PUTNAM TRAINING PROBLEMS, Exercises Let a and b two distinct integers, and n any positive integer. Prove that a n b n is divisible by a b.

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1 PUTNAM TRAINING PROBLEMS, 2008 (Last updated: October 29, 2008) Remark. This is a list of problems discussed durig the traiig sessios of the NU Putam team ad arraged by subjects. Miguel A. Lerma. Iductio... Prove that! > 2 for all 4. Exercises.2. Prove that for ay iteger, 2 2 is divisible by Let a ad b two distict itegers, ad ay positive iteger. Prove that a b is divisible by a b..4. The Fiboacci sequece 0,,, 2, 3, 5, 8, 3,... is defied as a sequece whose two first terms are F 0 = 0, F = ad each subsequet term is the sum of the two previous oes: F = F + F 2 (for 2). Prove that F < 2 for every Let r be a umber such that r + /r is a iteger. Prove that for every positive iteger, r + /r is a iteger..6. Fid the umber R() of regios i which the plae ca be divided by straight lies..7. We divide the plae ito regios usig straight lies. Prove that those regios ca be colored with two colors so that o two regios that share a boudary have the same color..8. A great circle is a circle draw o a sphere that is a equator, i.e., its ceter is also the ceter of the sphere. There are great circles o a sphere, o three of which meet at ay poit. They divide the sphere ito how may regios?.9. We eed to put cets of stamps o a evelop, but we have oly (a ulimited supply of) 5/c ad 2/c stamps. Prove that we ca perform the task if A chessboard is a 8 8 grid (64 squares arraged i 8 rows ad 8 colums), but here we will call chessboard ay m m square grid. We call defective a chessboard if oe of its squares is missig. Prove that ay 2 2 ( ) defective chessboard ca be

2 PUTNAM TRAINING PROBLEMS, tiled (completely covered without overlappig) with L-shaped tromios occupyig exactly 3 squares, like this... This is a modified versio of the game of Nim (i the followig we assume that there is a ulimited supply of tokes.) Two players arrage several piles of tokes i a row. By turs each of them takes oe toke from oe of the piles ad adds at will as may tokes as he or she wishes to piles placed to the left of the pile from which the toke was take. Assumig that the game ever fiishes, the player that takes the last toke wis. Prove that, o matter how they play, the game will evetually ed after fiitely may steps..2. Call a iteger square-full if each of its prime factors occurs to a secod power (at least). Prove that there are ifiitely may pairs of cosecutive square-fulls..3. Prove that for every 2, the expasio of ( + x + x 2 ) cotais at least oe eve coefficiet..4. We defie recursively the Ulam umbers by settig u =, u 2 = 2, ad for each subsequet iteger, we set equal to the ext Ulam umber if it ca be writte uiquely as the sum of two differet Ulam umbers; e.g.: u 3 = 3, u 4 = 4, u 5 = 6, etc. Prove that there are ifiitely may Ulam umbers. 2. Iequalities. 2.. If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca 2 ) 9a 2 b 2 c 2. ( ) Prove that! <, for = 2, 3, 4,..., If 0 < p, 0 < q, ad p + q <, prove that (px + qy) 2 px 2 + qy If a, b, c 0, prove that 3(a + b + c) a + b + c Let x, y, z > 0 with xyz =. Prove that x + y + z x 2 + y 2 + z Show that a 2 + b 2 + a b a 2 + b 2 (a + a2 + + a) 2 + (b + b2 + + b) Fid the miimum value of the fuctio f(x, x 2,..., x ) = x +x 2 + +x, where x, x 2,..., x are positive real umbers such that x x 2 x = Let x, y, z 0 with xyz =. Fid the miimum of S = x2 y + z + y2 z + x + z2 x + y.

3 PUTNAM TRAINING PROBLEMS, If x, y, z > 0, ad x + y + z =, fid the miimum value of x + y + z Prove that i a triagle with sides a, b, c ad opposite agles A, B, C (i radias) the followig relatio holds: a A + b B + c C π a + b + c (Putam, 2003) Let a, a 2,..., a ad b, b 2,..., b oegative real umbers. Show that (a a 2 a ) / + (b b 2 b ) / ((a + b )(a 2 + b 2 ) (a + b )) / 2.2. The otatio! (k) meas take factorial of k times. For example,! (3) meas ((!)!)! What is bigger, 999! (2000) or 2000! (999)? 2.3. Which is larger, or ? 2.4. Prove that there are o positive itegers a, b such that b 2 + b + = a (Ispired i Putam 968, B6) Prove that a polyomial with oly real roots ad all coefficiets equal to ± has degree at most (Putam 984) Fid the miimum value of ( 2 (u v) 2 + u2 9 v for 0 < u < 2 ad v > 0. ( ) ( ) ( ) Show that ) 2 < (Putam, 2004) Let m ad be positive itegers. Show that (m + )! m!! < (m + ) m+ m m Let a, a 2,..., a be a sequece of positive umbers, ad let b, b 2,..., b be ay permutatio of the first sequece. Show that a + a a. b b 2 b Let a, a 2,..., a ad b, b 2,..., b icreasig sequeces of real umbers, ad let x, x 2,..., x be ay permutatio of b, b 2,..., b. Show that a i b i a i x i. i= i=

4 PUTNAM TRAINING PROBLEMS, Prove that the p-mea teds to the geometric mea as p approaches zero. I other other words, if a,..., a are positive real umbers, the ( /p ( ) / lim a p p 0 k) = a k 3. Number Theory. k= 3.. Show that the sum of two cosecutive primes is ever twice a prime Ca the sum of the digits of a square be (a) 3, (b) 977? 3.3. Prove that there are ifiitely may prime umbers of the form Prove that the fractio ( 3 + 2)/( ) is i lowest terms for every possible iteger Let p(x) be a o-costat polyomial such that p() is a iteger for every positive iteger. Prove that p() is composite for ifiitely may positive itegers. (This proves that there is o polyomial yieldig oly prime umbers.) 3.6. Prove that two cosecutive Fiboacci umbers are always relatively prime Show that if a 2 + b 2 = c 2, the 3 ab. k= 3.8. Show that ca ever be a iteger for Let f() deote the sum of the digits of. Let N = Fid f(f(f(n))) Show that there exist 999 cosecutive umbers, each of which is divisible by the cube of a iteger. 3.. Fid all triples of positive itegers (a, b, c) such that ( + ) ( + ) ( + ) = 2. a b c 3.2. Fid all positive iteger solutios to abc 2 = a + b + c (USAMO, 979) Fid all o-egative itegral solutios (, 2,..., 4 ) to = The Fiboacci sequece 0,,, 2, 3, 5, 8, 3,... is defied by F 0 = 0, F =, F = F + F 2 for 2. Prove that for some k > 0, F k is a multiple of Do there exist 2 irratioal umbers a ad b greater tha such that a m = b for every positive itegers m,?

