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1 22 MATHEMATICAL MAYHEM Mathematcal Mayhem bega 988 as a Mathematcal Joual fo ad by Hgh School ad Uvesty Studets. It cotues, wth the same emhass, as a tegal at of Cux Mathematcoum wth Mathematcal Mayhem. All mateal teded fo cluso ths secto should be set to the Mayhem Edto, Nao Sato, Deatmet of Mathematcs, Yale Uvesty, PO Box Yale Stato, New Have, CT 06520{8283 USA. The electoc addess s stll mayhem@math.tooto.edu The Assstat Mayhem Edto s Cyus Hsa (Uvesty of Tooto). The est of the sta cossts of Ada Cha (Ue Caada College), Doy Cheug (Uvesty of Wateloo), Jmmy Chu (Eal Hag Secoday School), Davd Savtt (Havad Uvesty) ad Wa Lg Yee (Uvesty of Wateloo). A Combatoal Poof Sheds ad Slces I Issue 2, ths volume, we ssued the followg challege to ou audece. Poblem. Fd a combatoal oof of the followg detty: +, +, 2 (, ) =. 2 Dave Athu, of Tooto, Otao, was the st, ad so fa oly eso to gve a combatoal oof. Hs ze s a coy of \Rddles of the Shx", by Mat Gade; we t hs soluto hee. The oose stll sees a combatoal oof that oves the detty oe ste, ad so a addtoal ze wll be oeed fo such a soluto. Soluto by Dave Athu. Lemma. Let us cosde the umbe of ways of choosg two dstct sets, A ad B, each, wth elemets fom a set of +, elemets. +, Theeae 2, ways to choose the elemets that wll belog to the 2 uo, ad thee ae ways to choose the elemets of these that wll belog to A. Theefoe, the umbe of ways s +, 2. 2

2 222, ways to choose A, ad,, +, Howeve, we also ote thee ae ways to choose B fom the emag elemets. It follows that +, 2 +,, =. 2 Lemma 2. Let us cosde the umbe of ways of choosg two dstct sets, C ad D, such that C has elemet ad D has elemets, fom a set of elemets.,, Thee aeways to choose C st, ad ways, to choose D fom, the emag elemets, so the umbe of ways s,. Also, thee ae ways to choose D st, ad, ways to choose C fom the emag elemets, so t follows that, (, ) =. Theefoe, by lemmas ad 2, +, 2 +, = 2 Eatum, +, =(,) O age 69, Issue 3 of ths Volume, oblem 6 of the Qualfyg Roud of the 990 Swedsh Mathematcal Olymad, the dmesos of ectagle ABCD ae descbed as 3000 metes by 500 metes. Ths s a tyo: the dmesos should be 300 metes by 500 metes. Thas to Jm Totte fo the coecto. The Mayhem Poblems edtos ae: Ada Cha Doy Cheug Davd Savtt Mayhem Poblems Mayhem Hgh School Poblems Edto, Mayhem Advaced Poblems Edto, Mayhem Challege Boad Poblems Edto. Note that all coesodece should be set to the aoate edto see the elevat secto. I ths ssue, you wll d oly solutos the ext ssue wll featue oly oblems. We wamly welcome oosals fo oblems ad solutos. Wth the ew schedule of eght ssues e yea, we equest that solutos fom the last ssue be submtted tme fo ssue 4 of

3 223 Hgh School Solutos Edto: Ada Cha, 229 Old Yoge Steet, Tooto, Otao, Caada. M2P R5 H228. Vefy that the followg thee equaltes hold fo ostve eals x, y, ad z: () x(x, y)(x, z) +y(y,x)(y, z) +z(z,x)(z, y) 0. (Ths s ow as Schu's Iequalty.) () x 4 + y 4 + z 4 + xyz(x + y + z) 2(x 2 y 2 + y 2 z 2 + z 2 x 2 ). () 9xyz +4(xy + yz + zx), whee x + y + z =. (Ca you deve a geous method that allows you to solve the oblem wthout havg to ove all thee equaltes dectly?) Addtoal soluto by Muay S. Klam, Uvesty of Albeta, Edmoto, Albeta. () As dcated, ths s the secal case =of Schu's Iequalty: x (x, y)(x, z) +y (y,z)(y, x) +z (z,x)(z, y) 0 fo eal. Fo a smle oof, let x y z wthout loss of geealty. The fo 0, x (x, y)(x, z) y (y, z)(x, y) ad z (x, z)(y, z) 0. Fo 0, z (x, z)(y, z) y (y, z)(x, y) ad x (x, y)(x, z) 0. Thee s equalty f ad oly f x = y = z. It also follows that f s a eve tege, the x, y ad z eed ot be ostve. () Sce as ow 2(y 2 z 2 + z 2 x 2 + x 2 y 2 ), (x 4 + y 4 + z 4 ) = (x + y + z)(y + z, x)(z + x, y)(x + y, z) (elated to Heo's fomula fo the aea of a tagle), the equaltyeduces to xyz (y + z, x)(z + x, y)(x + y, z). () I tems of the elemetay symmetc fuctos T = x + y + z, T 2 = yz + zx + xy, ad T 3 = xyz, () becomes T 3 +9T 34T T 2, whch s the same as ().

