HMMT February 2016 February 20, 2016

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1 HMMT February 016 February 0, 016 Combnatorcs 1. For postve ntegers n, let S n be the set of ntegers x such that n dstnct lnes, no three concurrent, can dvde a plane nto x regons (for example, S = {3, 4}, because the plane s dvded nto 3 regons f the two lnes are parallel, and 4 regons otherwse). What s the mnmum such that S contans at least 4 elements? Proposed by: Answer: 4 For S 3, ether all three lnes are parallel (4 regons), exactly two are parallel (6 regons), or none are parallel (6 or seven regons, dependng on whether they all meet at one pont), so S 3 = 3. Then, for S 4, ether all lnes are parallel (5 regons), exactly three are parallel (8 regons), there are two sets of parallel pars (9 regons), exactly two are parallel (9 or 10 regons), or none are parallel (8, 9, 10, or 11 regons), so S 4 = 4.. Startng wth an empty strng, we create a strng by repeatedly appendng one of the letters H, M, T wth probabltes 1 4, 1, 1 4, respectvely, untl the letter M appears twce consecutvely. What s the expected value of the length of the resultng strng? Proposed by: Answer: 6 Let E be the expected value of the resultng strng. Startng from the empty strng, We have a 1 chance of not selectng the letter M; from here the length of the resultng strng s 1 + E. We have a 1 4 chance of selectng the letter M followed by a letter other than M, whch gves a strng of length + E. We have a 1 4 chance of selectng M twce, for a strng of length. Thus, E = 1 (1 + E) ( + E) (). Solvng gves E = Fnd the number of ordered pars of ntegers (a, b) such that a, b are dvsors of 70 but ab s not. Proposed by: Casey Fu Answer: 50 Frst consder the case a, b > 0. We have 70 = 4 3 5, so the number of dvsors of 70 s 5 3 = 30. We consder the number of ways to select an ordered par (a, b) such that a, b, ab all dvde 70. Usng the balls and urns method on each of the prme factors, we fnd the number of ways to dstrbute the factors of across a and b s ( ( 6 ), the factors of 3 s 4 ( ), the factors of 5 s 3 ). So the total number of ways to select (a, b) wth a, b, ab all dvdng 70 s = 70. The number of ways to select any (a, b) wth a and b dvdng 70 s = 900, so there are = 630 ways to select a and b such that a, b dvde 70 but ab doesn t. Now, each a, b > 0 corresponds to four solutons (±a, ±b) gvng the fnal answer of 50. (Note that ab 0.) 4. Let R be the rectangle n the Cartesan plane wth vertces at (0, 0), (, 0), (, 1), and (0, 1). R can be dvded nto two unt squares, as shown; the resultng fgure has seven edges.

2 How many subsets of these seven edges form a connected fgure? Proposed by: Joy Zheng Answer: 81 We break ths nto cases. Frst, f the mddle edge s not ncluded, then there are 6 5 = 30 ways to choose two dstnct ponts for the fgure to begn and end at. We could also allow the fgure to nclude all or none of the sx remanng edges, for a total of 3 connected fgures not ncludng the mddle edge. Now let s assume we are ncludng the mddle edge. Of the three edges to the left of the mddle edge, there are 7 possble subsets we can nclude (8 total subsets, but we subtract off the subset consstng of only the edge parallel to the mddle edge snce t s not connected). Smlarly, of the three edges to the rght of the mddle edge, there are 7 possble subsets we can nclude. In total, there are 49 possble connected fgures that nclude the mddle edge. Therefore, there are = 81 possble connected fgures. 