USA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year

Transcription

1 /3/27. Fill i each space of the grid with either a or a so that all sixtee strigs of four cosecutive umbers across ad dow are distict. You do ot eed to prove that your aswer is the oly oe possible; you merely eed to fid a aswer that satisfies the costraits above. (Note: I ay other USAMTS problem, you eed to provide a full proof. Oly i this problem is a aswer without justificatio acceptable.) Solutio Let row be the -th row with a horizotal word ad let colum j be the j-th colum with a vertical word. (I particular, the loe i the top left corer is ot cosidered its ow row.) We use the otatio R i C j to deote the square i row i ad colum j. For example, we are give that R C 2 is a, while R C is ot give to us. There are oly two squares that itersect exactly oe word, the top left ad bottom right. Every other square is part of exactly two words. Therefore, if the bottom right square is a, the total umber of s i all 6 strigs is odd. But we ow that there are 32 total s, so the bottom right square must be a. This leaves colum 7 as the oly possible positio for the strig. There are two possible locatios for the strig : colum 5 ad colum 6. However, if colum 5 were, the there would be 5 strigs i the grid of the form xx (rows 2, 4, 5, 6, ad 7). But we ow there are oly 4 such strigs, so the strig must be placed i colum 6.

2 Looig at the grid, we see that all strigs of the form xx are accouted for (rows 2, 4, 7, ad 8). Thus, looig at colum 3, we see that R 3 C 3 is, looig at row 5, we see that R 5 C 5 is, ad looig at row 6, we see that R 6 C 5 is. After placig these, looig at colum 5 tells us that R 4 C 5 is. Hece, colum 5 is. Sice row 6 is, row 5 must be. Sice row 5 is, colum 8 must be. The colum 5 ad row 6 are both of the form x, which meas R 3 C 2 must be. Hece row 3 is, ad colum 2 must be. Sice row 8 is, we see that row 2 must be, which meas that row 4 must be. Colums 2 ad 3 are both of the form x, so looig at row tells us that R C 4 must be a. Fially, there are oly 7 strigs that start with a i our grid so far, so R C is a. This completes the grid.

3 2/3/27. Fames is playig a computer game with fallig two-dimesioal blocs. The playig field is 7 uits wide ad ifiitely tall with a bottom border. Iitially the etire field is empty. Each tur, the computer gives Fames a 3 solid rectagular piece of three uit squares. Fames must decide whether to oriet the piece horizotally or vertically ad which colum(s) the piece should occupy (3 cosecutive colums for horizotal pieces, colum for vertical pieces). Oce he cofirms his choice, the piece is dropped straight dow ito the playig field i the selected colums, stoppig all three of the piece s squares as soo as the piece hits either the bottom of the playig field or ay square from aother piece. All of the pieces must be cotaied completely iside the playig field after droppig ad caot partially occupy colums. If at ay time a row of 7 spaces is all filled with squares, Fames scores a poit. Ufortuately, Fames is playig i ivisible mode, which prevets him from seeig the state of the playig field or how may poits he has, ad he already arbitrarily dropped some umber of pieces without rememberig what he did with them or how may there were. For partial credit, fid a strategy that will allow Fames to evetually ear at least oe more poit. For full credit, fid a strategy for which Fames ca correctly aouce I have eared at least oe more poit ad ow that he is correct. Solutio Drop a horizotal piece occupyig colums 3, 4, ad 5. The drop two horizotal pieces occupyig colums, 2, ad 3 ad colums 5, 6, ad 7. We illustrate this i the diagram below (the ew pieces dropped at this step are i gray). Sice the three pieces we just dropped spa every colum, oe of latter two pieces is guarateed to be our highest piece o the board at this poit. Without loss of geerality, we assume the piece o the right is ow our highest piece (as i the diagram above). We repeat this patter. Whe we drop the horizotal piece occupyig colums 3, 4, ad 5, it will be

4 stopped by the horizotal piece occupyig colums 5, 6, ad 7 that we dropped previously. Each of our successive pieces will the be stopped by this piece. Thus, our board loos lie this: At this poit, Fames ca simply drop oe vertical piece i the middle ad exclaim I have eared at least oe more poit, as desired. Note: If Fames repeatedly dropped vertical pieces i each colum, he would evetually ear a poit, but he caot ow whe that has happeed.

