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Carlisle Math Team Meet #1 – Category 3 M1C3 Number Theory Self-study Packet 1. Mystery: Problem solving 2. Geometry: Angle measures in plane figures including supplements and complements 3. Number Theory: Divisibility rules, factors, primes, composites 4. Arithmetic: Order of operations; mean, median, mode; rounding; statistics 5. Algebra: Simplifying and evaluating expressions, solving equations with 1 unknown including Identities. For current schedule or information, see http://www.imlem.org A Prime Number can be divided evenly only by 1, or itself. A Composite Number can be divided evenly by numbers other than 1 or itself. Number 1 2 3 4 5 6 7 8 9 10 Can be Evenly Prime, or Divided By Composite? (1 is not considered prime or composite) 1,2 Prime 1,3 Prime 1,2,4 Composite 1,5 Prime 1,2,3,6 Composite 1,7 Prime 1,2,4,8 Composite 1,3,9 Composite 1,2,5,10 Composite Factors "Factors" are the numbers you multiply to get another number: QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. “How many factors” problems How many factors are there of 24? Answer: You can list them, making pairs whenever you find a factor: 1x24, 2x12, 3x8, 4x6. So, there are 8 factors. In Meet#2, when we do prime factorization, we learn a short-cut for counting the factors of a number. PRIME NUMBER CHART TO 1000 2 31 73 127 181 3 37 79 131 191 5 41 83 137 193 7 43 89 139 197 11 47 97 149 199 13 53 101 151 211 17 59 103 163 223 19 61 107 167 227 23 67 109 173 229 29 71 113 179 233 “Distinct (different) prime factors” problems What are the distinct prime factors of 100? Answer: 2 and 5. That’s all! Distinct means we only list each prime factor once, even though there are many of them in the number. 100 is 2x5 x 2x5. List all the distinct prime factors of 100,000. 2 and 5. What are the distinct prime factors of 300,000,000? 2, 3, and 5. “Finding an unknown digit” problems The number 8R3 is a multiple of 9. What could R be? Answer: Any multiple of 9 has the sum of digits as 9, so 8+R+3 = 11+R = multiple of 9. What values of R (0 through 9) make this true? 11+7=18, so R=7. 873 is indeed a multiple of 9. Greatest / Smallest Primes in a Region What is the largest prime less than 30? Answer: 29 What is the smallest prime greater than 100? Answer: 101 (We need to quickly test numbers to see if they’re prime or composite.) The Divisibility Rules These rules let you test if one number can be evenly divided by another, without having to do too much calculation! A number is divisible by: 2 If: The last digit is even (0,2,4,6,8) Example: 128 is 129 is not 381 (3+8+1=12, and 12÷3 = 4) Yes 3 The sum of the digits is divisible by 3 4 The last 2 digits are divisible by 4 5 The last digit is 0 or 5 6 The number is divisible by both 2 and 3 7 217 (2+1+7=10, and 10÷3 = 3 1/3) No 1312 is (12÷4=3) 7019 is not 175 is 809 is not 114 (it is even, and 1+1+4=6 and 6÷3 = 2) Yes 308 (it is even, but 3+0+8=11 and 11÷3 = 3 2/3) No 672 (Double 2 is 4, 67If you double the last digit and subtract it from the rest of the number 4=63, and 63÷7=9) Yes and the answer is divisible by 7 or 0. 905 (Double 5 is 10, 9010=80, and 80÷7=11 3/7) (Note: you can apply this rule to that No answer again if you want) 109816 (816÷8=102) Yes 8 9 10 11 12 The last three digits are divisible by 8 216302 (302÷8=37 3/4) No The sum of the digits are divisible by 9 1629 (1+6+2+9=18, and again, 1+8=9) Yes (Note: you can apply this rule to that answer again if you want) 2013 (2+0+1+3=6) No The number ends in 0 If you sum every second digit and then subtract the other digits and the answer is divisible by 11 or 0 The number is divisible by both 3 and 4 220 is 221 is not 7392 ((7+9) - (3+2) = 11) Yes 25176 ((5+7) - (2+1+6) = 3) No 648 (6+4+8=18 and 18÷3=6, also 48÷4=12) Yes 916 (9+1+6=16, 16÷3= 5 1/3) No “Divisibility” problems: Is 123 is divisible by 3? Yes, because 1+2+3=6 and 6 is a multiple of 3. Is 765,000 divisible by 3? Yes, because 7+6+5+0+0+0 = 18 and 18 is divisible by 3. Is 212 is divisible by 4? Yes, because 12 is. Is 77728 divisible by 4? Yes, because 28 is (4*7). What are the prime factors of 30? Answer: 2, 3, and 5. NOT 6 or 15, as they aren’t prime. Is 91 composite? Answer: Yes! 91 = 7x13. This, and 1001, are often tricky. Category 3 Number Theory Meet #1, October 2004 1. How many pairs of primes have a sum of 24? Hint: List the primes up to 24: 2 3 5 7 11 __ __ __ __ Now look for two that add to 24. Keep looking. 2. If the five-digit number 5N82N is divisible by 18, what is the value of N? Hint: Any multiple of 18 is even. So, what could the low digit be? If it’s a multiple of 9, try the sum of the digits. 