PUTNAM 2016 SOLUTIONS. Problem A1. Find the smallest positive integer j such that for every polynomial p(x) with integer coefficients

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1 Math 7 Fall 5 PUTNAM 6 SOLUTIONS Problem A Find the smallest positive integer j such that for every polynomial px with integer coefficients and for every integer k, the integer the j-th derivative of px at k is divisible by 6 p j k = dj dx j px x=k We claim that j = 8 is the smallest such integer Write P x = a + a x + a x + + a d x d Then d P j k = ii i i j + a i k j i i=j It is easy to observe that the product of 8 consecutive inters is divisible by 5, 3 and 7, and therefore by 6 = Hence if j 8, each term in the above sum is divisible by 6, implying that p j k is divisible by 6 For j < 8, take px = x j Then p j k = j! is not divisible by 6 Problem A Given a positive integer n, let Mn be the largest integer m such that m > n m n Evaluate Mn lim n n The given inequality is equivalent to mn > m n+m n, or m 3n m+n n < Considering this as a quadratic inequality in m, we obtain 3n 5n n + < m < 3n + 5n n + which yields Mn = 3n + 5n n +

2 Math 7 Fall 5 where is the Guass floor function Then we have 3n + 5n n + Dividing this inequality by n and taking the limit yields Mn 3n + 5n n + Mn lim n n = Problem A3 Suppose that f is a function from R to R such that fx + f = arctan x x for all real x As usual, y = arctan x means π/ < y < π/ and tan y = x Find Solution : The given functional equation is fx + f fxdx = arctan x for x x Substituiting x and to respectively yield x f + f x x f + fx = arctan x = arctan x x for x,, for x, 3 Then yields fx = arctan x + arctan arctan x x for x, and hence fxdx = arctan x dx + arctan dx x arctan dx 4 x

3 Math 7 Fall 5 To calculate the first two integrals in 4, we simply note that for x, we have arctan π arctan x, which yields arctan dx = x For the third integral in 4, we set y = arctan function trick we obtain On the other hand, we have Hence we calculate the integral by π/ π arctan x dx = π arctan x dx x Then we have x = dy arctan = x π/ tan y tan y = + tan y π tan y + = tan 4 + y + dy tan y = π tan π/ 4 + y + dy = π/4 π/4 = x, so using the inverse tan y tan y dy + π 4 = π 4 By 4 we get fxdx = 3π 8 Problem A 4 Consider a m n rectangular region, where m and n are integers such that m, n > 4 This region is to be tiled using tiles of the two types shown: The dotted lines divide the tiles into squares The tiles may be rotated and reflected, as long as their sides are parallel to the sides of the rectangular region They must all fit within the region, and they must cover it without overlapping What is the minimum number of tiles required to tile the region? We claim that the minimum number of tiles required is mn Consider the following coloring of the region:

4 Math 7 Fall 5 Note that each tile can cover at most one black square There are mn black squares, so we need at least mn tiles to cover the region Now it remains to prove that we can cover the region with mn tiles We prove this by induction on m and n When m = n = 4, we have the following covering: The inductive step follows from the following covering of m or n rectangle: Problem A5 Suppose that G is a finite group generated by the two elements g and h, where the order of g is odd Show that every element of G can be written in the form g m h n g m h n g mr h mr with r G and m, n, m, n,, m r, n r {, } Here G is the number of elements of G

5 Math 7 Fall 5 We call an expression of an element of G as stated in the problem a long form, and define r to be its length For example, ghg h is a long form with length Let H be the cyclic subgroup of G generated by ghg h Let s be the smallest integer such that g s H, which must be odd as it divides the order of G Note that we have ghh = hgh, and therefore xyh = yxh for any x, y G 5 In particular, we can choose the coset representatives to be {g i h j : i s, j t } where t is the smallest integer such that h t g u H for some u It suffices to prove that every element in g i h j H can be expressed in a long form Note that if s = t =, we have G = H Since every element in H are clearly in a long form, the assertion is true We will now assume that st Using 5 we find g s i H = g hg h s i/ H if i is odd, g i H = ghgh i/ H if i is even h j H = ghgh s / gh jh Hence we have jh g hg h ghgh s i/ gh s / if i is odd, g i h j H = jh ghgh ghgh i/ s / gh if i is even In other words, every element x in g i h j H can be written in the form jghg g hg h ghgh s i/ gh s / h k if i is odd, x = jghg ghgh ghgh i/ gh s / h k if i is even for some integer k H Now it remains to check that these expressions have length at most G If i is odd, the length is s i + s j + k s + s t + k st + H st H = G, where the last inequality follows from st Similarly, we can show that the length is at most G when i is even

6 Math 7 Fall 5 Problem A6 Find the smallest constant C such that for every real polynomial P x of degree 3 that has a root in the interval [, ], P x dx C max P x x [,] No idea

7 Math 7 Fall 5 Problem B Let x, x, x, be the sequence such that x = and for n, x n+ = lne xn x n as usual, the function ln is the natural logarithm Show that the infinite series x + x + x + converges and find its sum Since e x x > for all x >, an easy induction shows that x n > for all n On the other hand, the given recursive equation can be written as x n = e xn+ e xn which yields x + x + + x n = e x ex n+ = e e xn+ for all n 6 In particular, the partial sum x + x + + x n is strictly increasing and bounded above, so the infinite series x + x + converges This implies that x n as n, and therefore 6 yields x + x + = e Problem B Define a positive integer n to be squarish if either n is itself a perfect square or the distance from n to the nearest square is a perfect square For example, 6 is squarish, because the nearest perfect square to 6 is 45 = 5 and 5 6 = 9 is a perfect square Of the positive integers between and, only 6 and 7 are not squarish For a positive integer N, let SN be the number of squarish integers between and N, inclusive Find positive constants α and β such that or show that no such constants exist SN lim N N α = β We claim that α = 3/4 and β = 4/3 We use the little-o notation For example, we say that a given function hn is on c if lim n hn n c =