4 224 () I homogeeous fom, the equalty s equvalet to 9xyz +(x+y+z) 3 4(x + y + z)(yz + zx + xy), whch s the same as (). Fo a geealzato of () to (ux + vy + wz)(vx + wy + uz)(wx + uy + vz) (y + z, x)(z + x, y)(x + y, z), whee u + v + w =ad 0 u; v; w, see []. Refeece. Klam, M.S., Iequaltes fo a tagle assocated wth gve tagles, Publ. Electotech. Fa. Se. Mat. Fz. Uv. Beogad, No. 330 (970),. 3{7. H237. The lettes of the wod MATHEMATICAL ae aaged at adom. What s the obablty that the esultg aagemet cotas o adjacet A's? Soluto by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. Fst, the total umbe of aagemets s 2! 3!2!2!. Now, to cout the umbe of aagemets wth o adjacet A's, we st aage the othe 9 lettes, fo a total of 9! 2!2! ways. Fo each such aagemet, we choose ay 3 of the 0 \saces" that ae betwee cosecutve, lettes, cludg those o the eds. Ths ca be 0 doe a total of 3 ways. Now, we set the 3 A's to the 3 saces that we have chose. Ths coesods to a uque aagemet wth o adjacet A's. The total umbe s the 9! 0. 2!2! 3 The equed obablty s the equal to 9!, 0 2!2! 3 2! 3!2!2! = 6.

5 225 H238. Johy s dazed ad cofused. Statg at A(0; 0) the Catesa gd, he moves ut to the ght, the uts u, 2 uts left, 3 uts dow, 4 uts ght, 5 uts u, ad cotues the same atte detely. If s a ostve umbe less tha, he wll be aoachg a ot B(x; y). Show that the legth of the le segmet AB s geate tha 7 0. Soluto by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. Let P =(x ;y )deote the ot whee Johy s at afte moves, = 0,, 2, :::. So, P 0 = A = (0;0), P = (;0), P 2 = (;), P 3 =(, 2 ;),etc. A smle atte eveals tself, such that fo all m 2, x 2m, = x 2m =, 2 + 4,+(,) m, 2m,2, y 2m = y 2m+ =, 3 + 5,+(,) m, 2m,. Hece, x = y =, (, 2 ) =,(, 2 ) = + 2, + 2. So, Ad sce <, AB 2 = x 2 + y 2 = + 2. AB = > 5 2 = > 7 0. H239. Fd all as of teges (x; y) whch satsfy the equato y 2 (x 2 +)+x 2 (y 2 + 6) = 448. Soluto. We have y 2 (x 2 +)+x 2 (y 2 + 6) = 448 =) 2x 2 y 2 +6x 2 +y 2,448 = 0 =) 2x 2 (y 2 +8)+y 2 +8 = (2x 2 + )(y 2 +8) = 456 = Sce x ad y ae teges, both 2x 2 + ad y 2 +8 ae ostve teges. Sce 2x 2 +s odd, t must equal oe of the odd factos of456, amely, 3, 9, ad 57. Checg each of these cases, we d that x =0,, ad 3 ae the oly solutos. If x =0, the y = 456, fo whch thee s o soluto. If x =, the y = 52, fom whch we obta y = 2.

