[{"data":1,"prerenderedAt":894},["ShallowReactive",2],{"layer:prime-and-composite:deepen":3},{"layer":4,"contentHash":873,"dependencyHashes":874,"approval":888,"releaseId":893},{"schemaVersion":5,"conceptId":6,"locale":7,"depth":8,"revision":5,"title":9,"subtitle":10,"summary":11,"objectives":12,"estimatedMinutes":18,"plate":19,"blocks":40,"sourceIds":868,"reviewStatus":869,"authoring":870},1,"prime-and-composite","en","deepen","Why it all works: proofs about primes","Unique factorisation, the square-root rule, the reasons behind divisibility tests, and Euclid’s endless primes","Prove that every number is built from primes in exactly one way, see a world where that fails, count factors from a factorisation, explain the square-root rule and every divisibility test, follow Euclid’s proof that primes never end, and prove facts about co-primes and twin primes.",[13,14,15,16,17],"Explain why every number above 1 has a prime factorisation and why it is unique (the Fundamental Theorem of Arithmetic).","Use index form to count factors, test for squares and cubes, and decide divisibility.","Explain why trial division and the sieve can stop at the square root.","Explain why the divisibility tests for 2, 3, 4, 5, 6, 8, 9, 10 and 11 work, and use a test for 7.","Retell Euclid’s proof that there are infinitely many primes, and prove simple facts about co-primes and twin primes.",55,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Go deeper",{"label":26,"value":27},"Reading time","≈ 55 minutes",{"label":29,"value":30},"Prior knowledge","Understand and Investigate layers",{"label":32,"value":33},"Chapters","11",{"label":35,"value":36},"Labs","Big trees, factor-count match, sieve to 200, rules sort",{"label":38,"value":39},"Headline proof","Euclid: the primes never end",[41,45,51,57,60,71,93,98,101,106,109,114,141,150,155,158,177,187,196,233,247,259,271,277,282,285,327,341,355,367,372,375,378,381,446,451,454,463,467,470,475,478,487,500,505,508,518,548,552,557,560,571,576,629,634,639,643,648,698,836,840,856],{"id":42,"type":43,"markdown":44},"intro-deepen","prose","In the earlier layers you **used** facts about primes: every number splits into primes; you only test up to the square root; digit sums test for 3 and 9; there are infinitely many primes. This layer asks the question a mathematician always asks next: **why?**\n\nYou will meet some of the most beautiful short arguments in mathematics, several of them over 2,000 years old. None needs more than multiplication, division and careful thinking. The aim is not to memorise these arguments but to understand them well enough to explain them to someone else, and to know exactly where each one would break if a small detail were changed.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"how-to-use-d","callout","observation","How to read a proof","Read each step and ask: \"Is this definitely true, for **every** number, not just the examples?\" If you are not sure, try a few numbers, then come back to the words. It is normal to read an argument two or three times. When you reach the end, try to retell it without looking.",{"id":52,"type":53,"title":54,"eyebrow":55,"navLabel":56},"ch1","chapter","Every number is built from primes","Chapter 01","1 Building blocks",{"id":58,"type":43,"markdown":59},"existence","**Claim:** every counting number bigger than 1 is either prime or a product of primes.\n\n**Why:** take any number n bigger than 1. Look at its **smallest** factor that is bigger than 1; call it p.\n\n- p must be **prime**. If p were composite, it would have a smaller factor bigger than 1, and that smaller factor would also divide n. But p was the smallest. Contradiction.\n- So n = p × m for some m. If m = 1, then n = p is prime and we are done. Otherwise, m is smaller than n, and we repeat the same step on m.\n- The numbers keep getting smaller (n, then m, then …), and a sequence of counting numbers cannot shrink forever. So the process stops, and when it does, we have written n as a product of primes.