5. Let a, b, c, d, e, f be ntegers selected from the set {1,,..., 100}, unformly and at random wth replacement. Set M = a + b + 4c + 8d + 16e + 3f. What s the expected value of the remander when M s dvded by 64? Proposed by: Evan Chen Answer: 63 Consder M n bnary. Assume we start wth M = 0, then add a to M, then add b to M, then add 4c to M, and so on. After the frst addton, the frst bt (defned as the rghtmost bt) of M s toggled wth probablty 1. After the second addton, the second bt of M s toggled wth probablty 1. After the thrd addton, the thrd bt s toggled wth probablty 1, and so on for the remanng three addtons. As such, the sx bts of M are each toggled wth probablty 1 - specfcally, the kth bt s toggled wth probablty 1 at the kth addton, and s never toggled afterwards. Therefore, each resdue from 0 to 63 has probablty 1 64 of occurrng, so they are all equally lkely. The expected value s then just Defne the sequence a 1, a... as follows: a 1 = 1 and for every n, { n f a n 1 = 0 a n = a n 1 1 f a n 1 0 A non-negatve nteger d s sad to be jet-lagged f there are non-negatve ntegers r, s and a postve nteger n such that d = r + s and that a n+r = a n + s. How many ntegers n {1,,..., 016} are jet-lagged? Proposed by: Pakawut Jradlok Let N = n+r, and M = n. Then r = N M, and s = a N a M, and d = r +s = (a N +N) (a M +M). So we are tryng to fnd the number of possble values of (a N + N) (a M + M), subject to N M and a N a M. Dvde the a nto the followng blocks : a 1 = 1, a = 0, a 3 = 1, a 4 = 0, a 5 = 3, a 6 =, a 7 = 1, a 8 = 0, a 9 = 7, a 10 = 6,..., a 16 = 0, and so on. The k th block contans a for k 1 < k. It s easy to see by nducton that a k = 0 and thus a k +1 = k 1 for all k 1. Wthn each block, the value a n + n s constant, and for the kth block (k 1) t equals k. Therefore, d = (a N + N) (a M + M) s the dfference of two powers of, say n m. For any n 1, t s clear there exsts an N such that a N + N = n (consder the n th

3 block). We can guarantee a N a M by settng M = m. Therefore, we are searchng for the number of ntegers between 1 and 016 that can be wrtten as n m wth n m 1. The pars (n, m) wth n > m 1 and n 10 all satsfy 1 n m 016 (45 possbltes). In the case that n = 11, we have that n m 016 so m 3, so m 5 (6 possbltes). There are therefore = 51 jetlagged numbers between 1 and Kelvn the Frog has a par of standard far 8-sded dce (each labelled from 1 to 8). Alex the sketchy Kat also has a par of far 8-sded dce, but whose faces are labelled dfferently (the ntegers on each Alex s dce need not be dstnct). To Alex s dsmay, when both Kelvn and Alex roll ther dce, the probablty that they get any gven sum s equal! Suppose that Alex s two dce have a and b total dots on them, respectvely. Assumng that a b, fnd all possble values of mn{a, b}. Proposed by: Alexander Katz Answer: 4, 8, 3 Ed. note: I m probably horrbly abusng notaton Defne the generatng functon of an event A as the polynomal g(a, x) = p x where p denotes the probablty that occurs durng event A. We note that the generatng s multplcatve;.e. g(a AND B, x) = g(a)g(b) = p q j x +j where q j denotes the probablty that j occurs durng event B. In our case, events A and B are the rollng of the frst and second dce, respectvely, so the generatng functons are the same: and so g(de, x) = 1 8 x x x x x x x x8 g(both dce rolled, x) = g(de, x) = 1 64 (x1 + x + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 ) where the coeffcent of x denotes the probablty of rollng a sum of. We wsh to fnd two alternate dce, C and D, satsfyng the followng condtons: C and D are both 8-sded dce;.e. the sum of the coeffcents of g(c, x) and g(d, x) are both 8 (or g(c, 1) = g(d, 1) = 8). The faces of C and D are all labeled wth a postve nteger;.e. the powers of each term of g(c, x) and g(d, x) are postve nteger (or g(c, 0) = g(d, 0) = 0). The probablty of rollng any gven sum upon rollng C and D s equal to the probablty of rollng any gven sum upon rollng A and B;.e. g(c, x)g(d, x) = g(a, x)g(b, x). Because the dce are far.e. the probablty of rollng any face s 1 8 we can multply g(a, x), g(b, x), g(c, x) and g(d, x) by 8 to get nteger polynomals; as ths does not affect any of the condtons, we can assume g(c, x) and g(d, x) are nteger polynomals multplyng to (x 1 + x x 8 ) (and subject to the other two condtons as well). Snce Z s a UFD (.e. nteger polynomals can be expressed as the product of nteger polynomals n exactly one way, up to order and scalng by a constant), all factors of g(c, x) and g(d, x) must also be factors of x 1 + x x 8. Hence t s useful to factor x 1 + x x 8 = x(x + 1)(x + 1)(x 4 + 1).