6 226 If x = 3, the y 2 +8=24,fom whch we obta y = 4. Theefoe, thee ae eght solutos, amely (; 2) ad (3; 4). Also solved by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. H240. Poosed by Alexade Tchtcheo, Booeld Hgh School, Ottawa, ON. A Pythagoea tle (a; b; c) s a tle of teges satsfyg the equato a 2 + b 2 = c 2. We say that such a tle s mtve f gcd(a; b; c) =. Let be a odd tege wth exactly me dvsos. Show that thee exst exactly 2, mtve Pythagoea tles whee s the st elemet of the tle. Fo examle, f =5, the (5; 8; 7) ad (5; 2; 3) ae the mtve Pythagoea tles wth st elemet 5. Soluto. We may wte uquely the fom = e e2 2 e ; whee, 2, :::, ae dstct odd mes, ad e, e 2, :::,e ae ostve teges. We see all mtve Pythagoea tles (; b; c). Fo such a tle, 2 + b 2 = c 2,oc 2,b 2 =(c+b)(c,b) = 2. Suose dvdes both c + b ad c, b fo some. The dvdes (c + b), (c, b) =2b, ad sce s odd, t follows that dvdes b. Smlaly, dvdes (c + b)+(c,b)=2c, so dvdes c. The (; b; c) fals to be a mtve Pythagoea tle, sce dvdes all thee umbes, so fo each, dvdes at most oe of c + b ad c, b. Ths mles that the factozato 2 =(c+b)(c, b), fo each, all the factos of must esde c + b o c, b. I othe wods, fo each, we ca mae oe of two choces of whee to lace all the factos of, fo a total of 2 factozatos. Howeve, half of them must be dscaded, sce c + b must be the geate umbe, ad c, b the lesse. (We caot have c + b ad c, b equal, sce they have deet me factos.) Each of the othe half, howeve, does lead to a uque soluto: If 2 = xy, whee x>yad x ad y ae elatvely me, the b =(x,y)=2ad c =(x+y)=2. Hece, thee ae 2, such Pythagoea tles. Also solved by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. Advaced Solutos Edto: Doy Cheug, c/o Coad Gebel College, Uvesty of Wateloo, Wateloo, Otao, Caada. N2L 3G6 <dccheug@uwateloo.ca> A22. Let A ad B be eal matces such that A 2 + B 2 = AB. Pove that f AB, BA s a vetble matx, the s dvsble by 3.

7 227 (Iteatoal Cometto Mathematcs fo Uvesty Studets) Soluto. Let S = A +!B, whee! s a mtve cube oot of uty. The we have that SS = (A +!B)(A +!B) = (A +!B)(A +!B) = A 2 +!BA +!AB + B 2 = AB +!BA +!AB =!(BA, AB), sce! + =,!. Also, det(ss) = det S det S s a eal umbe ad det!(ba, AB) =! det(ba, AB) 6= 0,so! must be a eal umbe. Hece, s dvsble by 3. A23. Show that the umbe of ostve tege solutos to the equato a + b + c + d =98, whee 0 <a<b<c<d, s equal to the umbe of ostve tege solutos to the equato +2q +3 +4s =98, whee 0 <,q,,s. I.Soluto by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. Suose that a, b, c, ad d ae ostve teges such that a < b < c < dad satsfy a + b + c + d =98. The lettg = d,c, q = c,b, = b,a, ad s = a, we see that, q,, ad s ae ostve teges satsfyg +2q+3+4s =(d,c)+2(c,b)+3(b,a)+4a = a+b+c+d = 98. Covesely, suose that, q,, ad s ae ostve teges satsfyg +2q+3+4s=98. The lettg a = s, b = + s, c = q + + s, ad d = + q + + s, we see that a, b, c ad d ae ostve teges satsfyg a<b<c<d, ad a + b + c + d = +2q+3+4s = 98. Ths establshes a bjecto betwee the solutos (a; b; c; d) ad (; q; ; s). II. Soluto. Fst cout the umbe of ways of dstbutg 98 detcal mables to 4 dstct boxes, wth the st box cotag the least umbe of mables, the secod box cotag the secod least umbe, ad so o. Ths value s equal to the umbe of solutos to the st equato wth the aoate codtos. Aothe way of coutg ths s to dstbute the same umbe h to each of the 4 boxes. The dstbute g to each of the secod, thd ad fouth boxes. Ths esues that the st box has the least. Dstbute f to each of boxes thee ad fou. Ths esues that the secod box s the secod least ad so o. The umbe of ways to do ths s equal to the umbe of solutos to the secod equato. A24. Show that ay atoal umbe ca be wtte as the sum of a te umbe of dstct ut factos. A ut facto s of the fom =, whee s a tege.

