SUMS PROBLEM COMPETITION, 200 SOLUTIONS Suppose that after n vsts to Aunt Joylene (and therefore also n vsts to Uncle Bruce Lnda has t n ten cent peces and d n dollar cons After a vst to Uncle Bruce she has 2d n twenty cent peces and t n + d n ffty cent peces So after her next vst to Aunt Joylene she has 2(t n + d n ten cent peces and 2d n + (t n + d n dollar cons Thus Hence so that t n+ 2t n + 2d n d n+ t n + 3d n t n+2 2t n+ + 2d n+ 2t n+ + 2(t n + 3d n 2t n+ + 2t n + 3(t n+ 2t n, t n+2 5t n+ + 4t n 0 ( Ths s a second order lnear recurrence equaton wth constant coeffcents The soluton s found by frst solvng the auxlary equaton λ 2 5λ + 4 0 The roots are λ 4 and λ 2 Then the general soluton to the recurrence relaton ( s Aλ n + Bλ n 2 A 4 n + B (2 We are told that t 0 0 and that d 0 Hence t 2t 0 + 2d 0 2 Usng t 0 0 and t 2, we qucly fnd that the constants A and B are 2/3 and 2/3, respectvely So t n ( 2 3 4 n 2 3 Smlarly, the d n satsfy the recurrence relaton d n+2 5d n+ + 4d n 0, and d n s gven by a formula (2 Snce d 0 and d t 0 + 3d 0 3, we qucly fnd that A 2/3 and B /3 ths tme Hence ( 2 d n 4 3 n + 3 Hence after n vsts to Aunt Joylene, Lnda has d n + 0 t n ( 4 n + 4 5 5 dollars 2 We consder more generally the case of n 3 delegates sttng around a round table, each puttng ther document on ther own seat and that of ther two mmedate neghbours There are 3 n dfferent ways for the thef to steal the documents, correspondng to the three choces he has at each seat So the probablty of obtanng a complete set s N n /3 n, where N n s the number of ways of stealng them whch result n a complete set We shall show that N n F n+ + F n + 2, where F n s the n-th Fbonacc number, defned by F, F 2, and F n+2 F n+ + F n for n, so that F 3 2, F 4 3, etc
In the partcular case n 2 ths gves N 2 F 3 + F + 2 233 + 89 + 2 324, so that the probablty of gettng a complete set s 324 3 2 4 3 8 4 656 Let us label the seats 0,, n Then on seat, we have documents, and +, these expressons beng understood modulo n Thus n front of seat 0 we have documents n, 0 and ; n front of seat we have documents 0, and 2, etc, untl n front of seat n we have documents n 2, n and 0 The ey to dervng the above formula for N n s the followng observaton: Suppose that the thef taes document from seat and he taes document from seat + Then he must tae document + from seat + 2, snce otherwse he would mss that document Contnung clocwse around the table, we see that he must tae document j from seat j for every j Seat number : + + 2 2 + + + 2 + + 2 + 3 Smlarly, f the thef taes document + from seat and he taes document + 2 from seat + Then he must tae document from seat Contnung ant-clocwse around the table, we see that he must tae document j + from seat j for every j If the thef taes document j from seat j for every j, or f he taes document j + from seat j for every j, then we shall say that he has commtted a specal theft When a thef steals the documents, let us wrte f( j f he steals document j from seat Then f : {0,, n } {0,, n } ( s a functon To say that he gets a complete set s to say that f s surjectve Snce the doman and codoman are of the same sze, ths s equvalent to f beng njectve, or to f beng a bjecton So the number N n s the number of bjectons ( such that f( {,, + } for all We call the bjectons f( and f( + specal Let N n denote the number of non-specal bjectons ( such that f( {,, +} for all Gven such a bjecton, let S f { {0,, n } : f( + } If S f, then f( + cannot equal + snce f s njectve, and t cannot equal + 2, snce otherwse f would be specal by the ey observaton above Hence f( + must be In partcular, + S f For the same reasons, f S f, then f( must be If nether nor s n S f, then f( must be For f( would mply that f( + by the above ey observaton, and so S f, e, f(, then mples that document would be mssed Conversely, suppose that S {0,, n } has the property that S + S (2 2
