Đang tải... (xem toàn văn)
Tạp chí toán học AMM của Mỹ
PROBLEMS AND SOLUTIONS Edited by Gerald A. Edgar, Doug Hensley, Douglas B. West with the collaboration of Paul T. Bateman, Mario Benedicty, Itshak Borosh, Paul Bracken, Ezra A. Brown, Randall Dougherty, Tam ´ as Erd ´ elyi, Zachary Franco, Chris- tian Friesen, Ira M. Gessel, Jerrold Grossman, Frederick W. Luttmann, Vania Mas- cioni, Frank B. Miles, Richard Pfiefer, Cecil C. Rousseau, Leonard Smiley, John Henry Steelman, Kenneth Stolarsky, Richard Stong, Walter Stromquist, Daniel Ull- man, Charles Vanden Eynden, and Fuzhen Zhang. Proposed problems and solutions should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before December 31, 2008. Additional information, such as gen- eralizations and references, is welcome. The problem number and the solver’s name and address should appear on each solution. An asterisk (*) after the num- ber of a problem or a part of a problem indicates that no solution is currently available. PROBLEMS 11376. Proposed by Bogdan M. Baishanski, The Ohio State University, Columbus OH. Given a real number a and a positive integer n,let S n (a) = an<k≤(a+1)n 1 √ kn −an 2 . For which a does the sequence S n (a) converge? 11377. Proposed by Christopher Hillar, Texas A&M University, College Station, TX and Lionel Levine, Massachusetts Institute of Technology, Cambridge, MA. Given a monic polynomial p of degree n with complex coefficients, let A p be the (n + 1) × (n +1) matrix with p(−i + j) in position (i, j),andletD p be the determinant of A p . Show that D p depends only on n, and find its value in terms of n. 11378. Proposed by Daniel Troy, Professor Emeritus, Purdue University Calumet, Hammond, IN. Let n be a positive integer, and let U 1 , ,U n be random variables defined by one of the following two processes: A: Select a permutation of {1, ,n} at random, with each permutation of equal probability. Then take U k to be the number of k-cycles in the chosen permutation. B: Repeatedly select an integer at random from {1, ,M} with uniform distribu- tion, where M starts at n and at each stage in the process decreases by the value of the last number selected, until the sum of the selected numbers is n.ThentakeU k to be the number of times the randomly chosen integer took the value k. Show that the probability distribution of (U 1 , ,U n ) is the same for both processes. 11379. Proposed by Oskar Maria Baksalary, Adam Mickiewicz University, Pozna ´ n, Poland, and G ¨ otz Trenkler, Technische Universit ¨ at Dortmund, Dortmund, Germany. Let A be a complex matrix of order n whose square is the zero matrix. Show that R(A + A ∗ ) = R(A) + R(A ∗ ),whereR(·) denotes the column space of a matrix ar- gument. 664 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 11380. Proposed by Hugh Montgomery, University of Michigan, Ann Arbor, MI, and Harold S. Shapiro, Royal Institute of Technology, Stockholm, Sweden. For x ∈ R,let x k = 1 k! k−1 j =0 (x − j).Fork ≥ 1, let a k be the numerator and q k the denominator of the rational number −1/3 k expressed as a reduced fraction with q k > 0. (a) Show that q k is a power of 3. (b) Show that a k is odd if and only if k is a sum of distinct powers of 4. 11381. Proposed by Jes ´ us Guillera, Zaragoza, Spain, and Jonathan Sondow, New York, NY. Show that if x is a positive real number, then e x = ∞ n=1 n k=0 (kx + 1) (−1) k+1 n k 1/n . 