all y 0, so this shows that fx) n+ fx) for all x 0, copleting the induction proof. In particular, we now have f) n f) = n for all n 0, which is obviously false. So it ust be the case that f n) x) < 0 for soe integer n 0 and soe x R. A6. Suppose that A, B, C, and D are distinct points in the Euclidean plane no three of which lie on a line. Show that if the squares of the lengths of the line segents AB, AC, AD, BC, BD, and CD are rational nubers, then the quotient area ABC) area ABD) is a rational nuber. NOTE: I don t believe this is quite the final wording of this proble. Solution. Let v, w, z be the displaceent vectors AB, #» AC, #» AD #» fro A to the points B, C, D respectively. Then the dot products v v = v, w w, v w) v w) are all rational, so v w = v v + w w v w) v w)) is also rational; siilarly, v z and w z are rational. Because A, B, and C are not collinear, v and w for a basis for R, so there are constants λ, µ R such that z = λv + µw. Then we have the syste of linear equations v z = v v) λ + v w) µ w z = w v) λ + w w) µ for λ and µ. The deterinant v v)w w) v w) of the atrix of this syste is positive because v and w are linearly independent) by the Cauchy-Schwarz inequality. Thus, because all coefficients of the syste are rational, λ and µ are also rational. Now we can rewrite the desired quotient as area ABC) area ABD) detv w) / = detv z) / = detv w) λ detv v) + µ detv w) =, µ a rational nuber. Note that µ 0 because A, B, D are not collinear.) The B section starts on the next page.)

6 B. Let P be the set of vectors defined by { } a P = 0 a, 0 b 00, and a, b Z. b) Find all v P such that the set P \ {v} obtained by oitting vector v fro P can be partitioned into two sets of equal size and equal su. Answer. The vectors v of the for with b even, 0 b 00. b) Solution. First note that if we add all the vectors in P by first suing over a) for fixed b, we get the su of for 0 b 00, which is = 3b) ) ) 303. Thus if the set P \ {v} is to be partitioned into two sets of equal ) 303 a su, the vector v ust have both coordinates even. For v =, b) this iplies that a is odd and b is even, so because v P, we have a =, b even, 0 b 00. It reains to show that this necessary condition on v is also sufficient. c Identify each of the vectors w = in P with the lattice point c, d). Given a d) particular vector v = in P with b even, there are 30 lattice points in P \ {v}. b) If we can nuber these points P, P,..., P 30 such that the su of the displaceent vectors P #» P, P #» 3 P 4,..., P #» 30 P 30 is zero, then we can partition P \ {v} into the set of points P i with i odd and the set of points P i with i even, and those sets will have equal size and equal su. To do so, we first partition the set P \ {v} into 4 rectangular sets of 4 3 lattice points, along with a single 5 3 rectangular set fro which one of the points in the iddle colun is issing. For the 4 3 rectangular sets, we can take three displaceent vectors pointing up and three displaceent vectors pointing down, as shown in the first diagra below. For the 5 3 rectangular set with a single point issing, there are up to syetry) two cases, depending on whether the issing point is on an edge or at the center the parity condition on b guarantees that it will be either on an edge or at the center). These are shown in the second and third diagras below. As an exaple, if the 5 3 rectangle is the set { } a 0 a, 0 b 4, and a, b Z b) and v = 0) is the issing point, then the second diagra corresponds to the partition of the rectangular set inus that point into the two subsets of equal size and equal su {0, 0), 0, ), 0, 4),, ),, 4),, ),, )} and {0, ), 0, 3),, 4),, ),, 3),, 3),, 0)}, which contain the starting and end points, respectively, of the displaceent vectors shown.

7 B. Let n be a positive integer, and let f n z) = n + n )z + n )z + + z n. Prove that f n has no roots in the closed unit disk {z C : z }. Solution. If z is a root of f n, then 0 = z)f n z) = n n j= zj, so n j= zj = n. n n On the other hand, when z, we have z j z j n, with equality only j= j= for z =. Because z = is not a root of f n, we are done. B3. Find all positive integers n < 0 00 for which siultaneously n divides n, n divides n, and n divides n. Answer. n =, 4, 6, 56. Solution. We first prove that for positive integers a and b, a divides b if and only if a divides b. If a divides b then we can write b = aq, and odulo a we then have b = a ) q q = 0. Conversely, suppose that a divides b. Let b = aq + r with 0 r < a; we then have b = r aq ) + r ). Because a divides b and aq, it ust also divide r. Because 0 r < a, it follows that r = 0, so a divides b. A positive integer n divides n if and only if n is a power of. So we ay assue that n = for soe nonnegative integer. Now n = divides n = if and only if divides, so if and only if is a power of. So we ay assue that = l, that is, n = l, for soe nonnegative integer l, and we have to find all n < 0 00 of this for for which n divides n. Note that n = l ), so n divides n = n ) = l ) if and only if l divides l, which is if and only if l divides l, that is, if and only if l is a power of. We can write l = k, so that n = k, and to finish we have to find the values of k 0 for which n < Suppose that n is a solution with n < Then we have = n < 0 00 < 4 ) 00 = 400. It follows that = l < 400 < 9, so l = k < 9 < 4 and thus 0 k 3. Conversely, 3 = 8 = 56 < 300 = 3 ) 00 < So the solutions are n = k with k = 0,,, 3; in other words, n is equal to, 4, 6 or 56. B4. Given a real nuber a, we define a sequence by x 0 =, x = x = a, and x n+ = x n x n x n for n. Prove that if x n = 0 for soe n, then the sequence is periodic. Solution. Let z be a coplex) root of the polynoial z az +. Then z 0, and z satisfies az = z +, so a = z + z. If the Fibonacci nubers F 0, F, F,... are