\n\nThis is exactly what a factor tree or division ladder does. The argument shows it **always** finishes.",{"id":61,"type":62,"title":63,"problem":64,"steps":65},"we-ladder-5040","worked_example","Smallest prime factor, again and again","Write 5,040 as a product of primes by always dividing by the smallest prime factor.",[66,67,68,69,70],"5,040 ÷ 2 = 2,520; ÷ 2 = 1,260; ÷ 2 = 630; ÷ 2 = 315. (315 is odd.)","315: digit sum 9, so ÷ 3 = 105; 105: digit sum 6, so ÷ 3 = 35.","35 ÷ 5 = 7, and 7 ÷ 7 = 1.","5,040 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 = **2⁴ × 3² × 5 × 7**.","5,040 = 1 × 2 × 3 × 4 × 5 × 6 × 7, a number with more factors than any smaller number. By the rule in Chapter 3 it has 5 × 3 × 2 × 2 = **60** factors.",{"id":72,"type":73,"component":74,"componentVersion":5,"config":75,"objective":87,"textAlternative":88,"help":89},"lab-tree-big","interactive","factor-tree",{"numbers":76,"showIndexForm":86},[77,78,79,80,81,82,83,84,85],360,1001,1729,2310,3600,4096,5040,7776,9999,true,"Factorise larger numbers with factor trees, choosing different first splits, and confirm the leaves never change.","This lab offers nine bigger numbers. Build each tree in more than one way (for example, start 3,600 as 60 × 60, then as 36 × 100, then as 2 × 1,800) and compare the prime leaves.\n\n- 360 = **2³ × 3² × 5** (24 factors)\n- 1,001 = **7 × 11 × 13** (8 factors)\n- 1,729 = **7 × 13 × 19** (8 factors)\n- 2,310 = **2 × 3 × 5 × 7 × 11** (32 factors)\n- 3,600 = **2⁴ × 3² × 5²** (45 factors)\n- 4,096 = **2¹²** (13 factors)\n- 5,040 = **2⁴ × 3² × 5 × 7** (60 factors)\n- 7,776 = **2⁵ × 3⁵** (36 factors)\n- 9,999 = **3² × 11 × 101** (12 factors)\n\n1,729 = 7 × 13 × 19 is Ramanujan’s famous taxi-cab number. 4,096 is 2 multiplied by itself 12 times. 2,310 is the product of the first five primes.",{"hints":90},[91,92],"Split off easy factors first: 10s, then 2s and 5s, then 3s.","For 9,999, try 9 first, then 11.",{"id":94,"type":53,"title":95,"eyebrow":96,"navLabel":97},"ch2","Only one way: the Fundamental Theorem of Arithmetic","Chapter 02","2 Unique factorisation",{"id":99,"type":43,"markdown":100},"fta","In Understand you saw that 60 always ends up as 2 × 2 × 3 × 5, however you start the tree. This is not luck. It is a theorem so important that it is called the **Fundamental Theorem of Arithmetic**:\n\n> Every counting number bigger than 1 can be written as a product of primes in **exactly one way**, apart from the order of the factors.\n\nChapter 1 proved the \"can be written\" part. The \"exactly one way\" part rests on a key fact about primes, first written down by **Euclid**:\n\n> **Euclid's lemma.** If a prime p divides a product a × b, then p divides a or p divides b (or both).\n\nFor example, 7 divides 42 × 5 = 210, and 7 divides 42. 3 divides 4 × 15, and 3 divides 15. This is **false** for composites: 6 divides 4 × 9 = 36, but 6 divides neither 4 nor 9. The 2 in 6 hides in the 4 and the 3 hides in the 9. A prime cannot be split up like that.\n\n**Why uniqueness follows:** suppose a number had two different prime recipes. Take any prime p in the first recipe. It divides the number, so by Euclid's lemma (used repeatedly) it divides one of the primes in the second recipe. But a prime's only factors are 1 and itself, so p must *be* that prime. Cancel it from both recipes and repeat. Every prime pairs off, so the recipes were the same all along.",{"id":102,"type":47,"variant":103,"title":104,"markdown":105},"aha-lemma","aha","Why a prime cannot \"split\"","Euclid’s lemma is the reason primes are called *prime*. A composite factor like 6 can divide a product by sending its 2 into one number and its 3 into the other. A prime has nothing to split: it must go whole into one side. Everything about unique factorisation grows from this one property.",{"id":107,"type":43,"markdown":108},"even-world","Uniqueness is not automatic. To see why it is special, imagine a world where the only numbers are the **even** numbers 2, 4, 6, 8, … In that world, call an even number an \"E-prime\" if it cannot be written as a product of two smaller even numbers.