Solutions to Problem Set 7

Transcription

1 8.78 Solutios to Problem Set 7. If the umber is i S, we re doe sice it s relatively rime to everythig. So suose S. Break u the remaiig elemets ito airs {, }, {4, 5},..., {, + }. By the Pigeohole Pricile, sice S cotais + elemets, it has to have two elemets from some air. These two umbers are cosecutive, ad thus corime. If oly elemets are chose, the result does t hold, because we ca choose the eve elemets.. Let S = {,..., + 9} be the set of 0 cosecutive ositive itegers. If some rime divides two elemets of the set, the it divides the differece of them, so it divides some atural umber betwee ad 9 (iclusive). So the oly ossibilities for are,, 5, 7. Now we ll call x S bad if it s ot corime to everythig else i S, ad good if it is. The strategy is to show that there are at most 9 bad elemets of S, so there must be at least oe good elemet, usig a iclusio-exclusio argumet. By the above reasoig, if x is bad, it has to be divisible by,, 5, or 7. Now there are exactly five elemets of S divisible by, so we ut these i the bad set. There are either four or three elemets of S divisible by (deedig o whether is divisible by or ot). But corresodigly, either two or oe of these elemets will be divisible by 6, ad hece already i the bad set. So we add two ew elemets to the bad set. The there are exactly two elemets of S divisible by 5, but oe of them is divisible by 0, so we ca oly add oe extra bad elemet. Fially, there s exactly oe elemet of S divisible by 7, ad it might or might ot be i the bad set already. This leaves us with at most ie elemets i the bad set. Therefore, there exists a good elemet of S.. The total umber of ossible ways to had back the coats is!. Now, the umber of ways to had back the coats so that erso i defiitely gets back their ow coat is ( )!. The umber of ways to had back the coats so that ersos i ad j both get back their coat is ( )!, ad so o. So by iclusio-exclusio, the total umber of ways o oe gets back their ow coat is! ( )! + ( )! ( )! + + () ( )!, which ca be re-writte as! () 0!! +!! + +.! Note that if is very large, this exressio is aroximately!/e. So the robability that o oe receives their coat back is aroximately /e 6.79%. αβ, 4. Oe ca try to rove this relatio by iductio, but it s easier usig the exlicit formula F = α β

2 where α > β are the roots of x x = 0. Notig that αβ =, we have ( α m β m ) ( α β ) ( α m β m ) ( α + β + ) F m F + F m F + = + α β α β α β α β [ = α m+ + β m+ α m β β m α (α β) +α m++ + β m++ α m β + β m α + ] [ = α m++ + β m++ + α m+ + β m+ ] (α β) [ = α m++ + β m++ α m+ β β m+ ] α (α β) = (α m+ α (α β) β m+ )( β) = F m+. Next, we eed to show F m F if m. Let s do this by iductio o k, where = mk. For k = 0 this is clear sice F 0 = 0 is divisible by F m. Now suose the iductive hyothesis is true for k. I the exasio F mk = F m(k)+m = F m(k )Fm + F m(k) F m+, the terms F m ad F m(k ) are both divisible by F, so F is divisible by F, comletig the iductio. m mk m 5. (a) We wat to show that r() = +r()+ +r(). If m = the the decomositio is just =. If m =, the the umber of decomositios is the umber of ways to choose m,..., m k such that = m + +m k, which is r(). Similarly, if m =, there are r() decomositios, ad so o. So r() = + r() + + r( ). Now sice r( ) = + r() + + r( ), we see that r() = ( + r() + + r( )) + r( ) = r( ). By iductio o, with the base case r() =, we must have r() =. (b) Note that for = m + + m k, we must have k. Now cosider ebbles i a row, betwee which there are saces. For each sace we ca either choose to lace a bar there or leave a emty sace. Each such set of choices bijectively corresods to a decomositio of. It follows that there are exactly choices. 6. Let f() be the umber of odd decomositios. The, as i art (a) of the revious roblem, { f( ) + f( ) + + f() if is eve f() = f( ) + f( ) + + f() if is odd. The recurrece f() = f()+f() follows immediately. Sice f() = = F ad f() = = F, we must have f() = F for all. 7. Let f() be the umber of all such sequeces, ad let g() be the umber of such sequeces which start with 0. The, by symmetry, g() is also equal to the umber of such sequeces startig with. Whe a sequece starts with, there are o further restrictios. So by cosiderig the first elemet of the sequece, we get the recurrece f() = g() + g() + f( ). By cosiderig the secod elemet of the sequece whe the first elemet is 0, we get Substitutig from the first equatio, g() = g( ) + f( ). f() f( ) f( ) + f( ) = + f( ),

