USA Mathematical Talent Search Round 2 Solutions Year 29 Academic Year

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1 athematical alent earch Year 29 caemic Year Important information: 1. You must show your work an prove your answers on all problems. If you just sen a numerical answer with no proof for a problem other than Problem 1, you will get no more than 1 point. 2. Put your name an ID# on every page you submit. 3. No single page shoul contain solutions to more than one problem. Every solution you submit shoul begin on a new page, an you shoul only submit work on one sie of each piece of paper. 4. ubmit your solutions by November 27, 2017, via one (an only one!) of the methos below: (a) Web: Log on to to uploa a PDF file containing your solutions. (No other file type will be accepte.) Dealine: 8 P Eastern / 5 P Pacific on November 27, 2017 (b) ail: P.O. Box 4499 New York, NY (olutions must be postmarke on or before November 27, 2017.) 5. Once you sen in your solutions, that submission is final. You cannot resubmit solutions. 6. Confirm that your aress in your Profile is correct. You can o so by logging onto an visiting the y pages. 7. Roun 2 results will be poste at when available. o see your results, log on to the website, then go to y. You will also receive an when your scores an comments are available (provie that you i item #6 above). hese are only part of the complete rules. Please rea the entire rules on.

2 athematical alent earch Year 29 caemic Year Each problem is worth 5 points. 1/2/29. Given a rectangular gri with some cells containing one letter, we say a row or column is eible if it has more than one cell with a letter an all such cells contain the same letter. Given such a gri, the hungry, hungry letter monster repeats the following proceure: he fins all eible rows an all eible columns an simultaneously eats all the letters in those rows an columns, removing those letters from the gri an leaving those cells empty. He continues this until no more eible rows an columns remain. Call a gri a meal if the letter monster can eat all of its letters using this proceure. In the 7 by 7 gri to the right, fill each empty space with one letter so that the gri is a meal an there are a total eight s, nine s, ten s, eleven s, an eleven s. ome letters have been given to you. You o not nee to prove that your answer is the only one possible; you merely nee to fin an answer that satisfies the constraints above. (Note: In any other problem, you nee to provie a full proof. Only in this problem is an answer without justification acceptable.) olution

3 athematical alent earch Year 29 caemic Year /2/29. Let b be a positive integer. Grogg writes own a sequence whose first term is 1. Each term after that is the total number of igits in all the previous terms of the sequence when written in base b. For example, if b = 3, the sequence starts 1, 1, 2, 3, 5, 7, 9, 12,.... If b = 2521, what is the first positive power of b that oes not appear in the sequence? olution First, we look for a pattern in how the sequence increases. It increases by 1 with each term until we hit b. hen, the sequence increases by 2 with each term until we hit b 2. In general, once the sequence has passe b k 1, it will increase by k each term until it passes b k. his means that, if b r 1 is in the sequence, b r will be in the sequence if an only if b r b r 1 is a multiple of r. Or, equivalently, iff b r b r 1 (mo r). ince b 0 is in the sequence by efinition, we re looking for the first power of b such that b r b r 1 (mo r). ince b = lcm(1, 2,..., 10) + 1, we know that b 1 (mo k) for 2 k 10. o, the first time this equivalence might fail is at r = 11. n inee it oes fail at r = 11, since (mo 11) an = (mo 11). hus, b 11 is the first power of b that oes not appear in the sequence. Note: It s easiest to gain an unerstaning of what s going on by trying b = 10. rying simpler problems that you unerstan better is an important problem solving strategy. In this case, this problem actually came from first trying the case for b = 10.

4 athematical alent earch Year 29 caemic Year /2/29. he tug-of-war team nees to pick a representative to sen to the national tug-of-war convention. hey on t care who they sen, as long as they on t sen the weakest person on the team. heir team consists of 20 people, who each pull with a ifferent constant strength. hey want to esign a tournament, with each roun planne ahea of time, which at the en will allow them to pick a vali representative. Each roun of the tournament is a 10-on-10 tug-of-war match. roun may en in one sie winning, or in a tie if the strengths of each sie are matche. how that they can choose a representative using a tournament with 10 rouns. olution Consier the following cyclic sequence of games. In each game, the top 10 players play against the bottom 10 in the list. Player 1 Player 2 Player 3 Player 4 Player 5 Player 6 Player 7 Player 8 Player 9 Player 10 Player 11 Player 12 Player 13 Player 14 Player 15 Player 16 Player 17 Player 18 Player 19 Player 20 Player 2 Player 3 Player 4 Player 5 Player 6 Player 7 Player 8 Player 9 Player 10 Player 11 Player 12 Player 13 Player 14 Player 15 Player 16 Player 17 Player 18 Player 19 Player 20 Player 1 Player 10 Player 11 Player 12 Player 13 Player 14 Player 15 Player 16 Player 17 Player 18 Player 19 Player 20 Player 1 Player 2 Player 3 Player 4 Player 5 Player 6 Player 7 Player 8 Player 9 Player 11 Player 12 Player 13 Player 14 Player 15 Player 16 Player 17 Player 18 Player 19 Player 20 Player 1 Player 2 Player 3 Player 4 Player 5 Player 6 Player 7 Player 8 Player 9 Player 10 We get each successive list by moving the player on the top of the previous list to the bottom of the next list. Note that there are a total of 11 games here, but the last game is exactly the same as the first, so we know the results of all 11 games by only playing the first 10. Note that if the first game is a tie, the secon game etermines which of player 1 an player 11 is stronger. o, without loss of generality, we ll assume the bottom team lost the first game. In each successive game, one player on each team switches spots. o, the first time the bottom team wins a game, we know that the player that switche to the bottom is stronger than the person she swappe with.

