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Number: Number Theory Skipton Girlsβ High School Objectives: Have an appreciation of properties of integers (whole numbers), including finding the Lowest Common Multiple, Highest Common Factor. Extra: Divisibility rules. For Teacher Use: Recommended lesson structure: Lesson 1: Introduction to Number Theory/Sums of primes+squares problems. Lesson 2: Prime Factorisation Lesson 3: LCM/HCF Lesson 4: Uses of Prime Factorisations Lesson 5+6: Divisibility Rules Extension: Divisibility of Terms/Within Equations Go > Go > Go > Go > Go > Go > Starter Key Skill: You should try to memorise these. List the following numbers in your books. The first 16 square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ? 225, 256 121, 144, 169, 196, The first 8 cube numbers: 1, 8, 27, 64, 125, 216, 343, 512 The prime numbers up to 40: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ? 37 The first 10 triangular numbers: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66 ? (e.g. 3 is a triangular number as you can form a triangle using 1 dot on the first row and 2 on the next) ? If you finish: A βperfectβ number is a number who factors (excluding itself) add up to itself. For example. The factors of 6 (excluding 6) are 1, 2, 3, and 1 + 2 + 3 = 6. Find the first perfect number after 6. Key Fact: All perfect numbers are triangular numbers. Solution: ππ = π + π +?π + π + ππ Other numbers you might want to investigate yourself: Tetrahedric numbers, Fibonacci numbers. Key Terms ! ! ? Integer: A whole number. Positive integer: An integer that ? is at least 1. Non-negative integer: An integer that ? is at least 0. Perfect square: For integers, just ?a square number. Divisor: Another word ? for factor. Composite: The opposite of prime: ? has other factors. Distinct integers: Numbers which ?are different! Key Note: A βperfect squareβ more generally refers to βsomething squaredβ, which can be an algebraic expression. For example π₯ + 1 2 is a βperfect squareβ, but is not necessarily a square number, e.g. if π₯ = 0.5 (as 1.52 = 2.25) Puzzles involving sums of primes/squares/β¦ Example: Goldbachβs Conjecture (as of current, unproven!) states that all even numbers greater than 2 are the sum of two primes. Key Tip: It often How many ways are there of expressing 100 as the sum of two primes? helps to write Solution: 6 (3 + 97, 11 + 89, 17 + 83, 29 + 71, 41 + 59, 47 + 53) out your ? numbers of interest (primes, squares, β¦) first. Further Example: The Indian mathematician Ramanujan once famously noted that the 1729 number of a taxi ridden by his friend Hardy: βis a very interesting number; it is the smallest integer expressible as a sum of two different cubes in two different waysβ. What is the smallest integer (not necessarily a square) that is expressible as the sum of two distinct squares in two different ways? (Hint: 1 is used in one of the sums) ππ = ππ + ππ = ππ + π (Side note: the smallest square number expressible as the sum of two squares in two different ways is πππ = π?π + πππ = πππ + πππ) Exercise 1 (Problems on provided sheet) 1 [JMC 2015 Q11] What is the smallest prime number that is the sum of three different prime numbers? A 11 B 15 C 17 D 19 E 23 Solution: D ? 