8 228 Soluto. We solve the oblem by ste-clmbg though seveal cases. Case I: Postve atoals [0; ]. Let be a ostve atoal umbe [0; ]. The we show that ca be wtte as a sum of dstct ut factos by ovdg a algothm that exlctly ds the ut factos, amely the geedy algothm. Let = a=b, educed tems. Let = db=ae. I othe wods, s the tege b=a ouded u, so b=a < b=a +. The = s the lagest ut facto whch s less tha o equal to a=b, ad whe we subtact = fom a=b, we obta 0 = a b, = a, b b. Sce b=a < b=a +,b a<b+a,o0a, b<a. So, the umeato 0 s less tha the umeato. If 0 s educble, the we educe, ad the umeato deceases stll. We ow subtact the lagest ut facto fom 0, ad so o. Sce the umeato deceases by at least each ste, the algothm must sto at some ot, fact afte at most a, stes. The s the sum of the ut factos oduced by the algothm. Fo examle, fo =6=7, we have that 6, 7 2 = 5, 5, =, 42 so that 6 7 = Case II: Postve atoals geate tha. P Let be a ostve atoal geate tha. Sce the hamoc sees = = dveges, thee exsts a ostve tege such that X = < + X If = P = =, the we have aexesso of as a sum of dstct ut factos, so assume that > P = =. Cosde the atoal s =, X = =., so that 0 < s < +. The s s a ostveatoal [0; ],sobycase I, s s the sum of dstct ut factos s = mx j= j.

9 229 Sce s < =( +), each j s at least +(othewse j =) = j = =) s =, cotadcto), so that = X = + mx j= j s a exesso of as a sum of dstct ut factos. Case III: Negatve atoals ad 0. If < 0, the,, by Cases I ad II, ca be wtte as the sum of dstct ut factos. Negate each tem to get such a sum fo. Fally, 0 ca be wtte as 0 =, 2, 3, 6. A25. Fo a xed tege 2, deteme the maxmum value of ++, whee, :::, ae ostve teges wth (Polsh Mathematcal Olymad) Soluto by Edwad T.H. Wag, Wlfd Laue Uvesty, Wateloo, Otao. Let T = P = 3 ad S = P =. We clam that 6 S +, 7 whee bxc deotes, as usual, the geatest tege less tha o equal to x. Ths maxmum s attaed f ad oly f b6=7c of the equal 2 ad the est equal. Fst ote that f 2 fo all, the T 8, acotadcto. Hece =fo at least oe, say =. If j 3 fo some j >, the we cosde T 0 obtaed fom T as follows: Relace ad j by 0 = 2ad 0 j = j, esectvely, ad leave all the othe uchaged. The clealy the value of S s uchaged. O the othe had, T 0 T, whch meas that ( j, ) j +.Rewtg ths, we obta ( j + )( j, 2) 0. Ths s tue as log as j 3. Hece T 0 7, ad to obta the maxmum value of S, we may assume, wthout loss of geealty, that =o 2 fo all. Suose amog all the, that thee aem2's ad, m 's. The T = 8m +(,m) 7, whch mles that m b6=7c. Clealy, the maxmum value of S s attaed whe m = b6=7c, whch case we obta S =2b6=7c +,b6=7c = + b6=7c. A26. Gve a cotuous fucto f : R! R satsfyg the codtos: f (000) = 999, f (x) f (f (x)) = fo all x 2 R. Deteme f (500). (Polsh Mathematcal Olymad)

10 230 Soluto. By the secod codto, f (000)f (f(000)) =, so we have that 999f (999) = o f (999) = =999. Sce f s a cotuous fucto, by the Itemedate Value Theoem, thee exsts aa2[999; 000] such that f (a) = 500. The f (a)f (f(a)) =, gvg 500f (500) =, sof(500) = =500. I fact, f (x) = =x fo all x 2 [=999; 999]. To comlete the fucto, fo ay x outsde ths age, set f (x) to ay value, wth the teval [=999; 999]. The fo ay x, =999 f (x) 999, ad so f (f (x)) = =f (x). Note: Because f (000) = 999 6= =000, f (x) ca eve equal 000. Challege Boad Solutos Edto: Davd Savtt, Deatmet of Mathematcs, Havad Uvesty, Oxfod Steet, Cambdge, MA, USA 0238 <dsavtt@math.havad.edu> C77. Let F deote the th Fboacc umbe, wth F 0 = ad F =. (The F 2 =2,F 3 =3,F 4 =5,etc.) (a) Pove that each ostve tege s uquely exessble the fom F a ++F a, whee the subscts fom a stctly ceasg sequece of ostve teges o a of whch ae cosecutve. (b) Let = 2 ( + 5), ad fo ay ostve tege, let f () equal the tege eaest to.pove that f = F a ++F a s the exesso fo fom at (a) ad f a 2 6=3, the f () =F a+ + +F a+. (c) Keeg the otato fom at (b), f a 2 = 3(so that a = ), t s ot always tue that the fomula f () =F a+ + + F a+ holds. Fo examle, f =4=F 3 +F =+3, the the closest tege to =6:47 ::: s 6, ot F 2 + F 4 =2+5=7. Fotuately, the cases whee the fomula fals, we ca coect the oblem by settg a =0 stead of a = : Fo examle, 4 = F 0 + F 3 = + 3as well, ad deed 6=F +F 4 = +5. Deteme fo whch sequeces of a ths coecto s ecessay. Soluto. (a) Suose we ow that evey ostve tege less tha F s uquely exessble the desed fom. (The base case = s vacuous.) If N satses F N < F +, the by ou ductve assumto, N, F s exessble the desed fom, say N, F = F a + + F a. Sce

