Are you ready f Beast Academy C? Befe beginning Beast Academy C, a student should be able to compute fluently with fractions and integers and be able to add and subtract decimals. The student should also have a basic understanding of statistics, greatest common fact, and least common multiple. A student ready f Beast Academy C should be able to answer at least 2 of the 6 problems below crectly. Step. The student should try to answer every question without a calculat and without help. Step 2. Check the student s answers using the solutions at the end of this document. Step. The student should be given a second chance on problems that he she answered increctly. Evaluate each expression below...2+.9= 2. 8.6.08=. 6 + 2 0 =. 2 =. = 6. 8 =. Order the numbers below from least to greatest..,,, 0.0 00.0 8. a. What is the greatest common fact (GCF) of 8 and 90? b. What is the least common multiple (LCM) of 8 and 90? a. GCF: b. LCM: 9. a. What is the GCF of 6,, and 60? b. What is the LCM of 6,, and 60? a. GCF: b. LCM:
Are you ready f Beast Academy C? F problems 0-2, use the given numbers to fill in the blanks so that each statement is true, and each fraction is in simplest fm. 0. Numbers:,, 9, 20. Numbers:,, 6, 2. Numbers:, 2, 20, 0 = 6 8 = + = Use the prime factization below to help you answer problems and. 9,600=2 2 9. What is the smallest positive integer that we can multiply 9,600 by to get a product that is a perfect square?.. What is the smallest positive integer that is not a fact of 9,600?.. A wheelbarrow contains five 6-pound pumpkins and some 9-pound pumpkins. If the weight of a pumpkin in the wheelbarrow is pounds, how many 9-pound pumpkins are in the wheelbarrow?. 6. Fill each empty white square below with a positive digit so that the clues given in the surrounding shaded squares give the crect,,, and f the row column they label. 2 6 8
Are you ready f Beast Academy C?..2+.9=6.282. 2. 8.6.08=.02.. 6 + 2 2 = 0 0 + 2 9 = 0 0 =6 0 =6 2 =92.. 2 = 0 6 = 0 0 6 0 = 9 0 =9 0.. We first divide and by their greatest common fact,. We then divide 2 and 8 by their greatest common fact, 6. Now that we have canceled all common facts, we compute the product. 2 2 = 2 = 2 2 6. =2. So, 8 =2 8 =2 8 =2 8 =0 =.. We write each fraction as a decimal. 0 =..0 =..0 00 Then, we compare the decimals:.0 <. <.0 <.. Writing the iginal numbers in der from least to greatest, we have.0,,.0, 0 00. We write each number as a mixed number with a denominat of. 0 = 00.0= = 00 00 = 0 0.0= Then, we compare the mixed numbers: < 00 < 0 < 0. Writing the iginal numbers in der from least to greatest, we have.0,,.0, 0 00. 8. a. We use the prime factizations of 8 and 90. 8=2 and 90= 2 2 The GCF of 8 and 90 is the product of all the prime facts they share. So, the GCF is 2 =6. b. To compute the LCM, we take the largest power of each prime in either number s prime factization. 8=2 and 90= 2 2 So, the LCM of 8 and 90 is 2 2 =20. 9. a. We use the prime factizations of 6,, and 60. 6=2 2 2 = 2 60= 2 2 The GCF of 6,, and 60 is the product of all the prime facts they share. So, the GCF is 2 =6. Solutions b. To compute the LCM, we take the largest power of each prime in the numbers prime factizations. 6=2 2 2 = 2 60= 2 2 So, the LCM of 6,, and 60 is 2 2 =0. 0. All fractions must be in simplest fm. Among our choices, only can be the numerat of the fraction with denominat 6.. The remaining numbers are, 9, and 20. Since the denominat of the product is not a multiple of, the fact of in the denominat of must cancel. So, 6 the numerat of the middle fraction is a multiple of. Among our choices, only and 9 are multiples of. Since all fractions are in simplest fm, 20 cannot be the numerat of 8. Therefe, 20 is the denominat of the middle fraction. Finally, we place and 9 in the empty numerats in the only way that makes a true statement. Check: 6 9 20 = 2 = 8. ü 2 The denominat of the product is =. So, the denominats of the two fractions that we multiply must include at least one multiple of and one multiple of. Among our choices, only is a multiple of, and only is a multiple of. These two numbers are therefe the denominats of the fractions that we multiply. Since all fractions are in simplest fm, cannot be the denominat of. We place and in the empty denominats as shown. The remaining numbers are and 6. All fractions are in simplest fm, so we place the and 6 in the empty numerats as shown. Check: = 2 = 6. ü 2 6 = 8 9 6 = 8 9 6 20 = 8 6 9 20 = 8 = = 6 2. Both the numerat and denominat of the middle fraction are empty. and 2 All fractions are in simplest fm. Among our four number choices, and 2 are the only pair with a GCF of. So, only and 2 can be the numerat and denominat of the middle fraction. + =