5 PUTNAM TRAINING PROBLEMS, The umbers ad are writte oe after the other (i decimal otatio). How may digits are writte altogether? 3.7. If p ad p 2 + 2, are primes show that p is prime Suppose > is a iteger. Show that is ot prime Let m ad be positive itegers such that m < + 2. Prove that m + 2 < Prove that the fuctio f() = + + /2 ( =, 2, 3,... ) misses exactly the squares Prove that there are o primes i the followig ifiite sequece of umbers: 00, 0000, , , (Putam 975, A.) For positive itegers defie d() = m 2, where m is the greatest iteger with m 2. Give a positive iteger b 0, defie a sequece b i by takig b k+ = b k + d(b k ). For what b 0 do we have b i costat for sufficietly large i? Let a = for. For each, let d deote the gcd of a ad a +. Fid the maximum value of d as rages through the positive itegers Suppose that the positive itegers x, y satisfy 2x 2 + x = 3y 2 + y. Show that x y, 2x + 2y +, 3x + 3y + are all perfect squares. 4. Polyomials. 4.. Fid a polyomial with itegral coefficiets whose zeros iclude Let p(x) be a polyomial with iteger coefficiets. Assume that p(a) = p(b) = p(c) =, where a, b, c are three differet itegers. Prove that p(x) has o itegral zeros Prove that the sum is irratioal (USAMO 975) If P (x) deotes a polyomial of degree such that P (k) = k/(k+) for k = 0,, 2,...,, determie P ( + ) (USAMO 984) The product of two of the four zeros of the quartic equatio is 32. Fid k. x 4 8x 3 + kx x 984 = 0

6 PUTNAM TRAINING PROBLEMS, Let be a eve positive iteger, ad let p(x) be a -degree polyomial such that p( k) = p(k) for k =, 2,...,. Prove that there is a polyomial q(x) such that p(x) = q(x 2 ) Let p(x) be a polyomial with iteger coefficiets satisfyig that p(0) ad p() are odd. Show that p has o iteger zeros (USAMO 976) If P (x), Q(x), R(x), S(x) are polyomials such that P (x 5 ) + xq(x 5 ) + x 2 R(x 5 ) = (x 4 + x 3 + x 2 + x + )S(x) prove that x is a factor of P (x) Let a, b, c distict itegers. Ca the polyomial (x a)(x b)(x c) be factored ito the product of two polyomials with iteger coefficiets? 4.0. Let p, p 2,..., p distict itegers ad let f(x) be the polyomial of degree give by Prove that the polyomial f(x) = (x p )(x p 2 ) (x p ). g(x) = (f(x)) 2 + caot be expressed as the product of two o-costat polyomials with itegral coefficiets. 4.. Fid the remaider whe you divide x 8 + x 49 + x 25 + x 9 + x by x 3 x Does there exist a polyomial f(x) for which xf(x ) = (x + )f(x)? It possible to write the polyomial f(x) = x 05 9 as the product of two polyomials of degree less tha 05 with iteger coefficiets? 4.4. Fid all prime umbers p that ca be writte p = x 4 + 4y 4, where x, y are positive itegers (Caada, 970) Let P (x) = x + a x + + a x + a 0 be a polyomial with itegral coefficiets. Suppose that there exist four distict itegers a, b, c, d with P (a) = P (b) = P (c) = P (d) = 5. Prove that there is o iteger k with P (k) = Show that ( + x + + x ) 2 x is the product of two polyomials. 5. Complex Numbers. 5.. Let m ad two itegers such that each ca be expressed as the sum of two perfect squares. Prove that m has this property as well. For istace 7 = , 3 = , ad 7 3 = 22 =

7 5.2. Prove that PUTNAM TRAINING PROBLEMS, si k = si 2 si 2 k=0 si Show that if x + /x = 2 cos a, the for ay iteger, x + /x = 2 cos a Factor p(z) = z 5 + z Fid a close-form expressio for si kπ. k= Cosider a regular -go which is iscribed i a circle with radius. What is the product of the legths of all ( )/2 diagoals of the polygo (this icludes the sides of the -go) (Putam 99, B2) Suppose f ad g are o-costat, differetiable, real-valued fuctios o R. Furthermore, suppose that for each pair of real umbers x ad y f(x + y) = f(x) f(y) g(x) g(y) g(x + y) = f(x) g(y) + g(x) f(y) If f (0) = 0 prove that f(x) 2 + g(x) 2 = for all x Give a circle of lights, exactly oe of which is iitially o, it is permitted to chage the state of a bulb provided that oe also chages the state of every dth bulb after it (where d is a divisor of strictly less tha ), provided that all /d bulbs were origially i the same state as oe aother. For what values of is it possible to tur all the bulbs o by makig a sequece of moves of this kid? 5.9. Suppose that a, b, u, v are real umbers for which av bu =. Prove that a 2 + b 2 + u 2 + v 2 + au + bv Geeratig Fuctios. 6.. Prove that for ay positive iteger ( ) ( ) ( ) ( ) = 2, 2 3 where ( ) a b = a! (biomial coefficiet). b!(a b)! 6.2. Prove that for ay positive iteger ( ) ( ) 2 + ( ) ( ) 2 = ( ) 2.

8 PUTNAM TRAINING PROBLEMS, Prove that for ay positive itegers k m,, k ( )( ) ( ) m m + =. j k j k j= Let F be the Fiboacci sequece 0,,, 2, 3, 5, 8, 3,..., defied recursively F 0 = 0, F =, F = F + F 2 for 2. Prove that F 2 = 2. = 6.5. Fid a recurrece for the sequece u = umber of oegative solutios of 2a + 5b = How may differet sequeces are there that satisfy all the followig coditios: (a) The items of the sequeces are the digits 0 9. (b) The legth of the sequeces is 6 (e.g ) (c) Repetitios are allowed. (d) The sum of the items is exactly 0 (e.g. 322) (Leigrad Mathematical Olympiad 99) A fiite sequece a, a 2,..., a is called p-balaced if ay sum of the form a k + a k+p + a k+2p + is the same for ay k =, 2, 3,..., p. For istace the sequece a =, a 2 = 2, a 3 = 3, a 4 = 4, a 5 = 3, a 6 = 2 is 3-balaced because a + a 4 = + 4 = 5, a 2 + a 5 = = 5, a 3 + a 6 = = 5. Prove that if a sequece with 50 members is p-balaced for p = 3, 5, 7,, 3, 7, the all its members are equal zero. 7. Recurreces. 7.. Fid the umber of subsets of {, 2,..., } that cotai o two cosecutive elemets of {, 2,..., } Determie the maximum umber of regios i the plae that are determied by vee s. A vee is two rays which meet at a poit. The agle betwee them is ay positive umber Defie a domio to be a 2 rectagle. I how may ways ca a 2 rectagle be tiled by domioes? 7.4. (Putam 996) Defie a selfish set to be a set which has its ow cardiality (umber of elemets) as a elemet. Fid, with proof, the umber of subsets of {, 2,..., } which are miimal selfish sets, that is, selfish sets oe of whose proper subsets are selfish Let a, a 2,..., a be a ordered sequece of distict objects. A deragemet of this sequece is a permutatio that leaves o object i its origial place. For