Defne f S : {0,, n } {0,, n } by settng { + f S, f S ( f S, f, S Then f f S s a non-specal bjecton (, and S { : f S ( +} Ths shows that the set of non-specal bjectons ( are n one to one correspondence wth the set of subsets S {0,, n } havng property (2 Hence N n s the number of such subsets Another way of expressng property (2 s to say that S contans none of the subsets {0, }, {, 2},, {n 2, n }, {n, 0} To evaluate N n, t s convenent to express N n n terms of the number A m of subsets S of {,, m} such that S + S for,, m (3 m Note that we are not usng arthmetc modulo n (or modulo m here Let us show that A m A m + A m 2 for all m 3 For let S be a subset of {,, m} satsfyng (3 m If m S, then m cannot be n S, and so S S {m}, where S {,, m 2} satsfes (3 m 2 On the other hand, f m S, then S s a subset of {,, m } satsfyng (3 m The set of subsets S of {,, m} satsfyng (3 m s therefore the unon of two dsjont subsets, one wth A m 2 elements, the other wth A m Hence A m A m + A m 2 for all m 3 Notce that A 2 F 3 (the subsets beng and {0} and A 2 3 F 4 (the subsets beng, {0} and {} It follows by nducton that A m F m+2 for m, 2, We now relate the numbers A m to what we really want: N n If S {0,, n } has the property (2 (whch s understood usng arthmetc modulo n, then ether n S or n S If n S, then 0 S and also n 2 S Hence S S \ {n } must be a subset of {,, n 3}, and t must satsfy (3 n 3 There are A n 3 F n such subsets If n S, then S s a subset of {0,, n 2} so that S + { + : S} s a subset of {,, n } satsfyng (3 n There are A n F n+ such subsets The set of subsets S of {0,, n } satsfyng (2 s therefore the unon of two dsjont subsets, one wth F n elements, the other wth F n+ Hence N n F n+ + F n for all n 2 Thus N n N n + 2 F n+ + F n + 2 for all n 2 3 Soluton We may assume our space has coordnate axes so that the orgn s one corner of the larger box B, and that B les n the frst octant Hence B s bounded by the 6 planes x 0, x l, y 0, y h, z 0 and z b Now suppose that P s one corner of the smaller box, wth coordnate vector p (p x, p y, p z Let p + u, p + v and p + w be the coordnate vectors of the vertces of B whch are joned by an edge to P Suppose that u has coordnates (u x, u y, u z, and smlarly for v and w Notce that some or all of u x, u y and u z mght be negatve (several entres were not correct because they ddn t consder ths possblty The length, breadth and heght of B are the lengths of the vectors u, v and w (n that order, say: l u u 2 x + u 2 y + u 2 z, b v By the trangle nequalty, v 2 x + v 2 y + v 2 z and h w w 2 x + w 2 y + w 2 z l u (u x, 0, 0 + (0, u y, 0 + (0, 0, u z (u x, 0, 0 + (0, u y, 0 + (0, 0, u z 3 u x + u y + u z
Smlarly, b v x + v y + v z and h w x + w y + w z The four vertces of B whch are not joned by an edge to P have coordnate vectors p + u + v, p + u + w, p + v + w and p + u + v + w Let us show that u x + v x + w x l ( To see ths, project the 8 vertces of B onto the x-axs In other words, loo at the eght numbers p x, p x + u x, p x + v x, p x + w x, p x + u x + v x, p x + u x + w x, p x + v x + w x, p x + u x + v x + w x These must all be between 0 and l, because B B mples that all ponts of B are between the two planes x 0 and x l So the dfference between any two of them must be n modulus at most l For example, (p x + u x + v x (p x + w x u x + v x w x must be at most l There are 8 possble choces of the sgns of u x, v x and w x If all are 0 