Are you ready f Beast Academy C? The remaining numbers are 20 and 0. Since each fraction is in simplest fm, we can only place 20 and 0 in the empty denominats as shown. Then, since the sum is less than and 0 we are adding only positive numbers, we know the middle fraction is also less than. So, we place the and 2 as shown. Check: 20 + 2 = 9 60 + 2 60 = 60 = 0. ü and 2 20 + = 0 20 + 2 = 0. If a number is a perfect square, then it can be written as the product of two identical groups of prime facts. So, in the prime factization of a perfect square, every prime has an even exponent. In the prime factization of 9,600=2 2 9, the primes,, and 9 each have odd exponents (, and 9 ). So, multiplying 9,600 by 9 gives a perfect square: 9,600 9=(2 2 9) 9 = 2 2 2 2 9 2 =(2 2 9) (2 2 9) =,980,980 =,980 2. Multiplying 9,600 by any integer less than 9 would result in a product that is not a perfect square. So, 9=99 is the smallest positive integer we can multiply 9,600 by to make a perfect square.. If an integer x is not a fact of 9,600, then either x has at least one prime fact that is not a fact of 9,600, the power of a prime fact x is larger than its power in 9,600. The smallest prime fact that is not in 2 2 9 is. The smallest power of 2 larger than 2 is 2 =2. The smallest power of larger than is 2 =9. The smallest power of larger than 2 is =2. The smallest power of larger than is 2 =9. The smallest power of 9 larger than 9 is 9 2 =6. Among the possibilities above, the smallest is 2 =9. So, 9 is the smallest positive integer that is not a fact of 9,600. Each integer from to 8 is a fact of 2 2 9: : 2 2 9= (2 2 9) 2: 2 2 9=2 (2 2 9) : 2 2 9= (2 2 9) : 2 2 9= (2 2 2 9) : 2 2 9= (2 9) 6: 2 2 9=6 (2 2 9) : 2 2 9= (2 2 9) 8: 2 2 9=8 (2 2 9) However, 9= 2 is not a fact 2 2 9. So, 9 is the smallest positive integer that is not a fact of 9,600.. Each 6-pound pumpkin is 8 pounds below the, and each 9-pound pumpkin is pounds above the. The five 6-pound pumpkins are a total of 8=0 pounds below the. To balance this, we need enough 9-pound pumpkins to equal 0 pounds above the. 6. 0 8 8 8 8 8 6 6 6 6 6 +0 +... + 9... 9 So, there are 0 =8 nineteen-pound pumpkins in the wheelbarrow. We write and solve an equation. Let n be the number of 9-pound pumpkins in the wheelbarrow. The five 6-pound pumpkins weigh a total of 6=0 pounds, and the n 9-pound pumpkins weigh a total of 9n pounds. So, the total weight of all the pumpkins is 0+9n. There are +n pumpkins in the wheelbarrow with an weight of pounds, so their total weight is (+n)=0+n. We have two expressions f the total weight of the pumpkins, so we write an equation: 0+9n=0+n. Solving f n, we get n=8. Therefe, there are 8 9-pound pumpkins in the wheelbarrow. Adding all of the numbers in a list and then dividing by the number of numbers gives us their. The of the numbers in the left column is, so the sum of these three numbers is =2. There is already a in this column, so the sum of the two remaining numbers is 2 =. The only way to make a sum of from two positive digits is 8+9. The is the difference between the largest and smallest numbers in a data set. In the top row, the is 2. The top row already contains a 6, so it cannot contain any digit larger than 6+2=8. 2 6 (8 and 9) 8 9 8 So, we place the 8 and 9 in the left column as shown. (8 and 9)
Are you ready f Beast Academy C? The is the number that appears most in a list. The of the three integers in the bottom row is 8, so this number must appear at least twice in the row. Since we already have a 9 in the bottom row, the two other squares in this row both contain 8. The is the number in the middle when we ar a list in der from least to greatest. The of the middle column is. Since this column already contains a 6 and 8, the remaining number in this column is. The of the numbers in the right column is, so the sum of these three numbers is =. There is already an 8 in this column, so the sum of the two remaining numbers is 8=. We can write as the sum of two digits in ways: +6, 2+, and +. The in the top row is 2, and the two digits already placed have a of 2. Therefe, the third number in this top row can only be 6,, 8. Only one of our three pairs in the right column contains a 6,, 8: +6. So, we place the and 6 in the right column as shown to complete the puzzle. ( and 6), (2 and ), ( and ) 6 ( and 6), (2 and ), ( and )