1. When the least common multiple of 8 and 20 is multiplied by the greatest common factor of 8 and 20, what is the result?


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1 Black Equivalent Fractions and LCM 1. When the least common multiple of 8 and 20 is multiplied by the greatest common factor of 8 and 20, what is the result? 2. The sum of three consecutive integers is a multiple of 89 and is a value between 600 and 900. What is the sum of the three integers? 3. How many positive integers less than or equal to 50 are multiples of 3 or 4, but not 5? 4. The house numbers in a housing development are the multiples of four starting at 4 and ending at 120. If metal digits cost 50 cents each, what is the total cost of all of the metal digits that are needed to number the houses in this development? 5. Explain the following statement: When rewriting fractions with a common denominator, the common denominator can be any common multiple of the two original denominators, however, using the least common multiple is the most convenient choice. 6. Multiple Days. For how many days during the year is the day a multiple of the month? 7. What is the least positive multiple of 72 that has exactly 16 positive factors? 8. Count Off. At a school, the children lined up in a long row and counted off. A hat was given to the sixteenth child in line and every sixteenth child after that. A noisemaker was given to the twentyfourth child and to every twentyfourth child after that. What were the positions in line of the first three children who received both a hat and a noisemaker? 9. Explain why the least common multiple of any two numbers is divisible by the greatest common factor of the numbers. 10. Unusual Age. My age is a multiple of 7. Next year, it will be a multiple of 5. When will this event occur again? 11. In Their Prime Jonathan is two years younger than his wife Hazel. Their current ages are both prime numbers. Next year Hazel's age will be a multiple of 11, and Jonathan's age will be the product of two consecutive numbers. How old are Jonathan and Hazel now? Remember to explain how you got your answer and show how you know you are right. Extra: Two prime numbers that differ by two (like Jonathan and Hazel's current ages) are called twin primes. They are always separated by a single even number. Find all the twin primes less than 100. Starting with the twin primes 5 and 7, what is the greatest common factor that those single even numbers all share? Explain how you found your answers. 1
2 12. Wooden Legs Wendy builds wooden dollhouse furniture. She uses the same kind of legs to make 3 legged stools and 4legged tables. She has a supply of 31 legs and wants to use them all to make stools and tables. Find all the possible ways she can use all 31 legs. Explain how you solved the problem and how you know you have found all solutions. Extra: Wendy sells her furniture to the local toy store. She gets $2 for each stool and $3 for each table. Of all the ways you found, which would earn her the most money? 2
3 Solutions 1. The greatest common factor (GCF) of 8 and 20 is 4, and the least common multiple (LCM) of 8 and 20 is 40. Their product is 4 x 40 = 160. Note that 8 x 20 = 160, also. In general, the GCF(a, b) x LCM(a, b) = a x b. 2. There are four multiples of 89 between 600 and 900: 7 x 89 = 623, 8 x 89 = 712, 9 x 89 = 801, and 10 x 89 = 890. The sum of any three consecutive integers is always a multiple of 3. Since 89 is prime and 9 is the only multiple of 3 in our list, the sum of the three consecutive integers can only be 801. Incidentally, the three consecutive integers are 266, 267 and There are 16 multiples of 3 less than 50 and 12 multiples of 4. We have counted the 4 multiples of 12 twice, however, so we must subtract 4. That s = 24 so far. Now we need to exclude any multiples of five we might have counted. Those would be the 3 multiples of 15 (15, 30 and 45) and the 2 multiples of 20 ( 20 and 40). That leaves = 19 integers less than 50 that are multiples of 3 or 4, but not We know that we are numbering houses from 4 through 120, but using only the multiples of four. We could list these out, but let s see if we can count them quickly. Since 4 = 4 x 1 and 120 = 4 x 30, we are working with the first 30 multiples of 4. Unfortunately, some are onedigit integers, others are twodigit integers and still others have three digits. Let s group them accordingly. (It helps to know that 100 is the first threedigit multiple of 4.) 1digit 2digit 3digit 4x1 = 4 4x3 = 12 4x25 = 100 4x2 = 8 4x24 = 96 4x30 = mult = 22 mult = 6 mult. The two onedigit multiples account for a total of two digits; there are 22 twodigit multiples, resulting in 22x2 = 44 digits; and there are six threedigit multiples yielding 6x3 = 18 digits. This is a total of = 64 digits, which will cost 64 x $0.50 = $ Any common multiple will work, however, the LCM is smallest. This makes multiplication easier. 3