2 [JMO 1999 A2] In how many different ways can 50 be written as the sum of two prime numbers? (Note: π₯ + π¦ and π¦ + π₯ do not count as different.) Solution: 4 ways (ππ + π, ππ ? + π, ππ + ππ, ππ + ππ) 3 [JMO 2009 A3] The positive whole numbers π, π and π are all different and π2 + π2 + π 2 = 121. What is the value of π + π + π? Solution: 17 ? Key Tip: Use your lists of numbers from the starter. Exercise 1 4 [JMC 2015 Q19] One of the following cubes is the smallest cube that can be written as the sum of three positive cubes. Which is it? A 27 B 64 C 15 D 216 E 512 Solution: D 5 [JMC 2006 Q20] The sum of three different prime numbers is 40. What is the difference between the two biggest of these numbers? A 8 B 12 C 16 D 20 E 24 Solution: E 6 ? Important Note: If three numbers sum to an even number, they canβt all be odd. But 2 is the only even number, so must be one of the numbers. [JMC 2010 Q22] Kiran writes down six different prime numbers, π, π, π, π , π‘, π’, all less than 20, such that π + π = π + π = π‘ + π’. What is the value of π + π? A 16 B 18 C 20 D 22 E 24 Solution: E 7 ? ? [TMC Regional 2009 Q9] 12345 can be expressed as the sum of two primes in exactly one way. What is the larger of the two primes? ? Solution: 12343. Note that odd = odd + even only. Thus one of the two primes must be 2. Exercise 1 N [JMO 2006 A9] The prime number 11 may be written as the sum of three prime numbers in two different ways: 2 + 2 + 7 and 3 + 3 + 5. What is the smallest prime number which can be written two different ways as the sum of the three prime numbers which are all different? Solution: 23 ? N [JMO 2014 B6] The sum of four different prime numbers is a prime number. The sum of some pair of the numbers is a prime number, as is the sum of some triple of the numbers. What is the smallest possible sum of the four prime numbers? ? Prime Factorisation To find the prime factorisation of a number is to express it as a product of prime numbers. ?5 30 = 2 × 3 × 12 = 22 × 3? Key Tip: While 12 = 2 × 2 × 3 is also correct, we can use βindex notationβ to group prime factors together that are the same. = 23 × ?3 × 5 120 20 4 2 6 5 2 2 3 We can use a βtreeβ to help us with the working. For each number, find two numbers the multiply to give it. If you get to a prime, we canβt branch out further, so we have a βleafβ. Itβs helpful to circle the leaves. Another quick example 2250 = 2 ? 2×3 ×5 2250 225 9 3 5 3 10 Possible Tree 5 5 2 ? 45 3 Check Your Understanding Using a tree, find the prime factorisation of 1350. When done, try coming up with more trees. What do you notice about the final result in each case? 1350 = 2 × 33? × 52 1350 1350 10 2 5 135 5 27 5 3 9 3 270 3 90 Trees Some Possible ?3 30 3 6 5 2 3 We always end up with the same leaves each time, and hence the same factorisation. Fundamental Law of Arithmetic/Unique Factorisation Theorem: Every positive integer can be uniquely expressed as a product of primes. Prime Factorising a number already in index form Sometimes you might have a number with powers, but the base (the big number) is not prime. How would you prime factorise this? What if a base was repeated? 103 = 2 × 5? 3 = 2 × 5 × 2 ×? 5 × 2 × 5 = 23 × 53 ? Quickfire Questions: 5 π ?π 6 =π ×π 21100 = ππππ ×? ππππ 1515 = πππ × π?ππ 704 = ππ × ππ?× ππ 55 × 52 = π π ? 720 × 720 = πππ? 2 3 × 24 = 2 × 2 × 2 × ?2 × 2 × 2 × 2 = 27 ? Bro Note: This is an example of a βlaw of indicesβ, which you will learn more about in Year 8. N: Working: 910 = 310 × 310 = 320 910 = πππ ? 810 = πππ ? 10050 = ππππ ×? ππππ 1212 = πππ × ?πππ 185 = ππ × π?ππ 12 × 2100 = π × ππππ ? Exercise 2 1 By drawing a tree of otherwise, find prime factorisations (in index form) for the following numbers. 28 = ππ × π 75 = π × ππ 1000 = ππ × ππ 378 = π × ππ × π 396 = ππ × ππ × ππ 1755 = ππ × ππ × ππ 432 = ππ × ππ ? ? ? ? ? ? ? 