捷克数学竞赛2010年试题答案1

The 20th Annual Vojtěch Jarník International Mathematical Competition Ostrava, 25th March 2010 Category I Problem 1 a) Is it true that for every bijection f : N→ N the series ∞∑ n=1 1 nf(n) is convergent? b) Prove that there exists a bijection f : N→ N such that the series ∞∑ n=1 1 n+ f(n) is convergent. (N is the set of all positive integers.) [10 points] Solution a) Yes. Applying the inequality, if 0 ≤ a1 ≤ · · · ≤ an and 0 ≤ b1 ≤ · · · ≤ bn and σ : {1, . . . , n} → {1, . . . , n} is a permutation, then n∑ j=1 ajbσ(j) ≤ n∑ j=1 ajbj , for every n we get n∑ j=1 1 jf(j) ≤ n∑ j=1 1 j2 ≤ ∞∑ j=1 1 j2 . Since the sequence (∑n j=1 1 jf(j) ) is increasing and bounded, it converges. b) No. We will construct a permutation f : N→ N such that the series ∞∑ n=1 1 n+ f(n) is convergent. Let f : N→ N be given in the following way: f(1) = 4 and for [(n!)2 + 1, ((n+ 1)!)2] ∩ N we put f((n!)2 + k) = [(n+ 2)!]2 − (k − 1) if 1 ≤ k < [(n+ 1)!]2 − 1− n−1∑ j=0 (−1)j [(n− j)!]2. and f([(n+ 1)!]2 − k) = [(n− 1)!]2 + k + 1 if 0 ≤ k ≤ 1 + n−1∑ j=0 (−1)j [(n− j)!]2. Then [(n+1)!]2∑ j=(n!)2+1 1 n+ f(n) ≤ ((n+ 1)!) 2 − (n!)2 (n!)2 + [(n+ 2)!]2 + 1 + (n!)2 − [(n− 1)!]2 [(n+ 1)!]2 + [(n− 1)!]2 + 1 < 1 (n+ 2)2 + 1 (n+ 1)2 . Thus we show that the sequence (∑n j=1 1 j+f(j) ) is bounded. Since it is increasing, it converges. � The 20th Annual Vojtěch Jarník International Mathematical Competition Ostrava, 25th March 2010 Category I Problem 2 Let A and B be two complex 2× 2 matrices such that AB −BA = B2. Prove that AB = BA. [10 points] Solution We may conclude that AB = BA if and only if 2 6= 0 in F (that is, charF 6= 2). If charF = 2, take B = ( 1 1 0 1 ) , A = ( 0 0 1 0 ) . Assume that charF 6= 2. Let B = ( a b c d ) , then B2 = ( a2 + bc b(a+ d) c(a+ d) d2 + bc ) . We have a2 + d2 + 2bc = traceB2 = traceAB − traceBA = 0. If B is invertible, then A = B(A+B)B−1, hence traceA = trace(B(A+B)B−1) = trace(A+B) = traceA+ traceB, so traceB = 0, d = −a, traceB2 = 2(a2 + bc) = 0. Since charF 6= 2, it implies a2 + bc = 0, hence B2 = 0 and AB = BA. If B is not invertible, then detB = ad− bc = 0, so (a+ d)2 = a2 + d2 + 2bc = 0, a+ d = 0, a = −d, a2 + bc = −ad+ bc = 0, so B2 = 0. � The 20th Annual Vojtěch Jarník International Mathematical Competition Ostrava, 25th March 2010 Category I Problem 3 Prove that there exist positive constants c1 and c2 with the following properties: a) For all real k > 1, ∣∣∣∫ 1 0 √ 1− x2 cos(kx) dx ∣∣∣ < c1 k3/2 . b) For all real k > 1, ∣∣∣∫ 1 0 √ 1− x2 sin(kx) dx ∣∣∣ > c2 k . [10 points] Solution Put f(x) = √ 1− x2. 1. Integrating by parts, we have∫ 1 0 f(x) · cos kxdx = [ f(x) · 1 k sin kx ]1 0 − ∫ 1 0 f ′(x) · 1 k sin kxdx . The first term is 0− 0 = 0. The second term is (−1/k) times∫ √1−1/k 0 f ′(x) · sin kxdx+ ∫ 1 √ 1−1/k f ′(x) · sin kxdx . (1) Here the first term equals [ −f ′(x) · 1 k cos kx ]√1−1/k 0 + ∫ √1−1/k 0 f ′′(x) · 1 k cos kxdx , whose absolute value is ≤ −2 k f ′ (√ 1− 1/k) = 2 k √ 1− 1/k√ 1/k < 2√ k . The absolute value of the second term in (1) is ≤ ∫ 1 √ 1−1/k |f ′(x)|dx = −[f(x)]1√ 1−1/k = 1√ k . Thus, we may choose c1 = 2 + 1 = 3. 