5 athematical alent earch Year 29 caemic Year o, we will have a representative if the bottom team ever wins a game. But, we know the bottom team wins the 11th game because it s the exact opposite of the first game! o, we must be able to fin a representative in at most 10 games. Note: While we expect it is the case, the authors o not know if 10 games is optimal.

6 athematical alent earch Year 29 caemic Year /2/29. Zan starts with a rational number 0 < a < 1 written on the boar in lowest terms. b hen, every secon, Zan as 1 to both the numerator an enominator of the latest fraction an writes the result in lowest terms. Zan stops as soon as he writes a fraction of the form n, for some positive integer n. If a starte in that form, Zan oes nothing. n+1 b s an example, if Zan starts with 13 14, then after one secon he writes = 7, then after two secons 8, then 9 = 3, at which point he stops (a) Prove that Zan will stop in less than b a secons. (b) how that if is the final number, then n Create PDF with GO2PDF for free, if you wish to remove n+1 this line, click here to buy Virtual PDF Printer n 1 n < a b n n + 1. olution (a) We go by inuction on b a. he base case b a = 1 is one by efinition, because Zan won t make any moves. uppose the fraction is a an the claim is true for all values b less than b a. Let m be the smallest positive integer with gc(a + m, b + m) > 1, an efine = gc(a + m, b + m). Note that we have m <, since otherwise we coul write m = + k an m = m woul be a smaller solution. he efinition of m means after m steps, we get a fraction whose numerator an enominator iffer by b a, which is an integer since = gc(a + m, b + m) = gc(a + m, [b + m] [a + m]) = gc(a + m, b a). By the inuction hypothesis, it takes at most b a 1 steps to finish after these m steps. o to complete the inuction we nee to show that m + b a 1 < b a. ince an b a are positive integers, we can say that ( ) b a ( 1) 1 0. Expaning an rearranging, this becomes + b a 1 b a. ince m <, this implies the inequality we wante to show.

7 athematical alent earch Year 29 caemic Year (b) We will assume that Zan s sequence has more than one term, since otherwise the result is obvious. First, we claim that Zan s sequence is strictly increasing. hat is, a+1 > a. b+1 b Clearing enominators, this is equivalent to b(a + 1) a(b + 1) = b a > 0, which is true since a < 1. his means that a < n, so it now suffices to show that b b n+1 n 1 < a. o that en, suppose towar a contraiction that a < k, where k < n. n b b k+1 Clearing the enominators in our conition, we get ak + a < kb. ing k to both sies an using that x < y x+1 y for integers x, y, this becomes ak + a + k + 1 kb + k. fter factoring the left sie an iviing both sies by (k + 1)(b + 1), we have a + 1 b + 1 k k + 1. his means that Zan s sequence is boune by n number is. o, we must have n+1 n 1 n < a b n n + 1. k, which is impossible if Zan s final k+1

8 athematical alent earch Year 29 caemic Year /2/29. here are n istinct points in the plane, no three of which are collinear. uppose that an B are two of these points. We say that segment B is inepenent if there is a straight line such that points an B are on one sie of the line, an the other n 2 points are on the other sie. What is the maximum possible number of inepenent segments? olution (REVIED base on input by Kevin Ren.) If n = 1, there are 0 inepenent segments. If n = 2, there is exactly 1 inepenent segment. If n = 3, there are exactly 3 inepenent segments (the eges of the triangle forme by the three points). If n 4, let H = H 1 H 2... H k be the convex hull of the given points an G 1, G 2,..., G l be all given points that are not vertices of H. We will nee two cases, first when H is a triangle, an secon when H is not a triangle. First assume the convex hull of the set of points is a triangle BC. No inepenent segment can have both of its enpoints in the interior of H. We claim that for any ege, say B, of this triangle there is at most one point X in the interior of BC such that both X an XB are inepenent. o that en, suppose that we have two points X an Y in the interior of BC such that X, BX an Y, BY are inepenent. First, we note that if Y an BX cross, then neither of them can be inepenent. o see this note that in this case XY B is a convex quarilateral. he segment that separate Y from the other points woul have to pass through X, XY, an B. he only such line is the iagonal BX, which is impossible because an Y are not on the same sie of that iagonal. imilarly, BX cannot be inepenent. o, Y an BX o not cross. Without loss of generality, Y is containe in the interior of XC. In this case, a segment that separates BY from, C, an X must pass through Y, CY, an XY. But that is impossible, because any such line passes through Y. his proves the claim.