11 23 N, F <F +, F = F,, we must have that a, 2, ad so the exesso N = F a + +F a +F s of the desed fom. It emas to show that ths s the uque exesso fo N as such a sum; obseve, by ducto, that t suces to show that ay such sum fo N must clude F. Gve a exesso N = F a + :::+F a of the ecessay fom, f we assume that a <, the we ow that a s at most,, soa, s at most, 3, ad so o. Thus, N s at most f s eve ad F + F 3 + F 5 + +F, F 2 + F 4 + +F, f s odd. Howeve, t s a staghtfowad ducto agumet to ove that fact fo ay ostve tege m, ad so we obta a cotadcto. F + F 3 + F 5 + +F 2m, = F 2m, F 2 + F 4 + +F 2m = F 2m+,, (b) Sce f () s the uque tege N such that jn, j < =2, we eed oly show that jf a+ + +F a+, (F a + +F a )j < 2. Recall that F = +, + 5, whee = 2 (, 5). Hece, we wat to ove that X, ( a+2, a+2 ), ( a+, a+ ) 5 <. 2 = Usg the fact that =,, ths becomes the same as showg that X =0 ( a+2 + a ) < 5 2, ad sce 2 +=, 5,weae futhe educed to showg that X a+ <. 2 =0

12 232 Let P P be the sum of the ostve tems the sum a+, ad let N be the sum of the egatve tems. The P s at most ad N s at least X = X = 2+ = 2 = 2, 2 < 0:62, 3 way that jp + Nj =2 s f P 2 +jnj., 2 >,0:39. I atcula, the oly If a 2 6=3, the ethe 2 o 4 s ot the sum fo P. Sce 2 > 0:38 ad 4 > 0:4, t follows that P < =2, sop<=2+jnj ad the desed fomula holds. (c) Keeg ou otato fom (b), f P + N > =2, the a =ad a 2 =3. Settg a 0 =0ad a0 = a fo >, we d that X =0 a0 + = (P, 2 )+(N +) = jp + N,j < 2, so the suggested coecto does deed wo whe ecessay. We would le to decde ecsely whe ths coecto s ecessay; that s, whe S = X =0 a0 + >, 2. Let us eta the assumto that a = (so that the deto of the a 0 maes sese) but do the assumto that a 2 =3. Ths wll tu out to be moe coveet, the ed. Obsevg, by dect calculato, that X =0 smallest (o-egatve) tege such that a =3(+). 3+ =,, let j be the 2 (If a 0 +2 =3(+)fo all, the ut j =,.) The S + 2 = j+ + X =j+2 a0 + = 2 3j+3 + X =j+2 a0 +. We wll attemt to show ( what wll amout to a clumsy vecato) that S +=2has the same sg as 2 3j+3, ad so the coecto s ecessay ecsely whe j s odd. Remembeg the deto of j ad the fact that the a ae ocosecutve, usg a 0 j+ = 3j,webea to thee cases: a0 j+2 = 3j +2, a 0 j+2 =3j+4,oa0 j+2 3j +5.