2. Integrating by parts, we have∫ 1 0 f(x) · sin kxdx = − [ f(x) · 1 k cos kx ]1 0 + ∫ 1 0 f ′(x) · 1 k cos kxdx . The first term is 1/k. The second term is (1/k) times∫ √1−1/k 0 f ′(x) · cos kxdx+ ∫ 1 √ 1−1/k f ′(x) · cos kxdx . (2) Here the first term equals [ f ′(x) · 1 k sin kx ]√1−1/k 0 − ∫ √1−1/k 0 f ′′(x) · 1 k sin kxdx , whose absolute value is ≤ −2 k f ′ (√ 1− 1/k) = 2 k √ 1− 1/k√ 1/k < 2√ k . The absolute value of the second term in (2) is ≤ ∫ 1 √ 1−1/k |f ′(x)|dx = −[f(x)]1√ 1−1/k = 1√ k . Thus, ∫ 1 0 f(x) · sin kxdx > 1 k ( 1− 3√ k ) . This proves the desired claim for k ≥ 3pi. The integral has a positive lower bound for k < 3pi as well, since∫ 1 0 f(x) · sin kxdx = ∫ 1 0 (−f ′(x)) · 1− cos kx k dx > 0 . � The 20th Annual Vojtěch Jarník International Mathematical Competition Ostrava, 25th March 2010 Category I Problem 4 For every positive integer n let σ(n) denote the sum of all its positive divisors. A number n is called weird if σ(n) ≥ 2n and there exists no representation n = d1 + d2 + · · ·+ dr , where r > 1 and d1, . . . , dr are pairwise distinct positive divisors of n. Prove that there are infinitely many weird numbers. [10 points] Solution The idea is to show that given a weird number, one can construct a sequence of weird numbers tending to infinity. We claim that for weird n and p a prime greater than σ(n) and coprime to n, the number pn is also weird. In fact, if 1 = d1, d2, . . . , dk = n are the positive divisors of n, the ones of pn are d1, d2, . . . , dk, pd1, . . . , pdk and they are pairwise distinct as (p, n) = 1. Suppose now that we have pn = di1 + · · ·+ dir + p(dj1 + · · ·+ djs) with ik, jl ∈ {1, . . . , k}. Then we have di1 + · · ·+ dir = p(n− dj1 − · · · − djs) . Note that n /∈ {dj1 , . . . , djs} as the representation must have more than only one summand and the assumption that n is weird implies n − dj1 − . . . − djs 6= 0. Hence as the right hand expression is divisible by p and non zero, so must be di1 + · · ·+ dir which is impossible as p > σ(n). It remains to find a weird number. A possible reasoning could be: look for a number n with σ(n) = 2n+ 4 that is not divisible by 3 and 4. Then the smallest possible divisors are 1, 2, 5 so that it will be impossible to represent 4, and hence n, as a sum of pairwise distinct divisors of n. Checking for numbers with three distinct prime factors 2, p, q yields σ(2pq) = 3(p+ 1)(q + 1) = 3pq + 3p+ 3q + 3 and hence we need 3pq + 3p+ 3q + 3 = 4pq + 4⇐⇒ (p− 3)(q − 3) = 8 . This equality is solved by p = 5 and q = 7 which yields the weird number n = 70. �