11382. Proposed by Roberto Tauraso, Universit ` a di Roma “Tor Vergata”, Rome, Italy. For k ≥ 1, let H k be the kth harmonic number, defined by H k = k j =1 1/j. Show that if p is prime and p > 5, then p −1 k=1 H 2 k k ≡ p −1 k=1 H k k 2 (mod p 2 ). (Two rationals are congruent modulo d if their difference can be expressed as a reduced fraction of the form da/b with b relatively prime to a and d.) SOLUTIONS A Group In Which Equations Are Solvable 11231 [2006, 567]. Proposed by Christopher J. Hillar, Texas A&M University, College Station, TX. Find a non-Abelian group G with the following property: for each positive integer n, each word W on the alphabet of n + 1 letters A 1 , ,A n and X, each list a 1 , ,a n of elements of G, and each b in G there exists a unique x in G such that W (a 1 , ,a n , x) = b . (Thus, in particular, ax 2 ax = b must have a unique solution x.) Solution by O. P. Lossers, Eindhoven University of Technology, Eindhoven, The Netherlands. We define such a group on R × R. The group operation, denoted ⊕,is defined by (a,α)⊕ (b,β)= (a + b,α+ e a β). This operation is associative: ((a,α)⊕ (b,β))⊕ (c,γ)= ((a + b) + c,(α + e a β) + e a+b γ) = (a +(b + c), α +e a (β + e b γ)) = (a,α)⊕((b,β)⊕(c,γ)). Clearly (0, 0) is an identity element. Inverses exist, with (a,α) −1 = (−a, −αe −a ).The operation is not commutative, since in general α +βe a = β +αe b . Assume that W contains at least one X. We seek a unique solution to W ((a 1 ,α 1 ), ,(a n ,α n ), (x,ξ))= (b,β). August–September 2008] PROBLEMS AND SOLUTIONS 665 If W has length l,then(b,β)is the result of l −1 applications of ⊕. Thus b is the sum of the first coordinates of the pairs in W ,andβ is the sum of l terms such that the ith term is the second coordinate of the ith pair in W times the exponential of the sum of the first coordinates of the previous pairs in W . We can thus write (b,β) as (mx + s,σ + μξ),wherem is the number of occur- rences of X in W,ands is an additive combination of a 1 , ,a n . Thus x = (b −s)/m, uniquely determined. Furthermore, μ is a sum of exponentials of additive combina- tions of x and a 1 , ,a n and hence is positive. Also, σ is a linear combination of α 1 , ,α n with coefficients depending on x and a 1 , ,a n .Sinceμ is positive, we must have ξ = (β −σ)/μ. Also solved by R. Bagby, R. Chapman (U. K.), M. Goldenberg & M. Kaplan, J. Lockhart, D. Spellman, R. Stong, C. T. Stretch (U. K.), T. Tam, N. Vonessen, BSI Problems Group (Germany), Szeged Problems Group “Fej ´ ental ´ atuka” (Hungary), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer. A Special Prime 11235 [2006, 568]. Proposed by Lenny Jones and Rachel Keller, Shippensburg Uni- versity, Shippensburg, PA. Find all primes p such that 128 p + 1 is composite and a divisor of 2 64 p − 1. Solution by Toni Ernvall and Anne-Maria Ernvall-Hyt ¨ onen, University of Turku, Fin- land. For a fixed prime p,letd be the smallest positive integer such that 2 d ≡ 1 (mod 128 p + 1). Suppose that 128 p + 1 is composite and is a divisor of 2 64 p − 1. Note that d | 64 p and d | φ(128p + 1),whereφ is Euler’s function. We divide the problem into cases according to whether p divides d. If p | d,thenp | φ(128 p + 1).Sincep does not divide 128 p + 1, there exists a prime q such that q ≡ 1 (mod p) and q | 128 p +1. Write q = rp + 1, so that 128 p +1 = (rp + 1)(sp + 1) = rsp 2 +rp +sp + 1 for some positive integer s,andrsp +r + s = 128. If p = 2, then there are no so- lutions, since 257 is prime. Hence p is odd, and r ≡ s ≡ 0 (mod 2).Letr = 2k and s = 2.Now2kp +k + = 64, and kp ≤ 31. We may assume that k ≤ .By straightforward calculations we notice that the only positive integer solution is p = 3, k = 1, and = 9, yielding 128 ·3 +1 = 385 = 5 · 7 ·11. However, 11 does not divide 2 64·3 − 1, so this solution does not satisfy the original conditions. Assume now that p does not divide d.Nowd | 64, and 128 p + 1 | 2 64 − 1. We compute 2 64 − 1 = (2 32 + 1)(2 16 + 1)(2 8 + 1)(2 4 + 1)(2 2 + 1)(2 + 1) = 641 · 6700417 ·65537 · 257 · 17 · 5 · 3. Checking through the possibilities gives 128 p + 1 = 65537 · 6700417. Thus p = 3430665851 is the only