3. How many of the positive factors of 660 are odd? Hint: List all 12 factor pairs, such as 1x660, 2x330, 3x220, 4x____ , … and count the odd factors. Answers 1. _______________ 2. _______________ 3. _______________ Solutions to Category 3 Number Theory Meet #1, October 2004 Answers 1. There are three (3) pairs of primes that have a sum of 24. They are: 19 + 5, 17 + 7, and 13 + 11. 1. 3 2. 6 3. 8 2. For the five-digit number 5N82N to be divisible by 18, it must pass the divisibility tests for both 2 and 9. Since there is an N in the units place (ones place), we know that N has to be even. To be divisible by 9, the sum of the digits must be a multiple of 9. Currently the sum of the digits is 5 + N + 8 + 2 + N = 15 + 2N. For N = 0, we get 15 + 2 0 = 15, which is not a multiple of 9. For N = 2, we get 15 + 2 2 = 19. For N = 4, we get 15 + 2 4 = 23. For N = 6, we get 15 + 2 6 = 27, which is a multiple of 9. This tells us that the five-digit number 56826 is divisible by 18. (56826 ÷ 18 = 3157) Note that this is the only single-digit value of N that will make 15 + 2N equal a multiple of 9. Thus, N = 6. 3. The factor pairs of 660 are listed below. The eight (8) odd factors are in bold. 1 660 2 330 3 220 4 165 5 132 6 110 10 66 11 60 12 55 15 44 20 33 22 30 Category 3 Number Theory Meet #1, October 2003 1. 9900 2 a 3b 5 c 7 d 11e Find the value of a b c d e . Hint: Prime factor 9900: Divide by 2 until you can’t. The number of 2’s is a. Now divide what’s left by 3 until you can’t. The number of 3’s is b, etc. 2. Find the sum of all possible values of the digit N such that the 5-digit number 318N4 is divisible by 12. Hint: 12 is divisible by 3 and 4, so 318N4 is too. 3 tells you about sum of digits. Is 14 divisible by 4? 24? 34? 44? 3. Melanie’s locker number is the product of the least pair of consecutive primes that have a difference of 6. What is Melanie’s locker number? Hint: Make a list of primes until you find two differing by 6. 2 3 5 7 11 __ __ __ __ __ __ Consecutive means ‘next to.’ Answers 1. _______________ 2. _______________ 3. _______________ Solutions to Category 3 Number Theory Meet #1, October 2003 Answers 1. 9900 2 2 32 5 2 7 0 111, so a b c d e = 2 2 2 0 17. 1. 7 2. 10 3. 667 2. A number that is divisible by 12 must be divisible by both 3 and to be divisble by 3, the sum of its 4. For the 5-digit number 318N4 digits must be a multiple of 3. 31 8 4 16, so N would have to be 2 or 5 or 8 to make a multiple of 3. This gives us 31824, 31854, and 31884 to consider. To determine which among these is divisible by 4, we need only check the last two digits of each number, since any multiple of 100 is divisible by 4. 24and 84 are multiples of 4, but 54 is not. Thus only 31824 and 31884 are divisible by 12 and the sum of all possible values of N is 2 + 8 = 10. 3. The least pair of consecutive primes with a difference of 6 is the pair 23 and 29. Their product is 667, so Melanie’s locker number must be 667 Category 3 Number Theory Meet #1, October, 2002 1. If the five-digit number 837A5 is known to be divisible by 15, what is the sum of all the possible values of the digit A? Hint: Any multiple of 15 is also a multiple of 3 and 5. Adding digits should give multiple of 3. 2. Twin primes are two primes that have a positive difference of 2, such as 11 and 13. What is the sum of the least twin primes greater than 50? Hint: Start listing primes > 50: 53 59 __ __ __ until you find Twins 3. Find the value of n, if n is a natural number, n is less than 100, n is the product of two primes, the sum of the digits of n is 10, and the positive difference between the two prime factors of n is a multiple of 13. Hint: Natural number = 1 2 3 … Sum of digits tells you 19,91 or 28,82, … Now check for product of 2 primes. Answers 1. _______________ 2. _______________ 3. _______________ Solutions to Category 3 Number Theory Meet #1, October, 2002 Answers 1. 12 2. 144 or 120 3. 82 1. To be sure a number is divisible by 15, it must be divisible by both 3 and 5. The five-digit number 837A5 is definitely divisible by 5, since its units digit is a 5. If it is to be divisible by 3, the sum of its digits must be a multiple of 3. The known digits have a sum of 8 + 3 + 7 + 5 = 23. This means that A can have the values 1, 4, or 7. The sum of these possible values is 1 + 4 + 7 = 12. 