\n\n- 4 = 2 × 2, 8 = 2 × 4 and 12 = 2 × 6 are not E-prime.\n- 2, 6, 10, 14, 18, 22, 26, 30, … (the numbers that leave remainder 2 when divided by 4) **are** E-prime: two even numbers multiply to a multiple of 4, so these can never be made.\n\nNow factorise 60 in the even world: **60 = 2 × 30** and **60 = 6 × 10**. Both are products of E-primes, and they are different! Also 180 = 2 × 90 = 6 × 30 = 10 × 18. In the even world, factorisation is **not unique**. Our ordinary numbers are special because Euclid's lemma holds for them.",{"id":110,"type":47,"variant":111,"title":112,"markdown":113},"why-not-one","nuance","The real reason 1 is not prime","If 1 counted as a prime, 12 = 2 × 2 × 3 = 1 × 2 × 2 × 3 = 1 × 1 × 2 × 2 × 3, and the theorem would need the clumsy words \"except for any number of 1s\". Mathematicians chose the definition that makes the most important theorem simplest. Definitions in mathematics are chosen for good reasons, not handed down.",{"id":115,"type":116,"itemId":117,"prompt":118,"check":119,"hints":135,"feedback":138},"prac-lemma","practice","prime-and-composite.deepen-euclid-lemma","A number n is not divisible by 7. Which of these could be divisible by 7?",{"kind":120,"options":121,"correct":134},"choice",[122,125,128,131],{"id":123,"label":124},"a","n × n",{"id":126,"label":127},"b","n × n × n",{"id":129,"label":130},"c","n × 14",{"id":132,"label":133},"d","None of them",[129],[136,137],"Use Euclid’s lemma: if 7 divides n × n, it divides n or n.","14 is 2 × 7.",{"correct":139,"incorrect":140},"Right. n × 14 = n × 2 × 7 is always divisible by 7. But 7 cannot divide n × n or n × n × n unless it divides n.","By Euclid’s lemma, if the prime 7 divided n × n it would have to divide n, which it does not. Only n × 14 contains a 7.",{"id":142,"type":62,"title":143,"problem":144,"steps":145},"we-power-2","Can a power of 2 ever end in 0?","Is there any number of the form 2 × 2 × 2 × … × 2 whose last digit is 0?",[146,147,148,149],"A number ending in 0 is a multiple of 10 = 2 × 5, so its prime factorisation must contain a **5**.","The prime factorisation of 2 × 2 × … × 2 is just 2s. By the Fundamental Theorem, that is its **only** factorisation.","So no 5 can ever appear, and **no power of 2 ends in 0**. The same argument shows no power of 6 = 2 × 3 ends in 5 or 0.","Without uniqueness this argument would fail: we could not rule out some other recipe that sneaks in a 5.",{"id":151,"type":53,"title":152,"eyebrow":153,"navLabel":154},"ch3","What a factorisation tells you","Chapter 03","3 Reading factorisations",{"id":156,"type":43,"markdown":157},"reading-pf","Once a number is written in index form, many questions become quick.\n\n**Which numbers divide it?** A number d divides n exactly when every prime in d appears in n **at least as many times**. So 12 = 2² × 3 divides 360 = 2³ × 3² × 5, but 16 = 2⁴ does not (360 has only three 2s).\n\n**How many factors?** Every factor of 360 is 2^a × 3^b × 5^c, where a can be 0, 1, 2 or 3 (**4 choices**), b can be 0, 1 or 2 (**3 choices**) and c can be 0 or 1 (**2 choices**). Different choices give different factors (because factorisation is unique!), so there are 4 × 3 × 2 = **24** factors. In general, **add 1 to each power and multiply**.\n\n**Is it a perfect square?** A square has every power **even**: 3,600 = 2⁴ × 3² × 5² = 60 × 60. That is why squares have an odd number of factors: each (even power + 1) is odd, and odd × odd × … is odd.\n\n**Is it a perfect cube?