9 PUTNAM TRAINING PROBLEMS, example, if the origial sequece is, 2, 3, 4, the 2, 4, 3, is ot a deragemet, but 2,, 4, 3 is. Let D deote the umber of deragemets of a -elemet sequece. Show that D = ( )(D + D 2 ) Let α, β be two (real or complex) umbers, ad defie the sequece a = α + β ( =, 2, 3,... ). Assume that a ad a 2 are itegers. Prove that 2 2 a is a iteger for every. 8. Calculus. 8.. Believe it or ot the followig fuctio is costat i a iterval [a, b]. Fid that iterval ad the costat value of the fuctio. f(x) = x + 2 x + x 2 x Fid the value of the followig ifiitely ested radical (Putam 995) Evaluate Express your aswer i the form (a + b c)/d, where a, b, c, d, are itegers (Putam 992) Let f be a ifiitely differetiable real-valued fuctio defied o the real umbers. If f( ) = 2, =, 2, 3, compute the values of the derivatives f (k) (0), k =, 2, 3,.... { 8.5. Compute lim } Compute lim 0 { k= ( + k ) } / (Putam 997) Evaluate ) ) (x ( x3 2 + x5 2 4 x7 + x x x 6 dx

10 PUTNAM TRAINING PROBLEMS, (Putam 990) Is 2 the limit of a sequece of umbers of the form 3 3 m (, m = 0,, 2,... )? (I other words, is it possible to fid itegers ad m such that 3 3 m is as close as we wish to 2?) 8.9. (Leigrad Mathematical Olympiad, 988) Let f : R R be cotiuous, with f(x) f(f(x)) = for all x R. If f(000) = 999, fid f(500) Let f : [0, ] R cotiuous, ad suppose that f(0) = f(). Show that there is a value x [0, 998/999] satisfyig f(x) = f(x + /999). 9. Pigeohole Priciple. 9.. Prove that ay ( + )-elemet subset of {, 2,..., 2} cotais two itegers that are relatively prime Prove that if we select + umbers from the set S = {, 2, 3,..., 2}, amog the umbers selected there are two such that oe is a multiple of the other oe (Putam 978) Let A be ay set of 20 distict itegers chose from the arithmetic progressio {, 4, 7,..., 00}. Prove that there must be two distict itegers i A whose sum if Let A be the set of all 8-digit umbers i base 3 (so they are writte with the digits 0,,2 oly), icludig those with leadig zeroes such as Prove that give 4 elemets from A, two of them must coicide i at least 2 places Durig a moth with 30 days a baseball team plays at least a game a day, but o more tha 45 games. Show that there must be a period of some umber of cosecutive days durig which the team must play exactly 4 games (Putam, 2006-B2.) Prove that, for every set X = {x, x 2,..., x } of real umbers, there exists a o-empty subset S of X ad a iteger m such that m + s +. s S 9.7. (IMO 972.) Prove that from te distict two-digit umbers, oe ca always choose two disjoit oempty subsets, so that their elemets have the same sum Prove that amog ay seve real umbers y,..., y 7, there are two such that 0 y i y j + y i y j Prove that amog five differet itegers there are always three with sum divisible by 3.

11 PUTNAM TRAINING PROBLEMS, Telescopig. 0.. (Putam 984) Express as a ratioal umber. k= 6 k (3 k+ 2 k+ )(3 k 2 k ) 0.2. (Putam 977) Evaluate the ifiite product Evaluate the ifiite series: =0 = Symmetries... A spherical, 3-dimesioal plaet has ceter at (0, 0, 0) ad radius 20. At ay poit of the surface of this plaet, the temperature is T (x, y, z) = (x + y) 2 + (y z) 2 degrees. What is the average temperature of the surface of this plaet?.2. (Putam 980) Evaluate π/2 0 dx + (ta x) Iclusio-Exclusio. 2.. How may positive itegers ot exceedig 000 are divisible by 7 or? 2.2. Imagie that you are goig to give kids ice-cream coes, oe coe per kid, ad there are k differet flavors available. Assumig that o flavor gets mixed, fid the umber of ways we ca give out the coes usig all k flavors Let a, a 2,..., a a ordered sequece of distict objects. A deragemet of this sequece is a permutatio that leaves o object i its origial place. For example, if the origial sequece is {, 2, 3, 4}, the {2, 4, 3, } is ot a deragemet, but {2,, 4, 3} is. Let D deote the umber of deragemets of a -elemet sequece. Show that ( D =!! + 2! + ) ( ).! 3. Combiatorics ad Probability.

12 PUTNAM TRAINING PROBLEMS, Peter tosses 25 fair cois ad Joh tosses 20 fair cois. What is the probability that they toss the same umber of heads? 3.2. From where he stads, oe step toward the cliff would sed a druke ma over the edge. He takes radom steps, either toward or away from the cliff. At ay step his probability of takig a step away is p, of a step toward the cliff p. Fid his chace of escapig the cliff as a fuctio of p Two real umbers X ad Y are chose at radom i the iterval (0, ). Compute the probability that the closest iteger to X/Y is odd. Express the aswer i the form r + sπ, where r ad s are ratioal umbers O the uit circle cetered at the origi (x 2 + y 2 = ) we pick three poits at radom. We cut the circle ito three arcs at those poits. What is the expected legth of the arc cotaiig the poit (, 0)? 3.5. I a laboratory a hadful of thi 9-ich glass rods had oe tip marked with a blue dot ad the other with a red. Whe the laboratory assistat tripped ad dropped them oto the cocrete floor, may broke ito three pieces. For these, what was the average legth of the fragmet with the blue dot? 3.6. We pick poits at radom o a circle. What is the probability that the ceter of the circle will be i the covex polygo with vertices at those poits? 4. Miscellay (Putam 986) What is the uits (i.e., rightmost) digit of? (IMO 975) Prove that there are ifiitely may poits o the uit circle x 2 +y 2 = such that the distace betwee ay two of them is a ratioal umber (Putam 988) Prove that if we pait every poit of the plae i oe of three colors, there will be two poits oe ich apart with the same color. Is this result ecessarily true if we replace three by ie? 4.4. Imagie a ifiite chessboard that cotais a positive iteger i each square. If the value of each square is equal to the average of its four eighbors to the orth, south, west ad east, prove that the values i all the squares are equal (Putam 990) Cosider a paper puch that ca be cetered at ay poit of the plae ad that, whe operated, removes precisely those poits whose distace from the ceter is irratioal. How may puches are eeded to remove every poit? 4.6. (Putam 984) Let be a positive iteger, ad defie f() =! + 2! + +!.