or all are 0, then ( holds because u x + v x + w x u x + v x + w x (p x + u x + v x + w x p x l If two of u x, v x and w x are 0 and one s 0, say u x, v x 0 and w x 0, then ( holds because u x + v x + w x u x + v x + ( w x u x + v x w x (p x + u x + v x (p x + w x l, as observed above Smlarly f one of u x, u y and u z s 0 and two are 0, then ( holds Smlarly, we have Hence u y + v y + w y b and u z + v z + w z h l + b + h ( u x + u y + u z + ( v x + v y + v z + ( w x + w y + w z ( u x + v x + w x + ( u y + v y + w y + ( u z + v z + w z l + b + h Soluton 2 The dea s to tae neghbourhoods of both boxes Let B d denote the set of ponts whch are at dstance at most d from a pont n B, and smlarly for B We compare vol(b d and vol(b d Let us llustrate the dea n 2 dmensons frst, where we compare areas 4
h l d The area of B d s easly seen to be B d lh + 2(l + hd + πd 2 Snce B B, we have B d B d for every d, and so lh + 2(l + hd + πd 2 l h + 2(l + h d + πd 2 Subtractng πd 2 from both sdes and dvdng by 2d, we have lh 2d + l + h l h 2d + l + h Ths s true for every d > 0 Now let d, and we get l + h l + h The proof n 3 dmensons s smlar One can easly show that vol(b d lbh + 2(lb + lh + bhd + (l + b + hπd 2 + 4 3 πd3 From vol(b d vol(b d we see that lbh + 2(lb + lh + bhd + (l + b + hπd 2 l b h + 2(l b + l h + b h d + (l + b + h πd 2 Dvdng by πd 2 and lettng d tend to nfnty, we fnd that l + b + h l + b + h 4 There were several correct solutons to the frst part of the queston For the second part, the equaton 4n(n + α(n 2 + x 2 can be wrtten x 2 n(n + y 2 for y 2α(n, and so s an example of Pell s equaton, whch s treated n many boos usng contnued fractons, etc, and ths approach was followed successfully by some entrants We gve nstead the followng elementary soluton (submtted by Van Mnh Nguyen, and hs soluton to the frst part usng smlar methods Suppose that 4n(n + a 2 + s a perfect square for n 0,, Of course a 0 has ths property, and so we assume that a Wrte 4n(n + a 2 + m 2 n, where m n 0 s an nteger Usng a a 2 and a 2, we get (2na+ 2 4n 2 a 2 +4na+ 4n 2 a 2 +4na 2 + m 2 n 4n(n+a 2 +a 2 ((2n+a 2, ( 5
and so 2na + m n (2n + a So we may wrte m n 2na + x n, where x n a for all n Hence 4n(n + a 2 + (2na + x n 2, (2 so that x n a + 4an x2 n 4an a as n (we use x n a to see that x 2 n/4an a 2 /4an a/4n here But x n s an nteger, and so x n a must hold f n s suffcently large For such n, (2 tells us that 4n(n + a 2 + (2na + a 2, from whch we see that a Now suppose that α(n an + b has the property that 4n(n + α(n 2 + s a perfect square for n 0,, Agan wrte 4n(n + α(n 2 + m 2 n, where m n 0 s an nteger There were several ncomplete solutons to ths part of the problem, n whch, wthout justfcaton, t was assumed that m n cn 2 + dn + e for some constants c, d and e The proof below essentally provdes the justfcaton for ths If a 0, then b by the frst part, and so assume that a Then α(n for all n, and so, replacng a by α(n n ( above, (2nα(n + 2 m 2 n ((2n + α(n 2 so that m n 2nα(n + x n, where x n α(n Replacng a by α(n n (2, we get 4n(n + α(n 2 + (2nα(n + x n 2, (3 Therefore and so x n α(n + x n n α(n n + 4α(nn 2 4α(nn x2 n 4α(nn, x 2 n 4α(nn 2 a as n (we use x n α(n to see that x 2 n/4α(nn 2 α(n/4n 2 0 here So we can wrte x n an + y n, where y n /n 0 as n Substtutng ths nto (3, we get and so 4n(n + α(n 2 + (2nα(n + an + y n 2, y n 4b a 4 y n 2n + b2 an by n an y2 n 4an 2 + 4an 2 Ths tends to (4b a/4 as n Snce y n s an nteger, we must have y n (4b a/n for suffcently large n For such n, 4n(n + α(n 2 + (2nα(n + an + (4b a/4 2, Expandng both sdes, and comparng constant terms and the coeffcents of n on both sdes, we get 8a(a 2b 0 and 6 a 2 8ab 6b 2 0 6
