MATH FIELD DAY Contestants Insructions Team Essay. 1. Your team has forty minutes to answer this set of questions.

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1 MATH FIELD DAY 2012 Contestnts Insructions Tem Essy 1. Your tem hs forty minutes to nswer this set of questions. 2. All nswers must be justified with complete explntions. Your nswers should be cler, grmmticlly correct, nd mthemticlly precise. 3. Your tem my turn in t most one nswer to ech question. Number the nswers nd submit them in order, strting ech problem on seprte pge. You re llowed to skip ny question nd proceed on to lter questions. Write on one side of the pper only, nd number the pges you turn in consecutively. Do not turn in the problem pckets. Write your school nme on ech pge. 4. Your tem my be orgnized in ny wy you choose. For exmple, you my designte one tem member to write every nswer, or you could designte different tem members to write different nswers. You my use the chlk or white bords in the room. 5. Only officil tem members my be in the room during the essy period. Tem members my not bring books, clcultors, or ny other mterils into the room. 6. A proctor outside the room will wrn you ten minutes before your time is up. 7. Give your nswers to the proctor t the end of the period to plce in the inner envelope. Plese erse ll bords in the room nd put ll scrtch pper in the trsh cn, leving the room net. 1
2 Pouring Problems We hve two jugs, one with cpcity of m qurts nd the other with cpcity of n qurts, for positive integers m nd n. We mesure mounts of wter by pouring from one jug to the other. We cn lso fill jug from nerby pool of wter or empty it into the pool. We cn tell when jug is full or empty but, becuse the jugs re unmrked nd oddly shped, we cnnot otherwise tell by looking t jug how much wter it contins. The pouring process is sequence of steps. It begins with both jugs empty t the strt of step 0. Ech step hs the following prts A nd B. Prt A. Pour s much wter s possible from the mjug to the njug. This leves the mjug empty or the njug full (or both). Prt B. If the mjug is empty, fill it from the pool. If the mjug is not empty, then the njug is full, nd empty the njug into the pool. We write the symbol u:v to show tht the mjug currently contins u qurts of wter nd the njug currently contins v qurts. We represent ech step of the pouring process by two symbols, one before prt A nd one before prt B. For exmple, when m = 3 nd n = 5, we pour from 3qurt jug to 5qurt jug. Figure 1 shows steps 0 through m + n = = 8 of the pouring process. Step 0. 0:0, 0:0 Step 1. 3:0, 0:3 Step 2. 3:3, 1:5 Step 3. 1:0, 0:1 Step 4. 3:1, 0:4 Step 5. 3:4, 2:5 Step 6. 2:0, 0:2 Step 7. 3:2, 0:5 Step 8. 3:5, 3:5 Figure 1 Step 0 strts with 0:0 becuse both jugs re initilly empty. Pouring s much wter s possible none from the 3jug to the 5jug gives the second symbol 0:0 in step 0. Becuse the 3jug is empty, we fill it from the pool nd strt step 1 with 3:0. Emptying the 3jug into the 5jug gives 0:3. Becuse the 3jug is empty, we fill it from the pool nd strt step 2 with 3:3. Filling the 5jug from the 3jug gives 1:5. Becuse the 3jug is not empty nd the 5jug is full, we empty the 5jug into the pool nd strt step 3 with 1:0. Emptying the 3jug into the 5jug gives 0:1. Becuse the 3jug is empty, we fill it from the pool nd strt step 4 with 3:1. Emptying the 3jug into the 5jug gives 0:4. Becuse the 3jug is empty, we fill it from the pool nd strt step 5 with 3:4. Filling the 5jug from the 3jug gives 2:5. Becuse the 3jug is not empty nd the 5jug is full, we empty the 5jug into the pool nd strt step 6 with 2:0. Emptying the 3jug into the 5jug gives 0:2. Becuse the 3jug is empty, we fill it 2