13 PUTNAM TRAINING PROBLEMS, Fid polyomials P (x) ad Q(x) such that for all. f( + 2) = P ()f( + ) + Q()f() 4.7. (Putam 974) Call a set of positive itegers cospiratorial if o three of them are pairwise relatively prime. What is the largest umber of elemets i ay cospiratorial subset of itegers through 6? 4.8. (Putam 984) Prove or disprove the followig statemet: If F is a fiite set with two or more elemets, the there exists a biary operatio o F such that for all x, y, z i F, (i) x z = y z implies x = y (right cacellatio holds), ad (ii) x (y z) (x y) z (o case of associativity holds).

14 PUTNAM TRAINING PROBLEMS, Hits...2. For the iductio step, rewrite 2 2(+) as a sum of two terms that are divisible by For the iductive step assume that step a b is divisible by a b ad rewrite a + b + as a sum of two terms, oe of them ivolvig a b ad the other oe beig a multiple of a b..4. Strog iductio..5. Rewrite r + + /r + i terms of r k + /r k with k..6. How may regios ca be itersected by the ( + )th lie?.7. Color a plae divided with of lies i the desired way, ad thik how to recolor it after itroducig the ( + )th lie..8. How may regios ca be itersected the by ( + )th circle?.9. We have = 5 ( 7)+2 6 = ( 2). Also, prove that if = 5x+2y 44, the either x 7 or y For the iductive step, cosider a defective chessboard ad divide it ito four 2 2 chessboards. Oe of them is defective. Ca the other three be made defective by placig strategically a L?.. Use iductio o the umber of piles..2. The umbers 8 ad 9 form oe such pair. Give a pair (, + ) of cosecutive square-fulls, fid some way to build aother pair of cosecutive square-fulls..3. Look at oddess/eveess of the four lowest degree terms of the expasio..4. Assume that the first m Ulam umbers have already bee foud, ad determie how the ext Ulam umber (if it exists) ca be determied. 2.. Oe way to solve this problem is by usig the Arithmetic Mea-Geometric Mea iequality o each factor of the left had side Apply the Arithmetic Mea-Geometric Mea iequality to the set of umbers, 2,..., Power meas iequality with weights 2.4. Power meas iequality. p p+q ad q p+q.

15 PUTNAM TRAINING PROBLEMS, This problem ca be solved by usig Mikowski s iequality, but aother way to look at it is by a appropriate geometrical iterpretatio of the terms (as distaces betwee poits of the plae.) 2.7. May miimizatio or maximizatio problems are iequalities i disguise. The solutio usually cosists of guessig the maximum or miimum value of the fuctio, ad the provig that it is i fact maximum or miimum. I this case, give the symmetry of the fuctio a good guess is f(,,..., ) =, so try to prove f(x, x 2,..., x ). Use the Arithmetic Mea-Geometric Mea iequality o x,..., x. x 2.8. Apply the Cauchy-Schwarz iequality to the vectors ( y+z, ad choose appropriate values for u, v, w Arithmetic-Harmoic Mea iequality. y z+x, 2.0. Assume a b c, A B C, ad use Chebyshev s Iequality. z x+y ) ad (u, v, w), 2.. Divide by the right had side ad use the Arithmetic Mea-Geometric Mea iequality o both terms of the left Note that! is icreasig ( < m =! < m!) 2.3. Look at the fuctio f(x) = (999 x) l (999 + x) The umbers b 2 ad (b + ) 2 are cosecutive squares Use the Arithmetic Mea-Geometric Mea iequality o the squares of the roots of the polyomial Thik geometrically. Iterpret the give expressio as the square of the distace betwee two poits i the plae. The problem becomes that of fidig the miimum distace betwee two curves Cosider the expressios P = ( ( 3 2) 4 that k < k, for k =, 2,.... k k Look at the biomial expasio of (m + ) m Arithmetic Mea-Geometric Mea iequality Try first the cases = ad = 2. The use iductio Take logarithms ad use L Hôpital. ) ( 2 ) ( 2 ad Q = 2 ) ( 4 ) ( ). Note 3.. Cotradictio.

16 PUTNAM TRAINING PROBLEMS, If s is the sum of the digits of a umber, the s is divisible by Assume that there are fiitely may primes of the form 4+3, call P their product, ad try to obtai a cotradictio similar to the oe i Euclid s proof of the ifiitude of primes Prove that ad are relatively prime Prove that p(k) divides p(p(k) + k) Iductio Study the equatio modulo Call the sum S ad fid the maximum power of 2 dividig each side of the equality!s = k=! k f() (mod 9) Chiese Remaider Theorem. 3.. The miimum of a, b, c caot be very large Try chagig variables x = a +, y = b +, z = c Study the equatio modulo Use the Pigeohole Priciple to prove that the sequece of pairs (F, F + ) is evetually periodic modulo N = Try a = 6, b = If p is a odd umber ot divisible by 3, the p 2 ± (mod 6) Sophie Germai s idetity: a 4 + 4b 4 = (a 2 + 2b 2 + 2ab)(a 2 + 2b 2 2ab) The umber is irratioal or a iteger If m + + /2, what ca we say about m? 3.2. Each of the give umbers ca be writte p (0 3 ), where p (x) = +x+x 2 + +x, =, 2, 3, Study the cases b k = perfect square, ad b k = ot a perfect square. What ca we deduce about b k+ beig or ot beig a perfect square i each case?

17 3.23. gcd(a, b) = gcd(a, b a). PUTNAM TRAINING PROBLEMS, What is (x y)(2x + 2y + ) ad (x y)(3x + 3y + )? 4.. Call x = ad elimiate the radicals Factor p(x) Prove that the sum is the root of a moic polyomial but ot a iteger Look at the polyomial Q(x) = (x + )P (x) x Use the relatioship betwee zeros ad coefficiets of a polyomial The ( )-degree polyomial p(x) p( x) vaishes at differet poits For each iteger k study the parity of p(k) depedig o the parity of k We must prove that P () = 0. See what happes by replacig x with fifth roots of uity Assume (x a)(x b)(x c) = p(x)q(x), ad look at the possible values of p(x) ad q(x) for x = a, b, c Assume g(x) = h(x)k(x), where h(x) ad k(x) are o-costat polyomials with itegral coefficiets. Prove that the ca be assumed to be positive for every x ad h(p i ) = k(p i ) =, i =,...,. Deduce that both are of degree ad determie their form. Get a cotradictio by equatig coefficiets i g(x) ad h(x)k(x). 4.. The remaider will be a secod degree polyomial. Plug the roots of x 3 x Fid the value of f() for iteger Assume f(x) = g(x)h(x), where g(x) ad h(x) have itegral coefficiets ad degree less tha 05. Look at the product of the roots of g(x) 4.4. Sophie Germai s Idetity We have that a, b, c, d are distict roots of P (x) Oe way to solve this problem is by lettig A = + x + + x ad doig some algebra. 5.. If m = a 2 + b 2 ad = c 2 + d 2, the cosider the product z = (a + bi)(c + di) = (ac bd) + (ad + bc)i The left had side of the equality is the imagiary part of e ik. k=0