Snce a > 0, the frst equaton tells us that a 2b Then the second equaton becomes 6 4b 2 0, so that b 2 Hence (a, b (4, 2 s the only soluton wth a > 0 Explctly, 4n(n + (4n + 2 2 + (8n 2 + 8n + 2 5 There were several dfferent solutons to ths problem Soluton (the shortest: Let A m,n (2n!(2m! n!m!(n+m! Then t s routne to chec that A m,n + A m,n 4A m,n Hence A m,n A m+,n + 4A m,n Snce A m,0 ( 2m m, the result s now a routne nducton on n Soluton 2 : The result s mmedate from the followng dentty (due to Szly, 895: (2n!(2m! n!m!(n + m! n n ( ( 2n 2m ( n m Ths dentty s seen by consderng ( x 2 n ( + x n ( x n, usng the bnomal theorem on the three expressons, and comparng coeffcents of x 2r on both sdes Soluton 3 : For each prme p, let n p (respectvely, d p denote the number of tmes that p (2n!(2m! dvdes the numerator (respectvely, the denomnator of n!m!(n+m! It s enough to show that d p n p for each p Now t s well nown that the number of tmes that p dvdes n! s n n n + p p 2 + p 3 +, the j-th term of the seres beng zero f n < p j So what we need to show s that m p j + j n p j + j and for ths t s suffcent to show that m + n j p j m n m + n p j + p j + p j 2m p j + j 2m 2n p j + p j 2n p j, j for each j In fact, t s easy to chec that m n m + n 2m + + 2n + for any nteger 6 To sum equals 2 log(2π To see ths, start by wrtng ( n n n(n + ζ(n ( n n n(n + 7 n ( n ( n n(n +
We next want to nterchange the order of ths double sum, and wrte ( n n(n + ( n ( n n(n + ( n ( The most common theorem allowng us to nterchange the order of double sums, and wrte has the condton ( a n, n ( n ( a n, n a n, < Ths condton s not satsfed n the present example, bascally because of the frst term / n n the seres for ζ(n So we only loo at the seres startng at 2 For all n 2, we have the estmate 2 n 2 n + 2 x n dx 2 n + n 2 n 2 n + 2 n 3 2 n, and so t s easy to see that s vald for our a n, s: Hence for these a n,, ( 2 a n, < a n, n ( n n(n + ( a n, ( ( a n, + a n, a n, + a n, + 2 ( a n, 2 ( a n, 2 ( a n, Ths justfes the nterchange of summatons n ( Now the double sum on the rght n ( equals ( ( 2 n + ( n n 8
The nner sum can be rewrtten Usng ( 2 n n + n ( n+ n ( n xn log, x vald f x <, we see that the expresson (2 equals n ( ( ( 2 log + + ( log + ( n (2 Now we sum ths from to N, say After tdyng up the last expresson, we see that we need to calculate the sum 2N + N (2 + ( log( + log( (3 Ths equals (we are usng partal summaton here N+ ( N N 2N+ 2( + log( (2+ log( 2N+(2N+ log(n+ 2 log( Now N log( log(n!, whch by Strlng s formula equals log( 2πe N N N+/2 a N, where a N as N So the expresson (3 equals ( 2N + (2N + log(n + 2 log( 2π N + ( N + log(n + log(an 2 (2N + log((n + /N log(2π + log(a N Usng the above seres for log(/( x, for example, we see that log((n + /N /N + O(/N 2, and so (2N + log((n + /N 2 as N The result follows 7 Let us call r j a j the weght of the strng a a r If r j a j, then the strng conssts of a sngle letter, whch s of the requred form Step We frst gve a procedure for transformng any strng s havng repeated letters nto a strng of smaller weght If s has weght W, then at most W applcatons of the procedure wll therefore produce a strng n whch there are no repeated letters Consder a strng s whch has some repeated letters Let m be the largest repeated letter n s Then we transform s nto a strng havng a par of m s wth no m + s between them For f there s a letter m + n the strng, use a successon of moves of type ( to get a strng s wth the m + at the rght hand end Snce there s no second m +, there are no m + s between any two m s 9