4 We thus have g(c, x)g(d, x) = x (x + 1) (x + 1) (x 4 + 1). We know that g(c, 0) = g(d, 0) = 0, so x g(c, x), g(d, x). It remans to dstrbute the remanng term (x + 1) (x + 1) (x 4 + 1) ; we can vew each of these 6 factors as beng assgned to ether C or D. Note that snce g(c, 1) = g(d, 1) = 8, and each of the factors x + 1, x + 1, x evaluates to when x = 1, exactly three factors must be assgned to C and exactly three to D. Fnally, assgnng x + 1, x + 1, and x to C results n the standard de, wth a = b = 8.. Ths gves us the three cases (and ther permutatons): g(c, x) = x(x + 1) (x + 1), g(d, x) = x(x + 1)(x 4 + 1). In ths case we get g(c, x) = x 5 + x 4 + x 3 + x + x and g(d, x) = x 11 + x 9 + x 7 + x 5 + x 3 + x, so the smaller de has faces 5, 4, 4, 3, 3,,, and 1 whch sum to 4. g(c, x) = x(x + 1)(x + 1), g(d, x) = x(x + 1)(x 4 + 1). In ths case we have g(c, x) = x 6 + x 5 + x 4 + x 3 + x + x and g(d, x) = x 10 + x 9 + x 6 + x 5 + x + x, so the smaller de has faces 6, 5, 4, 4, 3, 3, and 1 whch sum to 8. g(c, x) = x(x + 1) (x 4 + 1), g(d, x) = x(x + 1) (x 4 + 1). In ths case we have g(c, x) = x 9 + x 7 + x 5 + x 3 + x and g(d, x) = x 7 + x 6 + x 5 + x 3 + x + x, so the smaller de has faces 7, 6, 6, 5, 3,,, 1 whch sum to 3. Therefore, mn{a, b} s equal to 4, 8, or Let X be the collecton of all functons f : {0, 1,..., 016} {0, 1,..., 016}. Compute the number of functons f X such that ( ( ) ( ) ) max mn max(f(), g()) max mn(f(), g()) = 015. g X Proposed by: Answer: ( ) For each f, g X, we defne d(f, g) := Thus we desre max g X d(f, g) = 015. ( ) ( ) mn max(f(), g()) max mn(f(), g()) Frst, we count the number of functons f X such that g : mn max{f(), g()} 015 and g : mn max{f(), g()} = 0. That means for every value of, ether f() = 0 (then we pck g() = 015) or f() 015 (then we pck g() = 0). So there are A = functons n ths case. Smlarly, the number of functons such that s also B = g : mn max{f(), g()} = 016 and g : mn max{f(), g()} 1 Fnally, the number of functons such that s C = 017. g : mn max{f(), g()} = 016 and g : mn max{f(), g()} = 0 Now A + B C counts the number of functons wth max g X d(f, g) 015 and C counts the number of functons wth max g X d(f, g) 016, so the answer s A + B C = ( ).

5 9. Let V = {1,..., 8}. How many permutatons σ : V V are automorphsms of some tree? (A graph conssts of a some set of vertces and some edges between pars of dstnct vertces. It s connected f every two vertces n t are connected by some path of one or more edges. A tree G on V s a connected graph wth vertex set V and exactly V 1 edges, and an automorphsm of G s a permutaton σ : V V such that vertces, j V are connected by an edge f and only f σ() and σ(j) are.) Proposed by: Mtchell Lee Answer: 301 We decompose nto cycle types of σ. Note that wthn each cycle, all vertces have the same degree; also note that the tree has total degree 14 across ts vertces (by all ts seven edges). For any permutaton that has a 1 n ts cycle type (.e t has a fxed pont), let 1 a 8 be a fxed pont. Consder the tree that conssts of the seven edges from a to the seven other vertces - ths permutaton (wth a as a fxed pont) s an automorphsm of ths tree. For any permutaton that has cycle type + 6, let a and b be the two elements n the -cycle. If the 6-cycle conssts of c, d, e, f, g, h n that order, consder the tree wth edges between a and b, c, e, g and between b and d, f, h. It s easy to see σ s an automorphsm of ths tree. For any permutaton that has cycle type ++4, let a and b be the two elements of the frst two-cycle. Let the other two cycle consst of c and d, and the four cycle be e, f, g, h n that order. Then consder the tree wth edges between a and b, a and c, b and d, a and e, b and f, a and g, b and h. It s easy to see σ s an automorphsm of ths tree. For any permutaton that has cycle type , let a and b be the vertces n the -cycle. One of a and b must be connected to a vertex dstnct from a, b (follows from connectedness), so there must be an edge between a vertex n the -cycle and a vertex n a 3-cycle. Repeatedly