18 5.3. What are the possible values of x? PUTNAM TRAINING PROBLEMS, If ω = e 2πi/3 the ω ad ω 2 are two roots of p(z) Write si t = (e ti e ti )/2i Assume the vertices of the -go placed o the complex plae at the th roots of uity Look at the fuctio h(x) = f(x) + ig(x) Assume the lights placed o the complex plae at the th roots of uity, ζ, ζ 2,..., ζ, where ζ = e 2πi/ Hit Let z = a bi, z 2 = u + vi. We have z 2 = a 2 + b 2, z 2 = u 2 + v 2, R(z z 2 ) = au + bv, I(z z 2 ) =, ad must prove z 2 + z R(z z 2 ) Expad ad differetiate ( + x) Expad both sides of ( + x) ( + x) = ( + x) 2 ad look at the coefficiet of x Expad both sides of ( + x) m ( + x) = ( + x) m+ ad look at the coefficiet of x j Look at the geeratig fuctio of the Fiboacci sequece Fid the geeratig fuctio of the sequece u = umber of oegative solutios of 2a + 5b = The aswer equals the coefficiet of x 0 i the expasio of ( + x + x x 9 ) 6, but that coefficiet is very hard to fid directly. Try some simplificatio Look at the polyomial P (x) = a + a 2 x + a 3 x a 50 x 49, ad at its values at 3rd, 5th,... roots of uity. 7.. The subsets of {, 2,..., } that cotai o two cosecutive elemets ca be divided ito two classes, the oes ot cotaiig, ad the oes cotaiig The ( + )th vee divides the existig regios ito how may further regios? 7.3. The tiligs of a 2 rectagle by domioes ca be divided ito two classes depedig o whether we place the rightmost domio vertically or horizotally The miimal selfish subsets of {, 2,..., } ca be divided ito two classes depedig o whether they cotai or ot.

19 PUTNAM TRAINING PROBLEMS, Assume that b, b 2,..., b is a deragemet of the sequece a, a 2,..., a. How may possible values ca b have? Oce we have fixed the value of b, divide the possible deragemets ito two appropriate classes Fid a recurrece for a. 8.. u 2 = u Fid the limit of the sequece a = 2, a + = 2 + a ( ) Call the limit L. Fid some equatio verified by L Justify that the desired derivatives must coicide with those of the fuctio g(x) = /( + x 2 ) Compare the sum to some itegral of the form b a dx. x 8.6. Take logarithms. Iterpret the resultig expressio as a Riema sum Iterpret the first series is as a Maclauri series. Iterchage itegratio ad summatio with the secod series (do t forget to justify why the iterchage is legitimate.) 8.8. I fact ay real umber r is the limit of a sequece of umbers of the form 3 3 m. We wat r 3 3 m, i.e., r+ 3 m 3. Note that as If y f(r) what is f(y)? 8.0. Cosider the fuctio g(x) = f(x) f(x + /999). Use the itermediate value theorem. 9.. Divide the set ito subsets each of which has oly pairwise relatively prime umbers Divide the set ito subsets each of which cotais oly umbers which are multiple or divisor of the other oes Look at pairs of umbers i that sequece whose sum is precisely 04. Those pairs may ot cover the whole progressio, but that ca be fixed Prove that for each k =, 2,..., 8, at least 2 of the elemets give coicide at place k. Cosider a pair of elemets which coicide at place, aother pair of elemets which coicide at place 2, ad so o. How may pairs of elemets do we have? 9.5. Cosider the sequeces a i = umber of games played from the st through the jth day of the moth, ad b j = a j + 4. Put them together ad use the pigeohole priciple to prove that two elemets must be the equal.

20 PUTNAM TRAINING PROBLEMS, Cosider the fractioal part of sums of the form s i = x + + x i Cosider the umber of differet subsets of a te-elemet set, ad the possible umber of sums of at most te two-digit umbers Write y i = ta x i, with π x 2 i π 2 the x i s i the iterval ( π, π) 2 2 (i =,..., 7). Fid appropriate boxes for 9.9. Classify the umbers by their remider whe divided by Try to re-write the th term of the sum as A k B k. 3 k 2 k 3 k+ 2 k If you write a few terms of the product you will otice a lot of cacellatios. Factor the umerator ad deomiator of the th term of the product ad cacel all possible factors from k = 2 to k = N. You get a expressio i N. Fid its limit as N Write the th term as a sum of two partial fractios... Start by symmetrizig the give fuctio: f(x, y, z) = T (x, y, z) + T (y, z, x) + T (z, x, y)..2. Look at the expressio f(x) + f( π 2 x) Fid the umber of distributios of ice-cream coes without the restrictio usig all k flavors. The remove the distributios i which at least oe of the flavors is uused If P i is the set of permutatios fixig elemet a i, the the set of o-deragemets are the elemets of the P P 2 P. 3.. The probability of Joh gettig heads is the same as that of he gettig tails Cosider what happes after the first step, ad i which ways the ma ca reach the edge from there Look at the area of the set of poits verifyig the coditio The legths of the three arcs have idetical distributios The legths of the three pieces have idetical distributios Fid the probability of the polygo ot cotaiig the ceter of the circle. 4.. Compare to

21 PUTNAM TRAINING PROBLEMS, If cos u, si u, cos v, ad si v are ratioal, so are cos (u + v) ad si (u + v) Cotradictio Sice the values are positive itegers, oe of them must be the smallest oe. What are the values of the eighbors of a square with miimum value? 4.5. Try puches at (0, 0), (±α, 0),... for some appropriate α Note that f(+2) f(+) f(+) f() is a very simple polyomial i Start by fidig some subset T of S as large as possible ad such that ay three elemets of it are pairwise relatively prime Try a biary operatio that depeds oly o the first elemet: x y = φ(x).