3 from the pool nd strt step 7 with 3:2. Emptying the 3jug into the 5jug gives 0:5. Becuse the 3jug is empty, we fill it from the pool nd strt step 8 with 3:5. Pouring s much wter s possible none from the 3jug to the 5jug gives the second symbol 3:5 in step 8. Problem 1. Show steps 0 through m + n of the pouring process for m = 5 nd n = 8. Represent ech step by two symbols u:v, s in Figure 1. You do not need to describe the steps verblly s in the previous prgrph. Let y i be the totl number of qurts in the two jugs during step i of the pouring process. Tht is, y i is the sum u + v for the two symbols u : v representing tht step. For exmple, when m = 3 nd n = 5, Figure 1 shows tht y 2 = 6, since step 2 hs the symbols 3:3 nd 1:5 nd = 6 = Tht is, there re 6 qurts of wter in the two jugs combined during step 2. Consider the sequence y 1, y 2,..., y m+n (1) of numbers of qurts in steps 1 through m + n. Steps 1 through 8 of Figure 1 show tht (1) is the sequence when m = 3 nd n = 5. 3, 6, 1, 4, 7, 2, 5, 8 (2) Problem 2. Find the sequence (1) for m = 5 nd n = 8, s in (2). Use your nswer to Problem 1. Problem 5 shows tht (1) consists of the integers 1 through m + n in some order whenever m nd n hve no common fctor greter thn 1. The sequence (2) illustrtes this result for m = 3 nd n = 5, nd your nswer to Problem 2 should illustrte it for m = 5 nd n = 8. When m nd n hve no common fctor greter tht 1, the frction m n The next clim describes the frctions equl to m. n is in lowest terms. Clim I. Let m nd n be positive integers tht hve no common fctor greter thn 1. If nd b re positive integers such tht = m, then = mt nd b = nt for positive integer t. b n Clim I sys tht we get m by cnceling fctors of t in the numertor or denomintor of n. Problems 8 through 10 prove Clim I. We use Clim I until then to study the pouring b process. 3
4 Problem 3. Assume tht m nd n hve no common fctor greter thn 1. Let r nd s be positive integers such tht y r = y s nd r < s. () Let h be the number of times we fill the mjug from the pool nd let k be the number of times we empty the njug into the pool from the strt of step r to the strt of step s (tht is, when we perform prt B in steps r through s 1). Deduce from Clim 1 tht h = nt nd k = mt for positive integer t. (b) Conclude tht s r m + n. Problem 4. Why is y j n integer from 1 through m + n for every positive integer j? Problem 5. Assume tht m nd n hve no common fctor greter thn 1. Prove tht the sequence y 1,..., y m+n in (1) consists of the integers 1 through m + n in some order. Use Problems 3 nd 4. Problem 5 shows tht the pouring process gives every whole number of qurts from 1 through m + n when m nd n hve no common fctor greter thn 1. Problem 6. Assume tht m nd n hve no common fctor greter thn 1. Let s be positive integer such tht y s = m + n. () Let h be the number of times we fill the mjug from the pool nd let k be the number of times we empty the njug into the pool from the strt of step 0 to the strt of step s (tht is, when we perform prt B in steps 0 through s 1). Deduce from Clim I tht h = nt + 1 nd k = mt 1 for positive integer t. (b) Conclude from () tht s m + n. Problem 7. Assume tht m nd n hve no common fctors greter thn 1. () Deduce from Problems 5 nd 6 tht y m+n = m + n. (b) Why does y j = y m+n+j for every positive integer j? For m = 3 nd n = 5, Problem 7() sys tht y 8 = 8, s the lst entry in (2) shows. Likewise, the lst entry in your nswer to Problem 2 should illustrte Problem 7() for m = 5 nd n = 8. Problem 7 (b) sys tht the sequence y j repets every m + n steps when y 0 is set side. The three remining problems justify Clim I. They do not rely on fctoring integers into primes, nd neither should your nswers. 4
5 Problem 8. Let, b, c, nd d be positive integers such tht b = c d Let t be n integer. () If b = dt, prove tht = ct. (b) If b dt > 0, prove tht ct > 0 nd ct b dt = b. Let b nd d be positive integers. Let t be the lrgest integer such tht b dt 0. Since b d(t + 1) < 0, we hve 0 b dt < d. (3) For exmple, b = 25 nd d = 7 give t = 3 nd (3) < 7. Problem 9. Let nd b be positive integers. Let c nd d be positive integers such tht b = c d nd d is s smll s possible. Deduce from (3) nd Problem 8 tht = ct nd b = dt for positive integer t. Problem 10. Let m nd n be positive integers tht hve no common fctor greter thn 1. Let c nd d be positive integers such tht c d = m n nd d is s smll s possibe. Deduce from Problem 9 tht c = m nd d = n. Let b = m n for positive integers, b, m, nd n such tht m nd n hve no common fctor greter thn 1. By problem 10, we cn tke c = m nd d = n in Problem 9 nd see tht = mt nd b = nt for positive integer t. This proves Clim I. 5
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