applyng σ to ths edge leads to a cycle of length 4 n the tree, whch s mpossble (a tree has no cycles). Therefore, these permutatons cannot be automorphsms of any tree. For any permutaton that has cycle type 3 + 5, smlarly, there must be an edge between a vertex n the 3-cycle and a vertex n the 5-cycle. Repeatedly applyng σ to ths edge once agan leads to a cycle n the tree, whch s not possble. So these permutatons cannot be automorphsms of any tree. The only remanng possble cycle types of σ are and 8. In the frst case, f we let x and y be the degrees of the vertces n each of the cycles, then 4x + 4y = 14, whch s mpossble for nteger x, y. In the second case, f we let x be the degree of the vertces n the 8-cycle, then 8x = 14, whch s not possble ether. So we are lookng for the number of permutatons whose cycle type s not + + 3, 8, 4 + 4, The number of permutatons wth cycle type ++3 s ( ) ( 3) (!) = 110, wth cycle type 8 s 7! = 5040, wth cycle type s ( 1 8 ) 4 (3!) = 160, wth cycle type s ( 8 3) (!)(4!) = 688. Therefore, by complementary countng, the number of permutatons that ARE automorphsms of some tree s 8! = Krstoff s plannng to transport a number of ndvsble ce blocks wth postve nteger weghts from the north mountan to Arendelle. He knows that when he reaches Arendelle, Prncess Anna and Queen Elsa wll name an ordered par (p, q) of nonnegatve ntegers satsfyng p+q 016. Krstoff must then gve Prncess Anna exactly p klograms of ce. Afterward, he must gve Queen Elsa exactly q klograms of ce. What s the mnmum number of blocks of ce Krstoff must carry to guarantee that he can always meet Anna and Elsa s demands, regardless of whch p and q are chosen? Proposed by: Pakawut Jradlok Answer: 18 The answer s 18.

6 Frst, we wll show that Krstoff must carry at least 18 ce blocks. Let 0 < x 1 x x n be the weghts of ce blocks he carres whch satsfy the condton that for any p, q Z 0 such that p + q 016, there are dsjont subsets I, J of {1,..., n} such that α I x α = p and α J x α = q. Clam: For any, f x x 014, then x1 + + x x Proof. Suppose to the contrary that x +1 x1+ +x +. Consder when Anna and Elsa both demand x1+ +x + 1 klograms of ce (whch s possble as ( x1+ +x + 1 ) x x + 016). Krstoff cannot gve any ce x j wth j + 1 (whch s too heavy), so he has to use from x 1,..., x. Snce he s always able to satsfy Anna s and Elsa s demands, x x ( x1+ +x + 1 ) x x + 1. A contradcton. It s easy to see x 1 = 1, so by hand we compute obtan the nequaltes x 1, x 3, x 4 3, x 5 4, x 6 6, x 7 9, x 8 14, x 9 1, x 10 31, x 11 47, x 1 70, x , x , x 15 37, x , x , x And we know n 18; otherwse the sum x x n would not reach 016. Now we wll prove that n = 18 works. Consder the 18 numbers named above, say a 1 = 1, a = 1, a 3 =, a 4 = 3,..., a 18 = 799. We clam that wth a 1,..., a k, for any p, q Z 0 such that p + q a a k, there are two dsjont subsets I, J of {1,..., k} such that α I x α = p and α J x α = q. We prove ths by nducton on k. It s clear for small k = 1,, 3. Now suppose ths s true for a certan k, and we add n a k+1. When Krstoff meets Anna frst and she demands p klograms of ce, there are two cases. Case I: f p a k+1, then Krstoff gves the a k+1 block to Anna frst, then he consder p = p a k+1 and the same unknown q. Now p + q a a k and he has a 1,..., a k, so by nducton he can successfully complete hs task. Case II: f p < a k+1, regardless of the value of q, he uses the same strategy as f p+q a 1 + +a k and he uses ce from a 1,..., a k wthout touchng a k+1. Then, when he meets Elsa, f q a a k p, he s safe. If q a a k p + 1, we know q a k+1 a a k p + 1 ( a1+ +a k + 1 ) 0. So he can gve the a k+1 to Elsa frst then do as f q = q a k+1 s the new demand by Elsa. He can now supply the ce to Elsa because p + q a a k. Thus, we fnsh our nducton. Therefore, Krstoff can carry those 18 blocks of ce and be certan that for any p + q a a 18 = 396, there are two dsjont subsets I, J {1,..., 18} such that α I a α = p and α J a α = q. In other words, he can delver the amount of ce both Anna and Elsa demand.