2 Is 137776 odd or even? An odd number to any power is always odd. 3 Put in prime factorised form: 35 × 36 = πππ 22 × 33 × 24 × 35 = π π × π π 147 = ππ × ππ 3658 = ππ × πππ 5555 = πππ × ππππ 6 × 320 = π × πππ ? ? ? ? ? What are the factors of 25 ? Give your answers in index form. π, ππ , ππ ,? ππ , ππ , ππ Prime factorise the following: 10 000 = ππ × ππ 257 = πππ 1830 = πππ × πππ 46 × 86 = πππ 1 000 000100 = ππππ × ππππ ? ? N1 ? ? 4 5 ? ? ? Suppose 1 was considered to be a prime number. Explain why this violates the Fundamental Law of Arithmetic. For example, 6 could be expressed as π × π or π × π × π or π × π × π × π. But FLA states there is a unique factorisation for each integer. Thus 1 is not prime. ? N2 [TMC Regional 2012 Q4] Find the sum of all numbers less than 120 which are the product of exactly three different prime factors. Solution: 717 ? Starter List the factors of 35 35 divides for example by 32 because (keeping your factors in prime factorised form) You will learn in Year 8 that 30 = 1, so this is consistent with the pattern. 35 32 = 3×3×3×3×3 3×3 = 3 × 3 × 3 = 33 1, 31 , 32 , ?33 , 34 , 35 One number will be a factor of another if the prime factor(s) are the same but the powers are smaller (or equal). List a few multiples of 35 which only contains prime factors of 3. (keeping your multiples in prime factorised form) 35 , 36 , 3?7 , β¦ One number will be a multiple of another if the prime factor(s) are the same but the powers are greater (or equal). Lowest Common Multiple/Highest Common Factor Multiples of 8: Multiples of 12: 8, 16, 24, ?32, β¦ 12, 24, 36,? β¦ Lowest Common Multiple of 8 and 12: 24 ? For small numbers, we can list out multiples of the larger number until we see a multiple of the smaller number. Factors of 8: Factors of 12: 1, 2, 4, 8 ? 1, 2, 3, 4,?6, 12 Highest Common Factor of 8 and 12: 4? For small numbers, we can list out factors of each number and choose the greatest number which is common. Check Your Understanding ? πΏπΆπ 60,72 = πππ π»πΆπΉ 60,72 = ππ? Key Shortcut: Any multiple of 60 ends with a 0. Therefore the multiple of 72 must be x5, x10, β¦ Key Shortcut: Any number which goes into 60 and 72 must also go into their difference! (i.e. 12) πΏπΆπ 12,21 = ππ? π»πΆπΉ 12,21 = π ? But what about bigger numbers? 792, 378 Sometimes itβs not practical to use this method. Can we use the prime factorisation somehow? 3 2 2 ? 3 792 = × × 11 3? 378 = 2 × 3 × 7 But what about bigger numbers? 792 = 23 × 32 × 11 378 = 2 × 33 × 7 Method: βWhat wins what losesβ Alternative: Venn Diagram Method 792 = 23 × 32 × 11 378 = 2 × 33 × 7 792 Step 1: Align numbers so that each prime factor has its own column. 3 For HCF, out of 2 and 2, what factor is common to both? We saw from earlier that 2 is a factor of 2 and 23 . We effectively find what βlosesβ out of 2 and ππ . In this case 2. We see what βlosesβ in each column (where βnothingβ always loses against βsomethingβ) π»πΆπΉ 792,378 = π × ππ = ππ For LCM, what is both a multiple of 2 and 23 ? Again, from earlier, they both go into 23 , i.e. the one that βwinsβ. Repeating for the other numbers: πΏπΆπ 792,378 = ππ × ππ × π × ππ = ππππ 378 2 2 7 3 3 3 11 2 Step 1: Find the prime factors common to both numbers. Step 2: Fill in the remaining prime factors of each number. HCF is product of numbers in the intersection. π × π × π?