solution. Also solved by M. Avidon, R. Chapman (U. K.), P. Corn, S. M. Gagola Jr., O. P. Lossers ((Netherlands)), A. Nakhash, A. Stadler (Switzerland), R. Stong, M. Tetiva (Romania), D. B. Tyler, BSI Problems Group (Germany), Microsoft Research Problems Group, NSA Problems Group, and the proposers. An Equality for Parts in Partitions of n 11237 [2006, 655]. Proposed by Emeric Deutsch, Polytechnic University, Brooklyn, NY. Prove that the number of 2s occurring in all partitions of n is equal to the number 666 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 of singletons occurring in all partitions of n − 1, where a singleton in a partition is a part occurring once. (For example, partitions of 5 yield four 2s: one from (3, 2),two from (2, 2, 1) and one from (2, 1, 1, 1); partitions of 4 yield four singletons: one from (4), two from (3, 1) and one from (2, 1, 1).) Solution I by John H. Smith, Needham, MA. Let p(n) denote the number of partitions of n, with p(0) = 1and p(n) = 0forn < 0. The number of partitions of n with at least i occurrences of j is p(n − ij). Hence the total number of 2s in all partitions of n is ∞ i=1 p(n −2i), since a partition with exactly k occurrences of 2 is counted once in each of the first k terms. On the other hand, a partition of n with a singleton j is formed by appending j to a partition of n − j that has no j.Sincep(n − 2 j) partitions of n − j contain at least one j, the number of partitions of n with a singleton j is p(n − j ) − p(n − 2 j). Thus the total number of singletons in partitions of n − 1is ∞ j =1 ( p(n −1 − j) − p(n − 1 − 2 j)). This sum equals ( p(n − 2) − p(n −3)) + ( p(n −3) − p(n − 5)) + (p(n −4) − p(n − 7)) +···. Terms p(n − r) occur positively when r is even and with both signs when r is odd. Cancellation yields ∞ i=1 p(n −2i ), which is the formula obtained for the total number of 2s. Solution II by Albert Stadler, D ¨ ubendorf, Switzerland. Euler obtained the generating function for partitions: P(x) = ∞ n=1 p(n)x n = ∞ n=1 1 (1 − x n ) . The number of partitions of n with exactly k occurrences of 2 is the coefficient of x n in x 2k (1 − x 2 )P(x). Hence the total number of 2s over all partitions of n is the coefficient of x n in each expression below: ∞ k=1 kx 2k (1 − x 2 )P(x) = x 2 (1 − x 2 ) 2 (1 − x 2 )P(x) = x 2 (1 − x 2 ) P(x). On the other hand, the number of partitions of n − 1 containing the singleton k is the coefficient of x n−1 in x k (1 − x k )P(x). Hence the total number of singletons over all partitions of n − 1 is the coefficient of x n in x ∞ k=1 x k (1 − x k )P(x). Note that x ∞ k=1 x k (1 − x k )P(x) = x x 1 − x − x 2 1 − x 2 P(x) = x 2 (1 − x 2 ) P(x). Thus the coefficients are equal, as claimed. Also solved by M. Avidon, D. Beckwith, N. Caro (Brazil), R. Chapman (U. K.), J. Dalbac, P. P. D ´ alyay (Hun- gary), Y. Dumont (France), G. Keselman, M. Kidwell, S. C. Locke, O. P. Lossers ((Netherlands)), P. Massaro (Italy), R. McCoart, M. D. Meyerson, A. Nijenhuis, I. Novakovic & A. Milan (Serbia), R. Pratt, R. Stong, R. Tauraso (Italy), J. Taylor-Goodman, M. Tetiva (Romania), W. Watkins, BSI Problems Group (Germany), Szeged Problems Group “Fej ´ ental ´ atuka” (Hungary), GCHQ Problem Solving Group (U. K.), Microsoft Re- search Problems Group, NSA Problems Group, and the proposers. Sequences Built from the Golden Ratio 11238 [2006, 655]. Proposed by Aviezri S. Fraenkel, Weizmann Institute of Science, Rohovot, Israel. Let n be (n + 1)φ φ − nφ φ ,whereφ = (1 + √ 5)/2and August–September 2008] PROBLEMS AND SOLUTIONS 667 x denotes the integer part of x. Prove that the following hold for every positive integer n: (a) n is either 2 or 3; (b) (n +1)φ 2 φ − nφ 2 φ = 2 n − 1; (c) nφ + nφ 2 = nφ 2 φ ; (d) nφ 2 φ = nφ φ 2 + 1. Solution by Reiner