22 PUTNAM TRAINING PROBLEMS, Solutios.. We prove it by iductio. The basis step correspods to = 4, ad i this case certaily we have 4! > 2 4 (24 > 6). Next, for the iductio step, assume the iequality holds for some value of 4, i.e., we assume! > 2, ad look at what happes for + : ( + )! =! ( + ) > 2 ( + ) > 2 2 = 2 +. by iductio hypothesis Hece the iequality also holds for +. Cosequetly it holds for every For the basis step, we have that for = ideed 2 2 = 4 = 3 is divisible by 3. Next, for the iductive step, assume that ad 2 2 is divisible by 3. We must prove that 2 2(+) is also divisible by 3. We have 2 2(+) = = = (2 2 ). I the last expressio the last term is divisible by 3 by iductio hypothesis, ad the first term is also a multiple of 3, so the whole expressio is divisible by 3 ad we are doe..3. By iductio. For = we have that a b = a b is ideed divisible by a b. Next, for the iductive step, assume that a b is divisible by a b. We must prove that a + b + is also divisible by a b. I fact: a + b + = (a b) a + b (a b ). O the right had side the first term is a multiple of a b, ad the secod term is divisible by a b by iductio hypothesis, so the whole expressio is divisible by a b..4. We prove it by strog iductio. First we otice that the result is true for = 0 (F 0 = 0 < = 2 0 ), ad = (F = < 2 = 2 ). Next, for the iductive step, assume that ad assume that the claim is true, i.e. F k < 2 k, for every k such that 0 k. The we must prove that the result is also true for +. I fact: ad we are doe. F + = F + F < < = 2 +, by iductio hypothesis.5. We prove it by iductio. For = the expressio is ideed a iteger. For = 2 we have that r 2 + /r 2 = (r + /r) 2 2 is also a iteger. Next assume that > 2 ad that the expressio is a iteger for ad. The we have ( r + + r + ) = ( r + r ) ( r + r ) hece the expressio is also a iteger for +. ( r + r ),

23 .6. By experimetatio we easily fid: PUTNAM TRAINING PROBLEMS, Figure. Plae regios R() A formula that fits the first few cases is R() = ( )/2. We will prove by iductio that it works for all. For = we have R() = 2 = ( )/2, which is correct. Next assume that the property is true for some positive iteger, i.e.: R() = We must prove that it is also true for +, i.e., R( + ) = ( + )2 + ( + ) + 2 = So lets look at what happes whe we itroduce the ( + )th straight lie. I geeral this lie will itersect the other lies i differet itersectio poits, ad it will be divided ito + segmets by those itersectio poits. Each of those + segmets divides a previous regio ito two regios, so the umber of regios icreases by +. Hece: R( + ) = S() + +. But by iductio hypothesis, R() = ( )/2, hece: QED. R( + ) = = We prove it by iductio i the umber of lies. For = we will have two regios, ad we ca color them with just two colors, say oe i red ad the other oe i blue. Next assume that the regios obtaied after dividig the plae with lies ca always be colored with two colors, red ad blue, so that o two regios that share a boudary have the same color. We eed to prove that such kid of colorig is also possible after dividig the plae with + lies. So assume that the plae divided by lies has bee colored i the desired way. After we itroduce the.

24 PUTNAM TRAINING PROBLEMS, ( + )th lie we eed to recolor the plae to make sure that the ew colorig still verifies that o two regios that share a boudary have the same color. We do it i the followig way. The ( + )th lie divides the plae ito two half-plaes. We keep itact the colors i all the regios that lie i oe half-plae, ad reverse the colors (chage red to blue ad blue to red) i all the regios of the other half-plae. So if two regios share a boudary ad both lie i the same half-plae, they will still have differet colors. Otherwise, if they share a boudary but are i opposite half-plaes, the they are separated by the ( + )th lie; which meas they were part of the same regio, so had the same color, ad must have acquired differet colors after recolorig..8. The aswer is f() = The proof is by iductio. For = we get f() = 2, which is ideed correct. The we must prove that if f() = the f( + ) = ( + ) 2 ( + ) + 2. I fact, the ( + )th great circle meets each of the other great circles i two poits each, so 2 poits i total, which divide the circle ito 2 arcs. Each of these arcs divides a regio ito two, so the umber of regios grow by 2 after itroducig the ( + )th circle. Cosequetly f( + ) = = = ( + ) 2 ( + ) + 2, QED..9. We proceed by iductio. For the basis step, i.e. = 44, we ca use four 5/c stamps ad two 2/c stamps, so that = 44. Next, for the iductio step, assume that for a give 44 the task is possible by usig x 5/c stamps ad y 2/c stamps, i.e, = 5x + 2y. We must ow prove that we ca fid some combiatio of x 5/c stamps ad y 2/c stamps so that + = 5x + 2y. First ote that either x 7 or y 2 otherwise we would have x 6 ad y, hece = 42 < 44, cotradictig the hypothesis that 44. So we cosider the two cases:. If x 7, the we ca accomplish the goal by settig x = x 7 ad y = y + 6: 5x + 2y = 5(x 7) + 2(y + 6) = 5x + 2y + = O the other had, if y 2 the, we ca do it by settig x = x + 5 ad y = y 2: 5x + 2y = 5(x + 5) + 2(y 2) = 5x + 2y + = We prove it by iductio o. For = the defective chessboard cosists of just a sigle L ad the tilig is trivial. Next, for the iductive step, assume that a 2 2 defective chessboard ca be tiled with L s. Now, give a defective chessboard, we ca divide it ito four 2 2 chessboards as show i the figure. Oe of them will have a square missig ad will be defective, so it ca be tiled with L s. The we place a L coverig exactly oe corer of each of the other 2 2 chessboards (see figure). The remaiig part of each of those chessboards is like a defective chessboard ad ca be tiled i the desired way too. So the whole defective chessboard ca be tiled with L s... We use iductio o the umber of piles. For = we have oly oe pile, ad sice each player must take at least oe toke from that pile, the umber of tokes i

25 PUTNAM TRAINING PROBLEMS, Figure 2. A defective chessboard. it will decrease at each move util it is empty. Next, for the iductio step, assume that the game with piles must ed evetually. We will prove that the same is true for + piles. First ote that the players caot keep takig tokes oly from the first piles, sice by iductio hypothesis the game with piles evetually eds. So sooer or later oe player must take a toke from the ( + )th pile. It does ot matter how may tokes he or she adds to the other piles after that, it is still true that the players caot keep takig tokes oly from the first piles forever, so evetually someoe will take aother toke from the ( + )th pile. Cosequetly, the umber of tokes i that pile will cotiue decreasig util it is empty. After that we will have oly piles left, ad by iductio hypothesis the game will ed i fiitely may steps after that..2. The umbers 8 ad 9 are a pair of cosecutive square-fulls. Next, if ad + are square-full, so are 4( + ) ad 4( + ) + = (2 + ) For = 2, 3, 4, 5, 6 we have: ( + x + x 2 ) 2 = + 2x + 3x 2 + 2x 3 + x 4 ( + x + x 2 ) 3 = + 3x + 6x 2 + 7x 3 + ( + x + x 2 ) 4 = + 4x + 0x 2 + 6x 3 + ( + x + x 2 ) 5 = + 5x + 5x x 3 + ( + x + x 2 ) 6 = + 6x + 2x x 3 + A alterate proof based o properties of ordial umbers is as follows (requires some advaced settheoretical kowledge.) Here ω = first ifiite ordial umber, i.e., the first ordial after the sequece of atural umbers 0,, 2, 3,.... Let the ordial umber α = a 0 + a ω + a 2 ω a ω represet a cofiguratio of piles with a 0, a,..., a tokes respectively (read from left to right.) After a move the ordial umber represetig the cofiguratio of tokes always decreases. Every decreasig sequece of ordials umbers is fiite. Hece the result.

