Number Theory Factors & Divisibility LCM and GCF BASE number Modular Arithmetic Even & Odd Numbers

Divisibility Rules A number is divisible by 2n if and only if the n last digits of the number are divisible by 2n. A number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9. A number is divisible by 5n if and only if the last n digits are divisible by that power of 5n. A number is divisible by 11 if and only if the alternating sum of the digits is divisible by 11.

Example Sum of digits = 6+9+6+15=36 multiple of 3 & 9 Which of the following are factors of 123456780? 2, 3, 4,,5,6,8,9,10 Sum of digits = 6+9+6+15=36 multiple of 3 & 9 Ends with 10  multiple of 5 & 10 Last two digits dividable by 4 (22) multiple of 4 Answer: 2, 3, 4, 5, 6=2x3, 9, 10

Example Consider a five digit number ABCDE What is the smallest 5-digit number that is divisible by both 8 and 9? Consider a five digit number ABCDE The smallest will be A = 1, and we prefer the rest of the digits be 0. To be divisible by 9, we need sum of digits be multiple of 9. And the answer is: 10008

Example Note that CDCD = CD * 100 + CD = CD * 101 In the multiplication problem below, A, B, C and D are different digits. What is A + B? ABA X CD --------------- CDCD Note that CDCD = CD * 100 + CD = CD * 101 We get: 101 * CD = CDCD Hence ABA = 101; and A =1, B = 0 Answer: A + B = 1 + 0 = 1

Example 4 + 5 – 9 = 0 yes 9 + 3 – 8 – 5 = -1 no Which of the following is divisible by 11? 1) 495 2) 9835 3) 14806 4) 918291 4 + 5 – 9 = 0 yes 9 + 3 – 8 – 5 = -1 no 1 + 8 + 6 – 4 – 0 = 11 yes 9 + 8 + 9 – 1 - 2 - 1 = 22 yes

Factors Fundamental theorem of arithmetic Every positive integer has a unique prime factorization Example: Find all the prime factors of 120? 120 = 12 * 10 = 3 * 4 * 2 * 5 = 23 * 3 * 5 Answer: 2, 3, 5 Example: Find all the factors of 24? 24 = 23 * 3 Answer: 20, 21, 22, 23, and 3* 20, 3* 21, 3* 22, 3* 23

Sum of Factors Example: Find sum of factors of 24? 24 = 23 * 3 The list of factors are 1, 2, 4, 8, 3, 6, 12, 24 The sum = 1 + 2 + 4 + 8 + 3 + 6 + 12 + 24 = 60 Alternatively, we can calculate the sum: (20 + 21 + 22 + 23) * (30 + 31) = (1 + 2 + 4 + 8) * (1 + 3) = 15 * 4 = 60

Sum of Factors For a number X = al * bm * cn The sum of X’s factors is: (a0 + a1 + … + al) * (b0 + b1 + … + bm) * (c0 + c1 + … + cn) Example: Find the sum of factors of 5!? 5! = 5*4*3*2*1 = 120 = 23 * 3 * 5 The sum of its factors: (20 + 21 + 22 + 23) * (30 + 31) * (50 + 51) = (1 + 2 + 4 + 8) * (1 + 3) * (1 + 5) = 360

LCM and GCF Find the LCM and GCF of 84 and 140 84 = 2 * 2 * 3 * 7 140 = 2 * 2 * 5 * 7 LCM = 2 * 2 * 3 * 5 * 7 = 420 GCF = 2 * 2 * 7 = 28 Use Venn Diagram: 5 3 2, 2, 7 140 84

LCM and GCF Find the product of LCM and GCF of 45 & 105 45 = 3 * 3 * 5 45 = 3 * 3 * 5 105 = 3 * 5 * 7 LCM = 3 * 3 * 5 * 7= 315 GCF = 3 * 5 = 15 LCM * GCF = 315 * 15 = 4725 Note that: 45 * 105 = 4725 also!!!

Example The GCF for a pair of numbers is 18, and their LCM is 180. If one of the number is 90, what is the other number? Product of the pair = 18 * 180 Divide by one of the number, we got the other number: (18 * 180) / 90 = 18 * 2 = 36

Example How many integers between 1000 and 2000 have all three of the numbers 15, 20 and 25 as factors? A number with 15, 20 and 25 as factors must be divisible by their LCM. 15 = 3 x 5, 20 = 22 x 5, and 25 = 52 LCM(15, 20,25) = 22 x 3 x 52 = 300 Between 1000 and 2000, there are 3 numbers that are multiple of 300: 1200, 1500, 1800