Choose two m s n s wth no further m s between them, as near as possble and wth no m + s between them Then the strng s contans the substrng a m, a +,, a +l, a +l m ( for some l, and a +,, a +l are all not m or m + If l, then the two m s are adjacent, and we can delete one of them (a move of type (, resultng n a strng s of weght S m Step s done So l 2, and we have as substrng m, a +,, m If a + m+2 or f a + m 2, then we can nterchange a + and the frst m (a type ( move, resultng n a strng s havng repeated m s, no repeated letters greater than m, no m + between these m s, but wth the repeated m s closer together Repeat ths procedure as often as possble That s, contnue ths procedure untl the letter to the rght of the frst m s m, m, or m + By the ntal choce of par of m s, the letter to the rght of the frst m s m, and we have a substrng of the form ( n whch a + m If l 2, then s contans a substrng m, m, m But then a move of type (v turns ths substrng nto m, m, m, and s nto a strng s of weght S Step s done If l 3 and a +l m + 2 or a +l m 2, then we nterchange a +l and the rght hand m Contnue ths procedure untl the letter to the left of the second m s m, m, or m + Ths letter must be m for the same reasons that a + m Now f l 3, we have a substrng m, m, m, m, and so a move of type ( removes one of the m s, resultng n a strng s of weght S (m < S, and we have done Step If l 4, we loo at a +2 If a +2 m + 2 or f a +2 m 3, then we can mae two type ( moves to replace m, m, a +2 by a +2, m, m Repeat ths procedure as often as possble That s, contnue ths procedure untl the letter to the rght of m, m s m 2, m, m, or m + The cases m, m and m + are excluded as above, and so a +2 m 2 must hold Smlarly, f l 5, a +l 2 must be m 2 Contnung n ths way, we obtan a substrng ( n whch ether l 2j + s odd, and the substrng s or l 2j s even, and the substrng s m, m,, m j, m j,, m, m, m, m,, m j +, m j, m j +, m, m In the frst case, we can delete one of the m j s gettng a strng s of weght S (m j < S In the second case, we do a move of type (v to replace m j +, m j, m j + by m j, m j +, m j Ths gves a strng s of weght S < S So n all cases wth a repeated letter present we can transform the strng nto a strng of smaller weght Step 2 Now we show that f all the letters of the strng are dfferent, then we can reorder them so that these letters are n ncreasng order To start wth, we perform some moves of type ( to move the largest letter n the strng to the rght hand end We next am s to obtan a strng of the form a,, a r 2, a r, a r, (2 0
wth a r < a r and wth all of a,, a r 2 less than a r To do ths, suppose that a s the largest of the letters a,, a r Then n partcular, a a for +,, r, and, because a a r, we have a a r 2 for +,, r So by a successon of moves of type (, we can move a r to the left untl t s mmedately to the rght of a Then by a successon of moves of type (, we move the par a, a r to the rght hand end The strng s now of the form (2 Suppose that we have brought the strng nto the form a,, a r j+,, a r, (3 wth a r j+ < < a r and wth all of a,, a r j are less than a r j+ If the largest of a,, a r j s a, then as n the prevous step, all the letters a, < r j, are at most a r j+ 2 So we can move each of a r j+,, a r to the left untl they are mmedately to the rght of a Then by a successon of moves of type (, we move the bloc a, a r j+,, a r to the rght hand end The strng s now of the form (3, but wth j ncreased by We can contnue ths procedure untl the letters are all n ncreasng order 8 Ths s a result proved by JH Davenport n 935 In a 947 note, (Journal of the London Mathematcs Socety, Volume 22, 947, pages 00-0, he remars that the result was also proved by Cauchy n 83 Let m A and n B The