Martin, Teddington, U.K. Since n −1 <(nφ −1)/φ ≤ nφ /φ < n,wehave nφ /φ = n −1. With φ = 1 +1/φ, this implies nφ φ = nφ (1 + 1/φ) = nφ + n −1. Thus n = (n +1)φ − nφ + 1, and so (a) follows since 1 <φ<2. Note that nφ φ = nφ + nφ /φ ≤ nφ + n = n(φ +1) = nφ 2 . Thus nφ ≤ nφ 2 /φ ≤ nφ,andso nφ 2 /φ = nφ . Thus nφ 2 φ = nφ 2 + nφ 2 /φ = nφ 2 + nφ , which is claim (c). Moreover, it follows that nφ 2 φ = 2 nφ + n. Applying this to n +1andton and invoking n = (n + 1)φ − nφ + 1 yields (b). Finally, using nφ φ = nφ + n −1, (d) follows from nφ 2 φ = 2 nφ + n = nφ + nφ φ + 1 = nφ (1 + φ) + 1 = nφ φ 2 + 1. Also solved by M. R. Avidon, D. Beckwith, N. Caro & O. L ´ opez (Brazil), R. Chapman (U. K.), J. Christopher, P. P. D ´ alyay (Hungary), M. Goldenberg & M. Kaplan, G. Keselman, R. Lampe, J. H. Lindsey II, O. P. Lossers ((Netherlands)), A. Nijenhuis, S. Northshield, J. Rebholz, K. Schilling, A. Stadler (Switzerland), A. Stenger, R. Stong, BSI Problems Group (Germany), Con Amore Problem Group (Denmark), GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, and the proposer. Perfect Parity Patterns 11243 [2006, 759]. Proposed by Donald Knuth, Stanford University, Stanford, CA. An m × n matrix of 0s and 1s is a parity pattern if every 0 is adjacent (horizontally or vertically) to an even number of 1s and every 1 is adjacent to an odd number of 1s. It is perfect if no row or column is entirely zero. Thus, ⎛ ⎝ 11 00 11 ⎞ ⎠ , ⎛ ⎜ ⎝ 0011 0100 1101 0101 ⎞ ⎟ ⎠ , ⎛ ⎝ 01010 11011 01010 ⎞ ⎠ , ⎛ ⎜ ⎜ ⎜ ⎝ 01110 10101 11011 10101 01110 ⎞ ⎟ ⎟ ⎟ ⎠ are parity patterns of sizes 3 × 2, 4 × 4, 3 ×5, and 5 × 5, respectively; only the 4 ×4 and 5 × 5 patterns are perfect. (a) Determine the number c(n) of perfect parity patterns that have exactly n columns. (b) Alone among these examples, the 5 × 5 parity pattern is invariant under rotation by 90 degrees and under reflection across its central column. Thus it has eightfold symmetry. Prove that a perfect n ×n parity pattern with eightfold symmetry exists for all n of the form n = 3 · 2 k − 1 with k ≥ 1. Solution of part (a) by Robin Chapman, University of Bristol, Bristol, U. K. Let c(n) be the number of perfect parity patterns with n columns. To avoid a trivial addition of 668 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 1 to the formula, we require at least one row. We prove that c(n) = d|(n+1) μ n + 1 d 2 d−1 , where μ denotes the M ¨ obius function. Let M be an m × n matrix, with rows v 1 , ,v m .Let0 denote the all-zero vector of length n,andsetv 0 = v m+1 = 0. In binary arithmetic, the condition that M is a parity pattern is that the sum of entry (i, j) and its neighboring entries in M is 0 (that is, even). Equivalently, v i+1 + v i T n + v i−1 = 0 (∗) for 1 ≤ i ≤ m,whereT n is the n × n binary matrix whose (i, j ) entry is 1 if and only if |i − j|≤1. Note that (∗) is a recurrence that determines v i+1 from v i−1 and v i . Also, it is re- versible: from v i and v i+1 , it determines v i−1 . Fixing v 0 = 0,anyn-vector v 1 deter- mines a doubly infinite sequence (v i ) ∞ i=−∞ of vectors via (∗).SinceF n 2 is finite, some pair (v i ,v i+1 ) must repeat as (v i+p ,v i+1+p ) for some p.Now(∗) implies that v i = v i+p for all i, and hence the sequence is periodic. Call a parity pattern semi-perfect if it has no all-zero row (it may have all-zero columns). A semi-perfect parity pattern is determined by its first row v 1 ,whichis nonzero. Since the sequence is periodic, there is a first positive integer q such that v q = 0. The parity pattern with rows v 1 , ,v q−1 is the only semi-perfect parity pattern with first row v 1 .Asthereare2 n − 1 nonzero vectors of length n,thereare2 n − 1semi- perfect parity patterns with n columns. We now claim that in the sequence produced by v 0 and v 1 with