4 6. Multiple Days. Ninety times. Consider this on a monthbymonth basis. Month Number Number of Multiples We know that 72 = 2 3 x 3 2, which means that there are 4 x 3 = 12 factors. If the multiple of 72 included just one other prime factor different from 2 and 3, it would follow that this multiple m could be represented by m = (2 3 x 3 2 ) x p, and it would have 4 x 3 x 2 = 24 factors. Bringing a new prime factor into the picture creates too many factors for the multiple. Therefore, let s try to multiple (2 3 x 3 2 ) x 2 = 2 4 x 3 2, which has 5 x 3 = 15 factors. This doesn t satisfy our condition, so we re left with trying (2 3 x 3 2 ) x 3 = 2 3 x 3 3, which has 4 x 4 = 16 factors. Our multiple is 2 3 x 3 3 = Count Off. Number 48, 96, and 144 are the first three common multiples of 16 and Since the LCM is a multiple of the two numbers, both numbers are factors of the LCM. The GCF is a factor of both numbers and thus a factor of the LCM. 10. Mult. of In 1 year In 35 years Now 11. To find all the prime numbers between 1 and 100, I could get a hundreds chart with all the numbers that are NOT prime numbers crossed out. Then I started to look at the chart to see what prime numbers would fit the clues. I saw that both ages were prime numbers and Jonathan was two years older than Hazel. So, I needed to find two prime numbers that were two apart. Since there was more than one pair of two prime numbers, I decided to go to the next clue. I saw that next year, Hazel s age had to be a multiple of 11. Then this year, her age would be one year less than that multiple of 11. I did 2 x 11 = 22, and then 221 = 21, but 21 is not a prime number. Then I did 3 X 11 = 33, and then 331 = 32, but 32 is not a prime number. Then I did 4 X 11 = 44, and then 441 = 43, and 43 is a prime number. I decided to make Hazel s age 43, and see if the rest of the clues would work = 41, and 41 is a prime number. So, Jonathan could be 41 years old. The last clue was that next year, Jonathan s age will be the product of two consecutive numbers. If Jonathan was 41 this year, next year he would be x 7 = 42, and 6 and 7 are consecutive numbers. So the last clue worked. Since all the 4
5 clues worked, I know I am right. Extra: I made a chart of all the twin primes, and the number in between them. I looked at the chart I used for the first part to find them. Twin Primes Number between I started with 6 and found 1, 6, 2, and 3 for factors. Since 6 is the biggest factor, I started with 6 to see if it is a factor of all the other numbers. Next is 12, and 6 x 2 = 12. So 6 would work. Next is 18, and 6 x 3 = 18. So 6 would work. Next is 30, and 6 x 5 = 30. So 6 would work. Next is 42, and 6 x 7 = 42. So 6 would work. Next is 60, and 6 x 10 = 60. So 6 would work. Next is 72, and 6 x 12 = 72. So 6 would work. So, the greatest common factor that the even numbers between prime twins above 5 and 7 share is 6. I noticed that up to 10, there were only prime numbers multiplied by 6. There were 2, 3, 5, and 7. But the numbers that weren t prime weren t there. 4, 6, 8, and 9. I don t know why, but it was neat to see that. If you know why will you please tell me? 12. Key: x = incorrect data Tables Stools Combination 1x4=4 314=27/3=9 1 table, 9 stools 2x4=8 318=23 x 3x4= =19 x 4x4= =15/3=5 4 tables, 5 stools 5x4= =11 x 6x4= =7 x 7x4= =3/3=1 7 tables, 1 stool Explanation: Each table I multiplied by 4 and subtracted the product from 31, if the difference was a multiple of 3, I had an answer. The pattern is that tables go up by 3s because stools have 3 legs and stools go down by 4s because tables have 4 legs. The number of legs go up or down by 12 because 12 is 4x3 (4 legs times 3 legs). The combinations that work have to combine multiples of 4 and 3 to make 31. Extra: The first correct combination of numbers equals $21, the second equals $22 and the third equals $23,making it the largest number. 5
6 (1 table x $3 = $3) + (9 stools x $2 = $18) = $21 (4 tables x $3 = $12) + (5 stools x $2 = $10) = $22 (7 tables x $3 = $21) + (1 stool x $2 = $2) = $23 6
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