9 athematical alent earch Year 29 caemic Year herefore, we can boun the number of inepenent segments when the convex hull is a triangle. First, if there is an interior point X such that X, BX, an CX are all inepenent, then the other n 4 interior points are part of at most one inepenent segment, an each sie of the convex hull is an inepenent segment. his gives at most (n 4) = n + 2 inepenent segments. lternatively, if there is no interior point X such that X, BX, an CX are inepenent, then there are at most three interior points incient to two inepenent segments, one from each sie. his gives 6 inepenent segments. he remaining n 6 interior points are part of at most one inepenent segment, an each sie of the convex hull is an inepenent segment. o, we see that there are at most (n 6) = n + 3 inepenent segments in this case. ll together, we can say that there are at most n + 3 inepenent segments when the convex hull is a triangle. Next, we consier the case that the convex hull is not a triangle. Let H = H 1 H 2... H k be the convex hull of the given points an G 1, G 2,..., G l be all given points that are not vertices of H. If both of the enpoints of an inepenent segment are vertices of H then it must be a sie of H: here are at most k such inepenent segments.

10 athematical alent earch Year 29 caemic Year Next we look at inepenent segments with one enpoint an H i an the other a G j. If H i G j is inepenent, then G j lies in the interior of H i 1 H i H i+1 (H-inices run moulo k). ince no three such triangles have a common interior point, no G j is an enpoint of more than two inepenent segments. If G j is an enpoint of two inepenent segments, we say that it is a ouble. Let m be the number of oubles. If G j is a ouble, then it must lie in the interior of the intersection δ i of H i 1 H i H i+1 an H i H i+1 H i+2 for some i. uppose that some δ i contains two oubles P an Q. ince H i 1 H i H i+1 H i+2 is a convex quarilateral, the line P Q oes not intersect at least one of the segments H i 1 H i, H i H i+1, an H i+1 H i+2, say H i 1 H i. his means that (with P an Q s labels switche if nee be) H i 1 P QH i is a convex quarilateral. herefore, the segment H i 1 Q intersects the segment H i P in an interior point an H i P cannot be an inepenent segment, a contraiction. herefore, the number of oubles is at most the number of δs. ince there are l Gs an k δs, this means that the number of inepenent segments involving Gs is boune above by l + min{k, l}. ing in the eges of the convex hull, we see that the total number of inepenent segments is at most k + l + min{k, l} when the convex hull is not a triangle. ince k + l = n, this is at most n + n. 2 his gives us a boun of max(n + 3, n + n ). For n > 5, this boun is simply n + n 2 2 an is attaine when k = n an each G 2 j lies in the interior of δ j. o complete the proof, we just have to wrestle with the cases n = 4 an n = 5. If n = 4, there are at most 6 inepenent segments, achieve by placing a single point insie a triangular convex hull. Finally, we consier n = 5. If the convex hull is not a triangle, we know there are at most n + n = = 7 inepenent segments. 2 We claim that no interior point can form an inepenent segment with all three vertices in a triangular convex hull BC. o see this, note that there are exactly 2 vertices say X, Y in the interior of BC. he line XY intersects exactly two of the sies of BC, without loss of generality, assume that B is the thir sie. hen, XY B forms a convex quarilateral, as shown below.

11 athematical alent earch Year 29 caemic Year We know from earlier that neither Y nor BX can be inepenent in this case, so we have the claim. hus, there are at most = 7 inepenent segments for n = 5. his is achieve in the picture above (B, BC, C, X, XC, Y B, an Y C are inepenent). n hus the maximum value is n + for n 4, an 0 for n = 1, 1 for n = 2, an 3 2 for n = 3. (Problem propose by Nikolai Beluhov) Problems by Nikolai Beluhov, Billy wartworth, ichael ang, an taff. must be submitte by November 27, Please visit for etails about solution submission. c 2017 rt of Problem olving Initiative, Inc.