q defined as above, v i = 0 if and only if q | i. Suppose v k = 0.Fromv k+1 = v k T n + v k−1 , we obtain v k+1 = v k−1 . Applying the same computation to v k+j shows inductively that v k+j = v k−j for all j. Thus the sequence v j is symmetric about any k for which v k = 0.Since v 0 = 0 and v q = 0, it follows that v mq = 0 for all m.Sincev j = 0 for 0 < j < q,by our symmetry observation, v j = 0 for any j other than multiples of q. As a conse- quence, the number of rows in a parity pattern with first row v 1 must be one less than a multiple of q.LetM k (v 1 ) denote the parity pattern with kq −1rowsandfirstrowv 1 ; note that only k = 1 yields a semi-perfect parity pattern. Now consider an m ×n parity pattern M that is semi-perfect but not perfect; M has an all-zero column. Let the first all-zero column be in position r .Ifr = 1thenbythe definition of a parity pattern, the other columns are forced to be zero, so r > 1. The matrix N formed by the first r − 1 columns is a parity pattern with no all-zero column. Its transpose N T is a semi-perfect parity pattern. Hence N T = M 1 (w),where w is the transpose of the first column of M.NowM T = M k (w) for some integer k with k ≥ 2. By the preceding paragraph, n = kr − 1. As M is semi-perfect, M T has no zero column. Since every row of M T either equals a row of N T or is zero, it follows from the lack of all-zero columns in M T that N T has no all-zero column. Hence N is perfect. On the other hand, from a perfect parity pattern N with r − 1 columns and first column w, we can reverse this process to obtain a semi-perfect parity pattern M with kr − 1 columns: let M = M k (w) T ,whereN = M 1 (w) T .SinceN has no zero rows, neither does M. Thus M is semi-perfect and its first all-zero column is in position r. Counting semi-perfect parity patterns with n columns in groups according to the associated perfect parity pattern N gives r|(n+1),r>1 c(r − 1) = 2 n − 1. Setting j = n +1, a 1 = 1, and a r = c(r − 1) for r > 1, this becomes r|j a r = 2 j −1 . By M ¨ obius August–September 2008] PROBLEMS AND SOLUTIONS 669 inversion, a j = d|j μ( j/d)2 d−1 . This is equivalent to the claimed result for c(n) since c(n) = a n+1 . Solution of part (b) by Reiner Martin, Teddington, U. K. We recursively construct a perfect parity pattern P k with eightfold symmetry and 3 · 2 k − 1rows. Let P 1 be the 5 ×5 parity pattern in the problem statement. With a i, j denoting entry (i, j) in P k−1 , define entry (r, s) in P k to be b r,s ,where b 2i,2 j = a i, j , b 2i+1,2 j = a i, j + a i+1, j (mod 2), b 2i,2 j +1 = a i, j + a i, j +1 (mod 2), and b 2i+1,2 j +1 = 0, where a i, j is understood to be 0 if i or j equals 0 or 3 · 2 k−1 . It is straightforward to verify the parity condition for each type of entry; hence P k is indeed a parity pattern. Also, eightfold symmetry is preserved by the construction. A zero row in P k would imply two adjacent identical rows in P k−1 , which by induction does not occur. Thus P k is indeed perfect. Editorial comment. Parity patterns arise in the study of a popular puzzle called “Lights Out”, invented in the 1980s by Dario Uri and available as a Java applet at http://www.whitman.edu/mathematics/lights_out/. Klaus Supner called such pat- terns “even-parity covers” of the m × n grid; see Math. Intell. 