26 PUTNAM TRAINING PROBLEMS, I geeral, if ( + x + x 2 ) = a + bx + cx 2 + dx 3 +, the ( + x + x 2 ) + = a + (a + b)x + (a + b + c)x 2 + (b + c + d)x 3 +, hece the first four coefficiets of ( + x + x 2 ) + deped oly o the first four coefficiets of ( + x + x 2 ). The same is true is we write the coefficiets modulo 2, i.e., as 0 if they are eve, or if they are odd. So, if we call q (x) = (+x+x 2 ) with the coefficiets writte modulo 2, we have q (x) = + x + x 2 q 2 (x) = + 0x + x 2 + 0x 3 + x 4 q 3 (x) = + x + 0x 2 + x 3 + q 4 (x) = + 0x + 0x 2 + 0x 3 + q 5 (x) = + x + x 2 + 0x 3 + q 6 (x) = + 0x + x 2 + 0x 3 + We otice that the first four coefficiets of q 6 (x) coicide with those of q 2 (x), ad sice these first four coefficiets determie the first four coefficiets of each subsequet polyomial of the sequece, they will repeat periodically so that those of q (x) will always coicide with those of q +4. Sice for = 2, 3, 4, 5 at least oe of the first four coefficiets of q (x) is 0 (equivaletly, at least oe of the first four coefficiets of ( + x + x 2 ) is eve), the same will hold for all subsequet values of..4. Let U m = {u, u 2,..., u m } (m 2) be the first m Ulam umbers (writte i icreasig order). Let S m be the set of itegers greater tha u m that ca be writte uiquely as the sum of two differet Ulam umbers from U m. The ext Ulam umber u m+ is precisely the miimum elemet of S m, uless S m is empty, but it is ot because u m + u m S m. 2.. Usig the Arithmetic Mea-Geometric Mea Iequality o each factor of the LHS we get ( ) ( ) a 2 b + b 2 c + c 2 a ab 2 + bc 2 + ca 2 ( ) ( ) 3 3 a3 b c 3 a3 b 3 c 3 = a 2 b 2 c 2. Multiplyig by 9 we get the desired iequality. Aother solutio cosists of usig the Cauchy-Schwarz iequality: (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca 2 ) = ((a b) 2 + (b c) 2 + (c a) 2 ) (( b c) 2 + ( c a) 2 + ( a b) 2 ) (abc + abc + abc) 2 = 9a 2 b 2 c 2.

27 PUTNAM TRAINING PROBLEMS, This result is the Arithmetic Mea-Geometric Mea applied to the set of umbers, 2,..., : (+) < = = +. 2 Raisig both sides to the th power we get the desired result The simplest solutio cosists of usig the weighted power meas iequality to the p (weighted) arithmetic ad quadratic meas of x ad y with weights ad q : p+q p+q p p + q x + q p p + q y p + q x2 + q p + q y2, hece (px + qy) 2 (p + q)(px 2 + qy 2 ). Or we ca use the Cauchy-Schwarz iequality as follows: (px + qy) 2 = ( p p x + q q y) 2 ( { p} 2 + { q} 2) ( { p x} 2 + { q y} 2) = (p + q)(px 2 + qy 2 ). Fially we use p + q to obtai the desired result By the power meas iequality: a + b + c 3 } {{ } M (a,b,c) From here the desired result follows. ( a + b + c 3 ) 2 } {{ } M /2 (a,b,c) (Cauchy-Schwarz) 2.5. We have: x + y + z = (x + y + z) 3 xyz (xyz = ) (x + y + z)2 3 x 2 + y 2 + z 2. (AM-GM iequality) (power meas iequality) 2.6. The result ca be obtaied by usig Mikowski s iequality repeatedly: a 2 + b 2 + a b a 2 + b 2 (a + a 2 ) 2 + (b + b 2 ) a 2 + b 2 (a + a 2 + a 3 ) 2 + (b + b 2 + b 3 ) a 2 + b 2 (a + a a ) 2 + (b + b b ) 2

28 PUTNAM TRAINING PROBLEMS, Aother way to thik about it is geometrically. Cosider a sequece of poits i the plae P k = (x k, y k ), k = 0,...,, such that (x k, y k ) = (x k + a k, y k + b k ) for k =,...,. The the left had side of the iequality is the sum of the distaces betwee two cosecutive poits, while the right had side is the distace betwee the first oe ad the last oe: d(p 0, P ) + d(p, P 2 ) + + d(p, P ) d(p 0, P ) By the Arithmetic Mea-Geometric Mea Iequality = x x 2... x x + x x, Hece f(x, x 2,..., x ). O the other had f(,,..., ) =, so the miimum value is For x = y = z = we see that S = 3/2. We will prove that i fact 3/2 is the miimum value of S by showig that S 3/2. Note that ( ) 2 ( ) 2 ( ) 2 x y z S = + +. y + z z + x x + y Hece by the Cauchy-Schwarz iequality: ( xu S (u 2 + v 2 + w 2 ) + y + z yv z + x + Writig u = y + z, v = z + x, w = x + y we get S 2(x + y + z) (x + y + z) 2, zw x + y ) 2. hece, dividig by 2(x + y + z) ad usig the Arithmetic Mea-Geometric Mea iequality: S 2 (x + y + z) xyz = By the Arithmetic Mea-Harmoic Mea iequality: x + y + z = 3 3, x y z hece 9 x + y + z. O the other had for x = y = z = /3 the sum is 9, so the miimum value is Assume a b c, A B C. The 0 (a b)(a B) + (a c)(a C) + (b c)(b C) = 3(aA + bb + cc) (a + b + c)(a + B + C).