5 3/3/27. For >, let a be the umber of zeroes that! eds with whe writte i base. Fid the maximum value of a. Solutio We claim that the maximum value of a is. I biary, 2! is writte as, which eds with 2 zero. So a 2 =, ad a 2 2 =. So to coclude, it suffices to show that a is at most for all 2 2. Notice that the umber of zeroes that! eds with whe writte i base is the same as the maximum power of that divides!. Fix ad write = p m, where p is some prime dividig ad m is relatively prime to p. The a is at most the largest power of p dividig!. By Legedre s formula, the maximum power of p dividig! is ν p (!) =. So we have a j= < p j j= j= p j p j = The sum o the far right is geometric, ad sums to p. Thus, Dividig through by gives a < (p ). a < (p ). Therefore, if has ay prime divisor greater tha 2, or is divisible by the square of ay prime, a < 2. So we see that the maximum value of a is 2. j= p j.

6 4/3/27. Let ABC be a triagle with AB < AC. Let the agle bisector of BAC meet BC at D, ad let M be the midpoit of BC. Let P be the foot of the perpedicular from B to AD. Exted BP to meet AM at Q. Show that DQ is parallel to AB. Solutio B P Q D M A L C Let BP meet AC at L. By costructio, AP B = AP L = 9 ad P AL = P AB sice AP is the agle bisector from A. So AP B ad AP L are similar. Sice they share side AP, we see that AP B is cogruet to AP L. So P is the midpoit of BL. Sice M is the midpoit of BC, this implies that BP M is similar to BLC with ratio 2, ad P M is parallel to AC. Sice P M is parallel to AC, triagles DP M ad DAC are similar. Hece, = DM. To AC DC fiish, it suffices to show MDQ is similar to MBA, which meas that it suffices to show that MD = MQ. DB QA Sice AQL = P QM ad P M is parallel to AL, we see that QP M is similar to QLA. Thus MQ = MD ad we just eed to show that =. QA AL DB AL Notice that if a b = c d 2, the a = b 2a c d 2c AC 2P M =. Applyig this to AC MD CD 2MD. But 2P M = LC because BP M BLC with ratio 2. So AC 2P M = AC LC = AL. = DM DC gives us Similarly, we have CM = BD + DM, so CD = BD + DM + DM = BD + 2DM. Thus, MD CD 2MD = MD BD.

7 Puttig these together, we have as desired. AL = MD BD, Note: This problem ad solutio were proposed by Xie Guo-Xue.

8 5/3/27. Let a, a 2,..., a be a sequece of itegers. Iitially, a =, a 2 = ad the remaiig umbers are. After every secod, we perform the followig process o the sequece: for i =, 2,..., 99, replace a i with a i + a i+, ad replace a with a + a. (All of this is doe simultaeously, so each ew term is the sum of two terms of the sequece from before ay replacemets.) Show that for ay iteger M, there is some idex i ad some time t for which a i > M at time t. Solutio Fix a positive iteger M. Throughout this solutio we cosider idices modulo, so a 2 is really just a 2. Let a, deote a after secods, with a beig the origial sequece (a = a, ). We wat to show that the for some ad we have a, > M. We have a, = a + a 2 a,2 = a, + a 2, = (a + a 2 ) + (a 2 + a 3 ) = a + 2a 2 + a 3 a,3 = a,2 + a 2,2 = (a + 2a 2 + a 3 ) + (a 2 + 2a 3 + a 4 ) = a + 3a 2 + 3a 3 + a 4. Cotiuig this patter, we claim that ( ) a, = a + a + + ( ) a for all. We show this by iductio o. It s clearly true for =. Now suppose it s true for ad we ll show it s true for +. By iductio, we have a, = a, + a +, ( ( ) ( ) ) ( ( ) = a + a + + a a + + a (( ) ) (( ) ( )) = a + + a a But by Pascals idetity, ( ) ( r + ( r ) = r). Hece, we have the result. Of course, most of these terms are actually. Crossig out the s we have (( ) ( ) ) a i, = a i + a + a i 2 i ( ) ( ) =. i (mod ) 2 i (mod ) (( 2 i ) a 2 + ( 2 ( 22 i ) ) a +3 + ) ) a 22 + Where the last equality follows because a = ad a 2 =. Let ω be a th root of uity. Suppose toward a cotradictio that for all i ad, a i, < M. The by the triagle iequality a i, ω i < M. i=

9 However, we ca rewrite this sum as a i, ω i = i= = ω i= ( = i (mod ) ( ) ω ( ) ω i ( ) )ω +. = By the biomial theorem, we ca rewrite this as 2 i (mod ) ω ( ( + ω ) ω( + ω ) ) = (ω ω 2 )( + ω ). Taig magitudes, we have a i, ω i = ω ω 2 + ω. i= ( ) But + ω >, so we ca mae this absolute value arbitrarily large by cosiderig large eough. This cotradicts a i, ω i < M, i= ad we see that we must have some time t ad idex i for which a i > M at time t. ω i Note: This problem ca be geeralized. It s false if is replaced by 2 or 3, but true for ay other replacemet, by the same argumet as above. c 26 Art of Problem Solvig Foudatio