2. There are no twin primes in the fifties or the sixties. Thus the least twin primes greater than 50 are 71 and 73. Their sum is 144. Editor note: The original meet had 144 as the answer, but 59 and 61 are also twin primes, so the correct answer should be 120. Thanks to Zhiping You for pointing this out. 3. There are too many possible products of two primes to start with that clue. The natural numbers less than 100 with a sum of digits equal to 10 are 19, 28, 37, 46, 55, 64, 73, 82, and 91. Of these, four of these are the product of two primes: 46 = 2 • 23, 55 = 5 • 11, 82 = 2 • 41, and 91 = 7 • 13. Since the difference between the prime factors of 82 is 41 – 2 = 39 and 39 is a multiple of 13, n must be 82. Category 3 Number Theory Meet #1, October, 2001 1. What is the greatest possible difference of two primes if both of those primes are between 50 and 100? Hint: What is the lowest prime > 50, and the greatest prime < 100? Is 51 divisible by 3? 2. Use the following clues to find the value of n: n is a whole number n is divisible by 24 n is greater than 300 n is not divisible by 17 n is less than 400 n is divisible by 21 the sum of the digits of n is 12 Hint: If it’s divisible by 24 and by 21=3*7, then it’s also divisible by 24*7=___. 3. Find the sum of all the different prime factors of 364,000. Answers 1. _____________ Hint: Do the factors of 364 and 1000 separately. 10 = 2*5, and 2. _____________ 3. _____________ different prime factors of 1000 are still only 2 and 5. What are the prime factors of 364? After you divide by 2 a couple of times, is what’s left divisible by 7? Solutions to Category 3 Number Theory Meet #1, October, 2001 Answers 1. 44 2. 336 1. To get the greatest possible difference, we will want to find the prime closest to 100 and subtract the prime closest to 50. 99 9 11, so it’s not prime, and 51 3 17 , so it’s not prime either. Both 97 and 53 are prime, though, so our greatest difference is 97 53 , which is 44. 3. 27 2. That n is divisible by 24 and 21 are the most helpful clues 3 to start with. Since 24 2 3 and 21 3 7 , we know 3 that n is divisible by 2 , 3, and 7, which are relatively prime. 3 This means that n is a multiple of 2 3 7 168 . The only multiple of 168 that is between 300 and 400 is 2 168 336 , so that must be it. Checking the other clues, we know that 336 cannot be divisible by 17, since we created it by multiplying 2 2 3 7 and 17 is a different 3 prime. The final clue says that the sum of the digits is 12 and this is true of our answer 336. 3. 364,000 364 1000 and we know that 1000 23 53 . Now we need to look for other prime 2 2 factors in 364: 364 2 182 2 91 2 7 13 . So the different prime factors of 364,000 are 2, 5, 7, and 13. The sum of these numbers is 27. Category 3 Number Theory Meet #1, October, 2000 1. For how many positive integral values of n will be a whole number? Hint: 168/n is a whole number if n is a factor of 168. “positive integral values” means 1, 2, 3, … Try listing all the factors of 168. 2. What is the greatest integer that will always divide the product of four consecutive integers? Hint: Try 1*2*3*4 and make a guess, then check with 2*3*4*5 etc. Or, reason using divisibility rules. 3. Find the sum of all the positive integers less than 1000 that are both perfect squares and perfect cubes. Hint: Since 103=1000, there are only 9 perfect cubes<1000 so list and see which are also squares: 1 3=1, 23=8, 33=27, 43=64, 53=___, 63=___, 73=___, 83=___, 93=___. Don’t forget 1! Answers 1. _____________ 2. _____________ 3. _____________ Solutions to Category 3 Number Theory Meet #1, October, 2000 Answers 1. 16 2. 24 3. 794 1. The positive integral values of n that will make a whole number are the factors of 168. (By “positive integral values” we mean positive integers or whole numbers.) We can simply list all the factors of 168 in pairs: , , , , , , , and . Those are the 16 values of n that will make a whole number. Alternatively, there is a trick for figuring out the number of factors of a number without actually listing them. We express 168 in prime factors with exponents as follows: . We then raise each exponent by one and multiply them like this: . 2. The product of four consecutive integers is guaranteed to contain a multiple of 3, and two multiples of 2, one of which is also a multiple of 4. Being careful not to count the same factor of 2 as both a multiple of 2 and a multiple of 4, we can be certain of only three factors of 2 along with the one factor of 3. From these factors we obtain the product, so 24 will always divide the product of four consecutive integers. Some people will find this answer simply by multiplying the first four counting numbers: . 3. Numbers that are both perfect squares and perfect cubes are sixth powers, because. Since the cube root of 1000 is 10, we need only values under 10 for . Now if is under 10, then a must be 3 or less. The three sixth powers under 1000 are: , , and . Their sum is .