29 PUTNAM TRAINING PROBLEMS, Usig A + B + C = π ad dividig by 3(a + b + c) we get the desired result. - Remark: We could have used also Chebyshev s Iequality: ( ) ( ) a A + b B + c C a + b + c A + B + C Assume a i +b i > 0 for each i (otherwise both sides are zero). The by the Arithmetic Mea-Geometric Mea iequality ( ) / a a ( a + + a ), (a + b ) (a + b ) a + b a + b ad similarly with the roles of a ad b reversed. Addig both iequalities ad clearig deomiators we get the desired result. (Remark: The result is kow as superadditivity of the geometric mea.) 2.2. We have that! is icreasig for, i.e., < m =! < m! So 999! > 2000 = (999!)! > 2000! = ((999!)!)! > (2000!)! =... = 999! (2000) > 2000! (999) Cosider the fuctio f(x) = (999 x) l (999 + x). Its derivative is f (x) = l (999 + x)+ 999 x, which is egative for 0 x, because i that iterval x 999 x = l e < l (999 + x) x Hece f is decreasig i [0, ] ad f(0) > f(), i.e., 999 l 999 > 998 l Cosequetly > We have b 2 < b } 2 + {{ b + } < b 2 + 2b + = (b + ) 2. But b 2 ad (b + ) 2 are cosecutive a 2 squares, so there caot be a square strictly betwee them We may assume that the leadig coefficiet is +. The sum of the squares of the roots of x + a x + + a is a 2 2a 2. The product of the squares of the roots is a 2. Usig the Arithmetic Mea-Geometric Mea iequality we have a 2 2a 2 a 2. Sice the coefficiets are ± that iequality is ( ± 2)/, hece 3. Remark: x 3 x 2 x + = (x + )(x ) 2 is a example of 3th degree polyomial with all coefficiets equal to ± ad oly real roots The give fuctio is the square of the distace betwee a poit of the quarter of circle x 2 +y 2 = 2 i the ope first quadrat ad a poit of the half hyperbola xy = 9 i that quadrat. The tagets to the curves at (, ) ad (3, 3) separate the curves, ad both are perpedicular to x = y, so those poits are at the miimum distace, ad the aswer is (3 ) 2 + (3 ) 2 = 8.

30 2.7. Let PUTNAM TRAINING PROBLEMS, ( ) ( ) ( ) 3 2 P =, Q = ( 2 3 ) ( ) 4 5 ( ) We have P Q = 2. Also < 2 < 3 < 4 < < 2, hece 2P Q, so P 2 P Q =, ad from here we get P. 2 4 O the other had we have P < Q < Q, hece P 2 < P Q =, ad from 2 here P < The give iequality is equivalet to ( ) (m + )! m + m m = m m < (m + ) m+, m!! which is obviously true because the biomial expasio of (m + ) m+ icludes the term o the left plus other terms Usig the Arithmetic Mea-Geometric Mea iequality we get: { a + a a } a b b 2 b a2 a =. b b 2 b From here the desired result follows We prove it by iductio. For = the result is trivial, ad for = 2 it is a simple cosequece of the followig: 0 (a 2 a )(b 2 b ) = (a b + a 2 b 2 ) (a b 2 + a 2 b ). Next assume that the result is true for some 2. We will prove that is is true for +. There are two possibilities:. If x + = b +, the we ca apply the iductio hypothesis to the first terms of the sum ad we are doe. 2. If x + b +, the x j = b + for some j +, ad x + = b k for some k +. Hece: + a i x i = i= = a i x i + a j x j + a + x + i= i j a i x i + a j b + + a + b k i= i j

31 2.2. We have PUTNAM TRAINING PROBLEMS, (usig the iequality for the two-term icreasig sequeces a j, a + ad b k, b + ) a i x i + a j b k + a + b +. i= i j This reduces the problem to case. l ( k= a p k ) /p = l ( ) k= ap k. p Also, a k as p 0, hece umerator ad deomiator ted to zero as p approaches zero. Usig L Hôpital we get l ( ) k= lim ap k k= = lim ap k l a ( k k= p 0 p = l a ) / k = l a k. p 0 k= ap k From here the desired result follows. 3.. If p ad q are cosecutive primes ad p + q = 2r, the r = (p + q)/2 ad p < r < q, but there are o primes betwee p ad q (a) No, a square divisible by 3 is also divisible by 9. (b) Same argumet Assume that the set of primes of the form is fiite. Let P be their product. Cosider the umber N = P 2 2. Note that the square of a odd umber is of the form 4 +, hece P 2 is of the form 4 + ad N will be of the form Now, if all prime factors of N where of the form 4 +, N would be of the form 4 +, so N must have some prime factor p of the form So it must be oe of the primes i the product P, hece p divides N P 2 = 2, which is impossible That is equivalet to provig that ad are relatively prime for every. These are two possible ways to show it: - Assume a prime p divides = ( 2 + 2). The it must divide or Writig = 2 ( 2 + 3) + = ( 2 + )( 2 + 2) we see that p caot divide i either case. - The followig idetity ( 2 + )( ) ( 3 + 2) 2 = (which ca be checked algebraically) shows that ay commo factor of ad should divide, so their gcd is always. (Note: if you are woderig how I arrived to that idetity, I just used the Euclidea algorithm o the two give polyomials.) k=

32 PUTNAM TRAINING PROBLEMS, Assume p(x) = a 0 + a x + + a x, with a 0. We will assume WLOG that a > 0, so that p(k) > 0 for every k large eough otherwise we ca use the argumet below with p(x) istead of p(x). We have p(p(k) + k) = a i [p(k) + k] i. i=0 For each term of that sum we have that a i [p(k) + k] i = [multiple of p(k)] + a i k i, ad the sum of the a i k i is precisely p(k), so p(p(k) + k) is a multiple of p(k). It remais oly to ote that p(p(k) + k) p(k) for ifiitely may positive itegers k, otherwise p(p(x) + x) ad p(x) would be the same polyomial, which is easily ruled out for o costat p(x) This ca be proved easily by iductio. Base case: F = ad F 2 = are i fact relatively prime. Iductio Step: we must prove that if F ad F + are relatively prime the so are F + ad F +2. But this follows from the recursive defiitio of the Fiboacci sequece: F + F + = F +2 ; ay commo factor of F + ad F +2 would be also a factor of F, ad cosequetly it would be a commo factor of F ad F + (which by iductio hypothesis are relatively prime.) 3.7. For ay iteger we have that 2 oly ca be 0 or mod 3. So if 3 does ot divide a or b they must be mod 3, ad their sum will be 2 modulo 3, which caot be a square Assume the sum S is a iteger. Let 2 i be the maximum power of 2 dividig, ad let 2 j be the maximum power of 2 dividig! The!! 2 j 2i S = k2. j i k= For 2 the left had side is a eve umber. I the right had side all the terms of the sum are eve itegers except the oe for k = 2 i which is a odd iteger, so the sum must be odd. Hece we have a eve umber equal to a odd umber, which is impossible Sice each digit caot be greater tha 9, we have that f() 9 ( + log 0 ), so i particular f(n) 9 ( log ) < 9 ( ) = Next we have f(f(n)) 9 6 = 54. Fially amog umbers ot greater tha 54, the oe with the greatest sum of the digits is 49, hece f(f(f(n))) = 3. Next we use that f() (mod 9). Sice (mod 9), the (mod 9). We otice that the sequece 7 mod 9 for = 0,, 2,... is, 7, 4,, 7, 4,..., with period 3. Sice 4444 (mod 3), we have (mod 9), hece f(f(f(n))) 7