= ππ LCM is all numbers multiplied. ππ × ππ × π ? × ππ = ππππ More Examples 672 = 25 × 32 × 7 72 = 23 × 32 588 = 22 × 3 × 72 1232 = 24 × 7 × 11 π³πͺπ΄ πππ, ππ = ππ × ππ × π = ππππ π―πͺπ πππ, ππ = ππ × π = ππ Line numbers up: πππ = ππ × π × ππ ππππ = ππ × π × ππ ? (note that if thereβs a βdrawβ, both win and both lose) ? π³πͺπ΄ πππ, ππππ = ππ × π × ππ × ππ = πππππ π―πͺπ πππ, ππππ = ππ × π = ππ Check Your Understanding ?π 1936 = ππ × ππ 792 = ππ × ππ? × ππ ? π = πππππ πΏπΆπ = ππ × ππ × ππ π»πΆπΉ = ππ × ππ? = ππ If you finishβ¦ 3675 = π × ππ ×? ππ 875 = ππ ×?π πΏπΆπ = π × ππ × π?π = πππππ π»πΆπΉ = ππ × π ?= πππ Exercise 3 1 Find the LCM and HCF of the following pairs of numbers (using any suitable method). 6 and 8 HCF = 2, LCM = 24 13 and 5 HCF = 1, LCM = 65 12 and 15 HCF = 3, LCM = 60 21 and 35 HCF = 7, LCM = 105 ? ? ? ? 2 The K4 bus comes every 9 minutes. The K3 bus comes every 12 minutes. If they both come at 9am, at what time will they next arrive at the same time? 9:36am ? 3 Find the LCM and HCF of the following pairs, by prime factorising the numbers first. a) 36 and 378 LCM = 756, HCF = 18 b) 315 and 3675 LCM = 11025, HCF = 105 c) 72 and 66 LCM = 792, HCF = 6 d) 2880 and 792 LCM = 31680, HCF = 72 e) 375 and 325 LCM = 4875, HCF = 25 f) 252 and 2079 LCM = 8316, HCF = 63 ? ? ? ? ? ? N [JMC 2009 Q18] Six friends are having dinner together in their local restaurant. The first eats there every day, the second eats there every other day, the third eats there every third day, the fourth eats there every fourth day, the fifth every fifth day and the sixth eats there every sixth day. They agree to have a party the next time they all eat together there. In how many daysβ time is the party? 60 days ? Number: Divisibility Rules Divisibility Rules How can we tell if a number is divisible by: ! 2 3 4 5 6 ? Digits add up to multiple of 3. e.g: ? 1692: 1+6+9+2 = 18 οΌ Last two digits are divisible by 4. e.g. ? 143328 Last digit is 0 or 5. ? Number is divisible by 2 and 3 (so use ? tests for 2 and 3). Last digit is even. 7 Double the last digit and subtract it from the remaining number, and see if the result is divisible by 7. e.g: 2464 -> 246 β 8 = 238 -> 23 β 16 = 7. 8 Last three digits divisible by 8. 9 Digits add up to multiple of 9. 10 Last digit 0. 11 When you sum odd-positioned digits and subtract even-positioned digits, the result is divisible by 11. ? ? ? ? ? e.g. 47949: (4 + 9 + 9) β (7 + 4) = 22 β 11 = 11, which is divisible by 11. 12 Number divisible by 3 and by 4. ? Quickfire Divisibility 4 6 7 ? οΌ? ? 726 168 οΌ? οΌ? οΌ? 9196 οΌ? ? ? 252 οΌ? οΌ? οΌ? ? ? οΌ? 1001 ? ? οΌ? 91 216 οΌ? οΌ? ? 87912 οΌ? οΌ? ? 9 11 ? οΌ? ? ? ? οΌ? οΌ? ? ? οΌ? ? ? ? οΌ? οΌ? οΌ? Quickfire Mental Primes Apart from the obvious instant checks (divisibility by 2, 5), we usually only have to mentally check 3, 7 and 11 to have a good βguessβ that a number is prime. 3 91 101 234567 οΌ 131 781 751 221 7 οΌ? 11 Is it prime? No ? Yes ? ? ? ? ? οΌ ? ? ? ? ? Yes ? No! (13? × 17) No Yes No Advanced: 13 trick is βQuadruple last digit and add to remaining number. Is result divisible by 13? N For 221, what is the largest prime we would have had to test divisibility until weβd be certain it was prime? Up to πππ = ππ. ππ because all composite numbers have a factor (other than 1) up to the square root. ? Test Your Understanding Easier One: [JMO 1997 A5] Precisely, one of the numbers 234, 2345, 23456, 234567, 2345678, 23456789 is a prime number. Which one must it be? Solution: 23456789? Harder One: [JMC 2012 Q23] Peter wrote a list of all the numbers that could be produced by changing one digit of the number 200. How many of the numbers on Peterβs list are prime? A 0 B 1 C 2 D 3 E 4 Solution: A ?