8 given, as usual, by F 0 = 0, F =, and F n+ = F n + F n for n, then we see that x n = zfn + z Fn for n = 0,,. We now show by induction on n that this equation holds for all n. For the induction step, assuing the equation is correct for n, n, and n, we have x n+ = x n x n x n = zfn + z Fn )z F n + z F n ) z F n + z F n ) = zfn+f n + z Fn+F n ) + z Fn F n + z Fn F n ) z F n + z F n ) = zf n+ + z F n+, because F n F n = F n. Suppose that x n = 0 for soe n, say x = 0. Then z F =, so z is a dth root of unity, where d = 4F. Now note that the Fibonacci sequence odulo d is periodic: There are only finitely any specifically, d ) possibilities for a pair F i od d, F i+ od d) of successive Fibonacci nubers odulo d, and when a pair reoccurs, say F i od d, F i+ od d) = F j od d, F j+ od d) with i < j, it is straightforward to show by induction that F i+k = F j+k od d for all k, including negative k for which + z i + k 0. But then, because x n = zfn Fn is deterined by the value of F n odulo d, the sequence x n ) is also periodic, and we are done. B5. Let f = f, f ) be a function fro R to R with continuous partial derivatives f i x j that are positive everywhere. Suppose that f x f x 4 f x + f x ) > 0 everywhere. Prove that f is one-to-one. Solution. Consider the Jacobian atrix f f J = Jx, x ) = x x f f x x and the related syetric atrix f A = J + J T ) = x f + f ) x x f + f x x f x ) Fro the givens, the entries of A are positive everywhere, and the deterinant of A is also positive everywhere. So the eigenvalues of A are positive because their su tra) and their product are both positive), and so A is positive definite. That is, for any nonzero vector v, we have v T Av = v Av > 0, so v T J +J T )v = v T Jv+v T Jv) T > 0..

9 But the scalar v T Jv is its own transpose, showing that v T Jv > 0 at all points in R and for all nonzero vectors v. We now show that if P and Q are distinct points in R, then the dot product of the displaceent vectors P # Q» and fp # )fq)» is positive; it then follows that fp ) fq), showing that f is one-to-one. In doing so, we will identify points in R with vectors. In particular, the displaceent vector fp # )fq)» can be written as #» fp )fq) = fq) fp ) = fp + t P # Q)» t=0 d [ = fp + t #» ] P Q) dt dt = 0 0 JP + t #» P Q) #» P Q dt, using the ultivariable chain rule. Thus the dot product of this vector and P # Q» is #» P Q fp # )fq)» = P # Q» JP + t P # Q)» P # Q» ) dt = 0 0 v T JP + tv)v dt, where v = #» P Q. By our earlier observation, the integrand is positive for all t, and so the integral is also, copleting the proof. B6. Let S be the set of sequences of length 08 whose ters are in the set {,, 3, 4, 5, 6, 0} and su to Prove that the cardinality of S is at ost ) Solution. Note that the binary expansion of 08 is 08 = = , and consequently = 08 = We can interpret this as the probability that a rando variable X takes on a value in the set A = {,, 3, 4, 5, 6, 0}, if the possible values of the variable are all the positive integers, and the probability of taking on the value k is k. Note that k =, so this is a consistent assignent of probabilities.) Now consider a sequence X = X n ) 08 n= of independent rando variables that each take positive integer values k with probability k. For any given sequence s n ) S, the probability that X n = s n for all n with n 08) equals s s s 08 = s n = Therefore, the probability that the sequence X n ) will be in S is PX S) = S k=

10 On the other hand, this probability is less than the probability that each X n takes on a value in A; as we have seen, for any X n individually that probability is 08/048, and so because the X n are independent, the probability that they will all take on values in A is 08/048) 08. So we have the inequality ) S 3860 = PX S) <, 048 and the desired result follows.