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# Review Sheet for MATH 302A at UA

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Arizona taught by a professor in Fall. Since its upload, it has received 15 views.

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Date Created: 02/06/15

Math 302A Review sheet for Exam I Pattern Problems 0 What are the next two numbers in the sequence 1 72 3 74 5 76 Describe the pattern Solution 7 and 78 are the next two elements I will describe two smaller patterns that make up the whole pattern The rst pattern is if we ignore the negative signs of the sequence the terms go up by one each time The second pattern is that all of the even terms are negative and the odd terms are positive These two together make up the whole pattern 0 When Nelson Muntz wanted Lisa7s lunch money she decided she7d give it willingly and not tell Principal Skinner if he could guess how many of each coin she had in her pocket She told Nelson 1 have no paper money and only quarters dimes and pennies l have 27 coins in all a total of 237 and twice as many dimes as quarters77 Solution 12 pennies 10 dimes and 5 quarters Without using algebra we can use a guess check revise method Lets start with the most valuable coin quarters and see if we can come up with a range of possibilities for how many of these we can have The two hints well use here are the total amount in Lisa7s pocket and the fact that there are twice as many dimes as there are quarters So we have the following cases to think about number of quarters number of dimes cash value so far total number of coins 1 2 45 3 2 4 90 6 3 6 135 9 4 8 180 12 5 10 225 15 6 12 270 18 Notice that we can toss out the last situation since were already over 237 Next we observe that Lisa7s total cash value ends in 7 so this means she must have 2 7 12 17 22 or 27 pennies If she has 27 coins in all we can check each of the above cases in the table to see which case will leave us with the right number of coins The 5th case is the one that works since 15 coins are dimes and quarters that means 12 coins are left and they must be pennies Let7s check to see if this is right 52510101201 125 100 12 237 This gives us our answer 0 While in OZ Dorothy walks along many elds of owers She decides to pick some of them along the way At the rst eld she takes one At the second she takes three At the third she takes ve At the fourth she takes seven She continues this pattern of taking the next odd number of owers List how many owers she has total after each time she picks owers Do you notice a pattern How many owers will Dorothy have after picking from the 100th eld of owers Solution Dorothy7s ower total will be 1 4 9 16 25 and so on The pattern is that the total number of owers will always be a perfect square and after each eld the number of owers she will have is the next perfect square So after her 100th time picking she will have 1002 10 000 owers A cryptarithmetic problem is an arithmetic problem where the letters are used in place of numbers The object is to gure out what number each ofthe letters must represent In the following problem each letter stands for a different digit 0 through 9 no two letters stand for the same digit Determine which digit each letter represents if DRY RED ODOR Solution There are several possibilities Notice rst and foremost that O has to be either 0 or 1 Why If O is 0 then its not too hard to realize R 0 too but we told you that the letters represented distinct numbers So O is equal to 1 this lets us notice the following situations Case 1 Case ll Reason These cases come from D Y R D Y 10 R adding the ones place This is from adding the tens place and noticing if we borrowed from the tens RE11 RE10 earlier This is from adding the hundreds place and notic ing we had to borrow a one hundred when we were DR110D DR110D addinginthetensplace Notice in bot cases we have the equation DR1 10D R1 10 R 9 So now we have R 9 In Case ll this would force E 1 but O is already 1 so we must be in Case 1 this means E 2 and the only other condition we need to satisfy is D Y 9 So the possible solutions are DRY RED ODOR 693 926 1619 396 923 1319 495 924 1419 594 925 1519 Number System Problems 0 Convert the following numbers to base 10 numbers 7 5137 Solution 5137 5724174370 54973 255 4AD16 where the symbols from smallest to largest are 0 l 2 3 4 5 6 7 8 9 A B C D E 44D16 4162A16D160 4 162 10161 13160 425616013 101616013 1189 71100012 1100012 1254124o23022021120 252420 32161 49 75049 5049 592091490 581041 4054 409 0 Convert the following base 10 numbers to the speci ed base i 317 to base 8 464 lt 317 lt 564 317 4 464 61 78 lt 61 lt 88 61 4 78 5 So the answer is 4758 i 556 to base 12 where the symbols from smallest to largest are 0 1 2 3 4 5 6 7 8 9 A B 3144 lt556lt 4144 556 i 3144 124 1012 lt124lt 1112 1247120 4 Since the quantity 10 corresponds to the symbol A in base 12 we have 3A412 1025 to base 2 21 lt1025lt 211 102571024 1 So the answer is 100000000012 0 Add 567 627 witout converting to base 10 and then back to base 7 567 627 507 607 67 27 507 607 67 27 10757 67 67 27 107147 117 1407 117 1517 0 Review the different number systems presented in the Green Book You are expected to be able to interpret between these number systems and the base 10 system Addition and Subtraction Strategies 0 Analyze Joel7s strategy of addition applied to the example 4738 in other words apply rules of the base 10 system commutativity of addition and associativity of addition to break down Joel7s strategy piece by piece 4738 407308 407308 403078 403078 7015 70105 70105 o A student was asked to subtract 46 from 93 he solved the problem using the following reasoning I took away 40 from 90 and got 50 but 93 is 3 more than 90 so I get 53 and 46 is 6 more than 46 so I get 5977 ls this correct Why or why not Solution It is incorrect since 93 46 47 The mistake was when he added 6 instead of subtracting 6 Another argument would be that 93 50 43 and 93 40 537 so it should be that 93 46 is somewhere between 43 and 537 and to see that 93 46 is exactly 477 the student could build towards it by studying 93 49 and 93 41 and so on

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