Example A whole number larger than 2 leaves a remainder of 2 when divided by each of the numbers 3, 4, 5 and 6. What is the smallest such number? The smallest whole number that can be divided by each of 3, 4, 5 and 6 is LCM{3; 4; 5; 6} = 22 x3x5 = 60 The smallest whole number greater than 2 that leaves a remainder of 2 when divided by each of 3, 4, 5 and 6 is then: 60 + 2 = 62

Example Two farmers agree that pigs are worth 300 dollars and that goats are worth 210 dollars. When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary. (For example, a dollar debt could be paid with two pigs, with one goat received in change.) What is the amount of the smallest positive debt that can be resolved in this way? Let P be # of pigs, and G be # of goats. The difference will be: D = 300 * P – 210 * G D = 30 (10 * P - 7 * G) 30 is the GCF of 300 and 210, and D must be a multiple of 30, which can be achieved by P = 5, G = 7. Answer: 30

Base Numbers Examples of different base numbers: 123 = 1 * 102 + 2 * 10 + 1 = 12310 1 hr 2 m 3 sec = 1 * 602 + 2 * 60 + 1 = 12360 ‘D’ = 010001002 = 1 * 26 + 0 * 25 + 0 * 24 + 0 * 23 + 1 * 22 + 0 * 21 + 0 * 20 = 64 + 4 = 6810

Example Find the base 10 value of 7778 Answer: 7778 = 7 * 82 + 7 * 8 + 7 = 511 Find the base 8 representation of 5 * 86 + 2 * 83 + 1 Answer: 5 * 86 + 2 * 83 + 1 = 5 * 86 + 0 + 0 + 2 * 83 + 0 + 0 + 1 = 50020018

Example Find the base 2 value of 2510? 25 = 16 + 8 + 1 = 24 + 23 + 20 Answer: 2510 = 110012 Alternatively, we can calculate: 25 / 2 ----------- r 1 12 / 2 ----------- r 0 6 / 2 ----------- r 0 3 / 2 ----------- r 1 1 / 2 ----------- r 1 We get the answer: 110012

Example How many digits will it take to represent 24210 in base 3? Note that 24310 = 35 = 1000003 Also note that 24210 = 24310 - 110 We got 24210 = 1000003 - 13 = 22222 We need 5 digits in base 3.

Example What is the last digit of 10! in base-9? 10! = 10 * 9 * 8 * … * 1 which is divisible by 9 Therefore in base-9, the last digit must be 0.

Modular Arithmetic Detect repetition, and simply the problem with the remainder. Example: What is the units digit of 24682011 ? Note that 81 = 8; 82 = …4; 83 = …2; 84 = …6; 85 = …8; 86 = …4; … 2; …6; … The units digit repeats after every block of 4. 2011 / 4 = … --------- r 3 The last digit of 24682011 = last digit of 24683 Answer: 2

Modular Arithmetic Detect repetition, and simply the problem with the remainder. Example: What is the units digit of 24682011 ? Note that 81 = 8; 82 = …4; 83 = …2; 84 = …6; 85 = …8; 86 = …4; … 2; …6; … The units digit repeats after every block of 4. 2011 / 4 = … --------- r 3 The last digit of 24682011 = last digit of 24683 Answer: 2

Modular Arithmetic Strategy: Detect repetition, and simply the problem using modular arithmetic. Modular Arithmetic: X * Y (mod M) = X (mod M) * Y (mod M) X + Y (mod M) = X (mod M) + Y (mod M)

Examples What is the units digit of 24682011 ? Note that 81 = 8; 82 = …4; 83 = …2; 84 = …6; 85 = …8; 86 = …4; … 2; …6; … The units digit repeats after every block of 4. 2011 / 4 = … --------- r 3 The last digit of 24682011 = last digit of 24683 Answer: 2

Modular Arithmetic What is the remainder when 30 + 31 + 32 + 33 + … + 32009 is divided by 8? 30 / 8 -------- r1 31 / 8 -------- r3 32 / 8 -------- r1 33 / 8 -------- r3 34 / 8 -------- r1 (30 + 31 + 32 + 33 + … + 32009 )/ 8 = (1 + 3 + 1 + 3 + … + 3) /8 (mod 8) Note that there are 2010/2 = 1005 pairs of (1 + 3) Answer = 1005 * (1 + 3) / 8 = 4 (mod 8)

Even & Odd numbers even_number + even_number = even_number even_number + odd_number = odd_number odd_number + odd_number = even_number even_number * even_number = even_number even_number * odd_number = even_number odd_number * odd_number = odd_number

Example Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? 44 59 38

Example There are one odd and two even numbers showing Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers? 44 59 38 There are one odd and two even numbers showing There must be two odd number and one even number on the other side, all prime numbers! The only even prime number is 2 Hence we get the common sum: 59 + 2 = 61 And the other two numbers: 61 – 44 = 17; 61 – 38 = 23; Answer: (2 + 17 + 23) /3 = 14