11 No. 2 (Spring 1989) 49–53, and Theor. Comp. Sci. 230 (2000) 49–73. Also solved by D. Beckwith, S. M. Gagola, D. Serre (France), Microsoft Research Problems Group, and the proposer. Part (a) also solved by J. H. Lindsey II. Double Integral with Log and Sinh 11275 [2007, 164]. Proposed by Michael S. Becker, University of South Carolina at Sumter, Sumter, SC. Find ∞ y=0 ∞ x=y (x − y) 2 log((x + y)/(x − y)) xy sinh(x + y) dx dy. Solution I by David Beckwith, Sag Harbor, NY. Let I denote the required integral. Introduce the change of variables x = (z + t)/2, y = (z − t)/2 which gives I = 2 ∞ z=0 1 sinh z z t=0 t 2 log(z/t) z 2 − t 2 dt dz. In the inner integral, set t = zw to obtain the product of two known integrals, I = 2 ∞ 0 zdz sinh z 1 0 1 − 1 1 − w 2 log w dw = 2 π 2 4 π 2 8 − 1 = π 2 (π 2 − 8) 16 . Solution II by Hongwei Chen, Christopher Newport University, Newport News, VA. Let I denote the required integral. Substitute x = uy and interchange the order of integration to obtain I = ∞ u=1 (u − 1) 2 log((u + 1)/(u −1)) u ∞ y=0 y sinh(y(u + 1)) dy du. 670 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 Let I u denote the inner integral in the previous formula for I . Substituting t = e −(u+1)y gives I u = −2 (1 + u) 2 1 0 ln t t 2 − 1 dt = −2 (1 + u) 2 1 t=0 ∞ k=0 t 2k ln tdt = 2 (1 + u) 2 ∞ k=0 1 (2k + 1) 2 = π 2 4(1 + u) 2 . So I takes the form I = π 2 4 ∞ 1 (u − 1) 2 log((u + 1)/(u −1)) u(1 + u) 2 du. Finally, substitute s = (u − 1)/(u +1): I =− π 2 2 1 0 s 2 ln s 1 − s 2 ds =− π 2 2 1 0 ln s 1 − s 2 − ln s ds = π 2 (π 2 − 8) 16 . Editorial comment. Many other interesting substitutions were submitted. Also solved by R. Bagby, D. H. Bailey (USA) & J. M. Borwein (Canada), R. Chapman (U. K.), K. Dale (Norway), B. E. Davis, P. De (Ireland), N. Eklund, C. Fleming, W. Fosheng (China), M. L. Glasser, M. R. Gopal, R. Govindaraj & R Mythili & G. Anupama & G. Sudharsan (India), J. Grivaux (France), J. A. Grzesik, E. A. Herman, G. Keselman, O. Kouba (Syria), G. Lamb, D. Lovit, K. McInturff, O. Pad ´ e (Israel), P. Perfetti (Italy), T L. & V. R ˘ adulescu (Romania), O. G. Ruehr, H Z. Seiffert (Germany), A. Stadler (Switzerland), R. Stong, R. Tauraso (Italy), E. I. Verriest, M. Vowe (Switzerland), FAU Problem Solving Group, GCHQ Problem Solving Group (U. K.), Microsoft Research Problems Group, University of Sharjah Problem Solving Group (United Arab Emirates), and the proposer. A Matrix Inequality 11296 [2007, 252]. Proposed by Said Amghibech, Quebec, Canada. Let A beareal symmetric positive definite matrix, with entries a i, j ,1≤ i, j ≤ n.LetA k be the k × k matrix in the upper left corner of A. Show that n k=1 (det A k ) 1/k ≤ 1 + 1 n n Tr(A) where Tr(A) denotes the trace of A. Solution by Joanna Chachulska and Wojciech Matysiak, Polytechnica Warszawska, Warsaw, Poland. Hadamard’s inequality says det A k ≤ a 1,1 a 2,2 ···a k,k .Carleman’s inequality says n k=1 x 1 x 2 ···x k 1/k ≤ 1 + 1 n n n k=1 x k for x k ≥ 0. Our result follows. Note: Carleman’s inequality is usually stated in its limit form, with infinite sums and the number e replacing the term (1 +1/n) n by monotonicity. However, standard proofs of the inequality prove the finite form stated above. For example, see pp. 146–152 of G. P ´ olya, Mathematics and Plausible Reasoning, Princeton Univ. Press, Princeton, NJ, 1954. Also solved by D. Chakerian, R. Chapman (U. K.), J P. Grivaux (France), E. A. Herman, C. J. Hillar, F. Holland (Ireland), D. Jesperson, O. P. Lossers (Netherlands), M. Omarjee, H J. Seiffert (Germany), M. Tetiva (Romania), NSA Problems Group, and the proposer. August–September 2008] PROBLEMS AND SOLUTIONS 671 . should be sent in duplicate to the MONTHLY problems address on the inside front cover. Submitted solutions should arrive at that address before December 31, 2008. Additional information, such. denotes the column space of a matrix ar- gument. 664 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 115 11380. Proposed by Hugh Montgomery, University of Michigan, Ann Arbor, MI, and Harold. one X. We seek a unique solution to W ((a 1 ,α 1 ), ,(a n ,α n ), (x,ξ))= (b,β). August–September 2008] PROBLEMS AND SOLUTIONS 665 If W has length l,then(b,β)is the result of l −1 applications of