3 which whe rearraged becomes f() f( ) f( ) = 0. The characteristic olyomial has roots ±, so f() = A( + ) + B( ). We ca easily calculate f() =, f() = 7 to solve for A ad B, obtaiig f() = ( + ) + + ( ) +. Sice < < 0, f() must be the iteger closest to ( + ) Usig the exlicit formula for F, (( + ) ( ) ) 5 5 F = 5 [ ] = Reducig mod ad otig that ( ),..., are all divisible by, we get 5 F 5 (mod ), usig Fermat s Little Theorem ad Euler s criterio. By quadratic recirocity, ( 5 ) = ( 5 ), so F ( 5 ) (mod ). Therefore, { (mod ) if ± (mod 5) F (mod ) if ± (mod 5). Now let s comute [ ] F + = Note that ( + ) = ( + ) (mod ), but ( + ),..., ( + ) are all divisible by. Oe way to see this is to use the rule ( ar r ) + + a 0 a r a 0 (mod ). b r r + + b 0 b Now, usig Fermat s Little Theorem, r b 0 F + ( + 5 ) (mod ). Whe ± (mod 5), ( 5 ) = ad thus F + (mod ). Whe ± (mod 5), ( 5 ) = ad thus F + 0 (mod ). Fially, if ± (mod 5), the F = F + F 0 (mod ), so by Problem 4, Therefore, is a eriod. F + = F F + F F F 0 + F F (mod ). 9. (a) The subset { +,..., } has size ad roerty P. Now if S has size + the cosider the odd art of every elemet of S (if x = k y with y odd, the y is the odd art of x). There are ossible odd arts (amely,,..., ) ad + itegers i S. Therefore, two elemets must have the same odd art. So we have x, x S with x = k y ad x = l y. Sice either k < l or k > l, oe of x, x must divide the other.

4 (b) The same roof shows that o subset of + elemets ca have roerty P. As for a subset of elemets with roerty P, the subset {, +,..., } works. (c) As i art (a), we write each elemet i the form x = k y, where y is relatively rime to. Now there are ( + )/6 multiles of i the set {,, 5,..., }, so there are ( + )/ ossible choices for y, settig a uer boud for the size of S. To show this boud is attaiable, we form S by omittig elemets,, 5,..., ( + )/. Sice ( + ) ( ) + + >, o two elemets of S ca divide each other. 0. (a) Cosider ay arethesizatio of x 0 x +. The left-most symbol i the exressio is either ( or x 0. If it s x 0 the the exressio is x 0 (some arethesizatio of the roduct x x + ), ad the umber of ways this ca occur is C. If the left-most symbol is ( the the ) which airs with it falls after x i for some i, ad we must have (some arethesizatio of x 0 x i ) (some arethesizatio of x i+ x + ). The umber of ways that this ca occur is C i C i. So C + = C i C i. (b) The coefficiet of z i zc(z) is the coefficiet of z i C(z), which is equal to C 0 C + C C + + C C 0 = C. To match u whe = 0, we add. So C(z) = + zc(z). (c) Solvig the quadratic equatio zc(z) C(z) + = 0, we get C(z) = ± 4z. z i=0 Now C(z) is a olyomial i z, so the mius sig must be take i order to cacel out the costat term i the umerator. Now the coefficiet of z i C(z) is half the coefficiet of z + i ( 4z) / : [z / ] C(z) = ( ( + 4) + ) = ( 4) ( + )! + () + 5 ( ) = + ( ) ( + )! = ( + )!( 4 6 ) = ()! ( + )!! =. + 4

5 MIT OeCourseWare htt://ocw.mit.edu 8.78 Theory of Numbers Srig 0 For iformatio about citig these materials or our Terms of Use, visit: htt://ocw.mit.edu/terms.