13 233 To beg wth, the st case, X =j+3 a0 + < as desed. Smlaly, the secod case, X =j+3 a0 + < < Fally, the thd case, X =j+2 ad we ae doe. < 3j+5 X 2 = 3j+4 = j+3 = 2 3j+3 + a0 j+2 +, 3j+7 X 2 = 3j+6 =0 2 3j+3 + 3j+5 = 2 3j+3 + a0 j+2 +. a0 + < 3j+6 X 2 = 3j+5 < 2 3j+3, =0 Thus, we have show: If a =, the f j s the smallest o-egatve tege such that a +2 6=3(+), the the coecto s ecessay fad oly f j s odd. Rema: I the soluto of at (b), we eve fully used the fact that we wee usg the eesetato fom at (a). I atcula, the oof of (b) actually showed that f = F a + + F a ad the a ae dstct ostve teges, the as log as ad 3 ae ot both amog the a,weca coclude that f () =F a+ + +F a+. Ufotuately, as should ot be a suse, ot evey tege ca be exessed ths way, =4beg the smallest examle. Poblem of the Moth Jmmy Chu, studet, Eal Hag S.S. Poblem. A ectagula we ac, P QRS, holds ve ows of detcal bottles (Fgue ). The bottom ow cotas eough oom fo thee bottles (A, B, ad C) but ot eough oom fo a fouth bottle. The secod ow, cosstg of just two bottles (D ad E), holds B lace somewhee betwee A ad C, ad ushes A ad C to the sdes of the ac. The thd ow,

14 234 cosstg of thee bottles (F, G, ad H), les o to of those two bottles, ad F ad H est agast the sdes of the ac. The fouth laye holds two bottles (I ad J) ad the fth laye cotas thee bottles (K, L ad M). Pove that the fth ow s efectly hozotal egadless of how A, B ad C ae ostoed. P K L M S P K L M S F D I G E J H F I D G E J H Q A B C R Q A B C Fgue. Fgue 2. R Soluto. All the bottles ae detcal, ad so F ad K ae the same dstace away fom the wall (Fgue 2). Sce FK s vetcal, we oly eed to show that \FKL =90. The dstace betwee the cetes of touchg bottles s costat; t s the damete of a bottle. Theefoe, IF, IK, ad IL ae all equal, ad hece I s the ccumcete of tagle FKL.Iode fo \FKLto be a ght agle, the fom oetes of ght agle tagles, the ccumcete I s also the md-ot of FL. Hece, we wll show that I s the md-ot of FL. Note that the fou quadlateals GDF I, GILJ, GJHE, GEBD ae all homb (they have sde legth of a bottle damete). So,!,! FI = BE ad,!,,! IL = EH. Futhemoe, sce EB, EC, ad EH ae all equal, E s the ccumcete of tagle BCH. Howeve, we ow that BCH s a ght tagle. Hece, E s the md-ot of BH. Thus, I s the md-ot of FL, ad t follows that \FKL =90. Smlaly, \HML s a ght agle. Theefoe, the to ow s efectly hozotal, QED.

15 235 Fou Ways to Cout Jmmy Chu studet, Eal Hag Secoday School Poblem. Evaluate +2 2 whee s a ostve tege Soluto. Let the gve sum be equal to S. Now, S = = ,,2,3 0 = +(,) +(,2) +(,3) Addg the st ad thd equatos ad dvdg by 2, weobta S = = 2 2 = 2,.,. Soluto 2. We clam that = 2,. We wll ove ths by mathematcal ducto. The base case =s tval. Now we assume, fo some =, that = 2,. 2 3 Coyght c 999 Caada Mathematcal Socety

16 236 The, (+) = (+), = (2+) = = 2 +2(2, ) = (+ )2. Hece, the clam s tue fo = +, ad by the cle of mathematcal ducto, fo all ostve teges. The so that Soluto 3. Let But f (x) = X =0 f 0 (x) = by the Bomal Theoem. The x = + f 0 () = X = X = X = x,, f (x) = ( + x) f 0 (x) = ( + x),,. x. so that Hece, f 0 () = 2,. X = = 2,.

17 237 Soluto 4. Cosde a set of eole. We wsh to cout the umbe of deet teams that ca be fomed, wth the codto that thee s oe ad oly oe leade. Oe way s to d the total umbe of team membes st, ad the select a leade fom those chose. Suose you choose membes. (Note that,.) Thee ae such subsets. I each of these subsets, thee aeossble leades., Hece, the total umbe of teams that ca be fomed wth membes s. Theefoe, the total umbe of teams wth ay umbe of membes s meely the sum X. = Aothe way s to choose the leade st, ad the est of the membes aftewads. The leade ca be chose ways. The membes ca be chose out of the othe, eole ay way, ad thee ae2, ways of dog so. Hece the total umbe of teams s 2,. Thus, X = = 2,. Commet. Ths questo aeas the stagest of laces, ad t s leasat to see fou vey deet, yet equally elegat, solutos.