8 Math 7 Fall 5 Let k be the nearest perfect square to N, ie, k / N k + / For each positive integer j, the number of squarish integers in the interval [j /, j + / ] is j + j + Hence k SN = j + j + + ɛ where ɛ is the number of squarish integers in the interval [k /, N] Note that j= ɛ k + k + k + Hence we may write k SN = j + j + + k + ɛ k j= k = j + k + ɛ k j= k = j + on 4/3 7 j= Now let l be the positive integer such that l k < l + Then k j = l l l + lk l j= k = jj + j + lk l j= k = jj + + lk l = j= l ll 3 + l l + lk l Note that l N N k + / and l > k /, so l = N /4 + on /4 On the other hand, we have k l < l + l = l + Hence we can write k j = 3 N 3/4 + on 3/4 Thus we deduce from 7 that SN = 4 3 N 3/4 + on 3/4, which yields α = 3/4 and β = 4/3 as desired j=

9 Math 7 Fall 5 Problem B 3 Suppose that S is a finite set of points in the plane such that the area of triangle ABC is at most whenever A, B, C are in S Show that there exists a triangle of area 4 that together with its interior covers the set S Choose P, P, P 3 S with the maximum area of P P P 3 Let l be the line passing through P which is parallel to P P 3, and m be the reflection of l to P P 3 We similarly define the lines l, l 3, m, m 3 m 3 m l 3 l Q 3 P Q P P 3 m Q l Let Q be the intersection of l and l 3, and similarly define Q and Q 3 Then one easily sees that m i passes through Q i for i =,, 3 as in the figure The three points P, P, P 3 are the midpoints of the sides of the triangle Q Q Q 3, so we have Area Q Q Q 3 = 4Area P P P 3 The area of the triangle Q Q Q 3 is at most 4 as the area of the triangle P P P 3 is at most Let P be a point in S Since Area P P P 3 Area P P P 3, P must lie between l and m Similarly, P must lie between l and m, and also between l 3 and m 3 Hence P must lie in the triangle Q Q Q 3 This implies that the triangle the set S is covered by the triangle Q Q Q 3, whose area is at most 4 as seen above Problem B 4 Let A be a n n matrix, with entries chosen independently at random Every entry is chosen to be either or, each with probability / Find the expected value of deta A t as a function of n, where A t is the transpose of A

10 Math 7 Fall 5 Let B = A A t Let a i,j resp b i,j denote the i, j-entry of A resp B We know that B is anti-symmetric, ie, b i,j = a i,j a j,i for i, j n In particular, b i,i = for i =,,, n, and if i j, the probability distribution of b i,j is given as follows: with probability / b i,j = with probability /4 8 with probability /4 Furthermore, two random variables b i,j and b i,j are independent unless i = j and j = i The expected value of detb is given by EdetB = E sgnσb,σ b,σ b n,σn σ S n = sgnσe b,σ b,σ b n,σn 9 σ S n where S n is the group of permutations of {,,, n} We claim that E b,σ b,σ b n,σn = unless σ is a product of n transpositions, ie every element in {,,, n} has order under the permutation σ Suppose that there exists an element i {,,, n} whose order under σ is not equal to If i is a fixed point ie, the order is then b i,σi = and therefore E b,σ b,σ b n,σn = Now suppose that the order of i under σ is greater than Then σ i i, so b i,σi is independent of b j,σj for j i This yields E b,σ b,σ b n,σn = E b,σ b,σ b i,σi b i+,σi+ b n,σn E b i,σi = where the second equality follows from the fact that E b i,σi = Hence we establish the claim Now we assume that σ is a product of n transpositions Then we may write σ = i j i j i n j n Observe that b i,j, b i,j,, b in,j n are independent variables We also have E b i k,j k = by the probability distribution given in 8 for k =,,, n Hence we calculate E b n,σ b,σ b n,σn = E k= n b ik,j k b jk,i k = E k= b i k,j k = n k= E b ik,jk = n

11 Math 7 Fall 5 where the second equality follows from b jk,i k = b ik,j k On the other hand, sgnσ = n as σ is a product of n transpositions, so we find sgnσe b,σ b,σ b n,σn = n n = n We deduce from 9 that EdetB = N n where N is the number of permutations in S n which are a product of n transpositions One easily finds that N = n n 3 3 = n! nn 4 = n! n n!, so we conclude that EdetB = n! n n! Problem B5 Find all functions f from, to, with the following property: if x, y, and x y x 3, then fx fy fx 3 No idea Problem B6 Evaluate k k= k n= k n + We rewrite the given sum as k k= k n= k n + = k= n= k k k n + Note that every integer m > can be uniquely written in the form m = r s + where r, s are nonnegative integers with r odd We consider the terms in such that k n + = m Such terms appear when k = r s n and n =,,, s For such terms, we have k = if and only if n = s In other words, k k k n + = r s n m if n s r if n = s m

12 Math 7 Fall 5 Hence the sum of these terms is r s m r s n m = rm n= Thus the sum in can be computed by s n= m> s n = rm m = m s = r s m = m m = m m