proof s by nducton on n If n, then B {b}, say, and A + B A + b A A + B because the map x x + b s a bjecton of F p If n 2, wrte B {b, b 2 } Suppose that A+B < m+2 Then A+B m, and so A + B m, because A + B A + b, whch has m elements So A + B A + b, and smlarly A + B A + b 2 Thus A + b A + b 2 Hence A A + c, where c b 2 b 0 But then A (A+c+c A+2c, and A (A+2c+c A+3c, etc So for 0,, p, A A + c Pc any a 0 A Then the p elements a 0 + c, 0 p are dstnct and all n A So A F p So A + b F p too, and snce A + B A + b, we have A + B F p, so that A + B p Now suppose that n > 2, and that the result has been proved for any subsets A and B wth B < n Now let A, B F q, wth A m and B n Form C A + B, and let l C If l p, we are done, and so suppose that l < p Let b, b B be dstnct, and apply the prevous paragraph to C and {b, b } We see that C + {b, b } has at least l + elements Snce C + b and C + b both have only l elements, we see that C + b s not contaned n C + b So there s an element d C + b such that d C + b So d b C, but d b C Fx ths d, and order the elements of B so that B {b,, b n }, wth d b c C for,, r and d b j C for j r +,, n Here 0 < r < n because the above b s one of the b s, r, and the above b s one of the b j s, j r + Now f j r +, and f a A, then a + b j cannot equal any c, r For otherwse a + b j c d b, so that a + b d b j C,
contrary to the defnton of C So f we form B {b r+,, b n }, then A + B C \ {c,, c r } Let l A + B Then the last ncluson shows that l l r On the other hand, B n r < n, and so by the nducton hypothess we have l m + (n r Hence from whch we see that l m + n 9 For N {0,, }, let l r l m + (n r, P q ( q q + q The frst few P s are P 0 0, P, P 2 q q α, P 3 α 2 + and P 4 α 3 + 2α These are certanly polynomals n α wth coeffcents n N A routne calculaton shows that for, 2,, αp + P P + So we can prove that P s always a polynomal n α wth coeffcents n N by nducton: the asserton s true for 0, Assume that n and that P 0,, P n are all polynomals n α wth coeffcents n N Then P n+ αp n + P n s also a polynomal n α wth coeffcents n N We can exhbt the polynomals explctly We clam that P n n 2 0 ( n α n 2 ( We chec that by nducton too If n 0, then the rght hand sde s zero (snce 0 x s by conventon 0, for any x, as s P 0 If n, then the rght hand sde of ( s 0 ( α 2 0 because ( 0 0 So ( holds for n too Suppose that ( holds for n 0,,, where Then P + αp + P 2 ( α α 2 + 0 2 0 2 0 ( ( α 2 + α 2 + 2 2 0 2 2 0 ( 2 ( 2 2 ( α 2 2 α 2 2 α 2 (2
We have used 2 2 2 here, and n the last equaton replaced by + If ( /2, then the coeffcents of α 2 n the two sums n (2 add to gve ( ( ( + If ( 0, then there s a coeffcent of α 2 only n the frst sum n (2, and t s ( 0 0 0 If 2 2, e, f s odd, then we have shown that 2 ( P + α 2, (3 0 whch s (, wth n + If 2m s even, then for m 2 there s a coeffcent of α 2 n (2 only n the second sum there, and t s ( ( 2m m m ( m m ( m Hence the coeffcent of α 2 n (2 s ( for all between 0 and m (+ 2, and so (3 holds n ths case too Ths completes the nducton proof that ( holds for all n 0 We frst show that f x, y X satsfy x 2 x and y 2 y, then xy yx must hold To see ths, let a y(xy x Then and a 2 y(xy xy(xy x y(xy x a, (xya(xy xyy(xy xxy xy(xy xy xy, a(xya y(xy xxyy(xy x y(xy xy(xy x y(xy x a So a (xy by unqueness But a aaa, so agan by unqueness, a a (xy xy Thus (xy 2 a 2 xy Smlarly, (yx 2 yx, so (xy(yx(xy xyxy xy, and smlarly (yx(xy(yx yx So agan by unqueness, yx (xy xy Now let x, y X, and note that (uu 2 uu and (u u 2 u u for all u X So usng the last paragraph, and (xy(y x (xy x(yy (x xy x(x x(yy y xy, (y x (xy(y x y (x x(yy x y (yy (x xx yx, so that, fnally, agan by unqueness, (xy y x m 3