** Every power is a multiple of 3: 7,776 = 2⁵ × 3⁵ is not a cube, but 2⁶ × 3³ = 1,728 = 12 × 12 × 12 is.",{"id":159,"type":160,"items":161},"formulas-count","formulas",[162,165,168,171,174],{"expression":163,"caption":164},"n = p^a × q^b × r^c","Index form with different primes p, q, r",{"expression":166,"caption":167},"factors = (a+1)(b+1)(c+1)","Each prime’s power can be chosen from 0 up to its power in n",{"expression":169,"caption":170},"360 = 2³ × 3² × 5","(3+1)(2+1)(1+1) = 24 factors",{"expression":172,"caption":173},"p prime: 2 factors","p = p¹, so (1+1) = 2. This is the definition of prime again",{"expression":175,"caption":176},"p²: 3 factors","Squares of primes (4, 9, 25, 49, …) have exactly 1, p and p²",{"id":178,"type":62,"title":179,"problem":180,"steps":181},"we-count-3600","How many factors does 3,600 have, and how many are odd?","Use 3,600 = 2⁴ × 3² × 5².",[182,183,184,185,186],"All factors: (4 + 1)(2 + 1)(2 + 1) = 5 × 3 × 3 = **45**.","45 is odd, which confirms that 3,600 is a perfect square (60 × 60).","An **odd** factor contains no 2 at all, so the power of 2 must be 0: only 1 choice.","Odd factors: 1 × 3 × 3 = **9**. They are the factors of 3² × 5² = 225: 1, 3, 5, 9, 15, 25, 45, 75, 225.","Even factors: 45 − 9 = **36**.",{"id":188,"type":62,"title":189,"problem":190,"steps":191},"we-divides-720","Reading divisibility from index form","N = 2⁴ × 3² × 5 = 720. Without dividing, decide whether N is divisible by 12, 16, 25 and 45.",[192,193,194,195],"12 = 2² × 3. N has four 2s (at least two) and two 3s (at least one). **Yes.**","16 = 2⁴. N has exactly four 2s. **Yes.**","25 = 5². N has only one 5. **No.**","45 = 3² × 5. N has two 3s and one 5. **Yes.**",{"id":197,"type":73,"component":198,"componentVersion":5,"config":199,"objective":227,"textAlternative":228,"help":229},"lab-match-count","match-pairs",{"prompt":200,"mode":201,"pairs":202},"Match each prime factorisation with the number of factors it has.","connect",[203,206,209,212,215,218,221,224],{"a":204,"b":205},"2⁵","6 factors",{"a":207,"b":208},"2 × 3 × 5","8 factors",{"a":210,"b":211},"2² × 3²","9 factors",{"a":213,"b":214},"3 × 7","4 factors",{"a":216,"b":217},"2² × 3 × 5","12 factors",{"a":219,"b":220},"2⁴ × 3","10 factors",{"a":222,"b":223},"2⁶","7 factors",{"a":225,"b":226},"2 × 3 × 5 × 7","16 factors","Use the add-one-and-multiply rule to match factorisations with their number of factors.","This connect game pairs 8 prime factorisations with 8 factor counts. Use the rule: add 1 to each power and multiply.\n\n- 2⁵ (= 32) → 6 factors\n- 2 × 3 × 5 (= 30) → 8 factors\n- 2² × 3² (= 36) → 9 factors\n- 3 × 7 (= 21) → 4 factors\n- 2² × 3 × 5 (= 60) → 12 factors\n- 2⁴ × 3 (= 48) → 10 factors\n- 2⁶ (= 64) → 7 factors\n- 2 × 3 × 5 × 7 (= 210) → 16 factors\n\nNotice that 2⁶ = 64 and 2² × 3² = 36 are squares, so their factor counts, 7 and 9, are odd. 2 × 3 × 5 × 7 = 210 has 2 × 2 × 2 × 2 = 16 factors even though it is not very big.",{"hints":230},[231,232],"2⁵ has powers 0 to 5 of 2 to choose from: 6 choices.","A prime to the power 1 contributes a factor of 2 to the count.",{"id":234,"type":116,"itemId":235,"prompt":236,"check":237,"hints":241,"feedback":244},"prac-count-720","prime-and-composite.deepen-factors-of-720","720 = 2⁴ × 3² × 5. How many factors does 720 have?",{"kind":238,"answer":239,"tolerance":240},"number",30,0,[242,243],"Add 1 to each power: 4 + 1, 2 + 1, 1 + 1.","Multiply the results.",{"correct":245,"incorrect":246},"(4 + 1)(2 + 1)(1 + 1) = 5 × 3 × 2 = 30 factors.","Add 1 to each power and multiply: 5 × 3 × 2 = 30.",{"id":248,"type":62,"title":249,"problem":250,"steps":251,"help":257},"we-root2","Unique factorisation proves √2 is not a fraction","Show that no fraction a⁄b (whole numbers) can satisfy (a⁄b) × (a⁄b) = 2.",[252,253,254,255,256],"Suppose it could. Then a × a = 2 × b × b.","Count the 2s in the prime factorisation of each side. Squaring doubles every power, so a × a has an **even** number of 2s, and so does b × b.","So the right-hand side, 2 × b × b, has an even number of 2s plus one more: an **odd** number.","But the Fundamental Theorem says a × a and 2 × b × b, being the same number, must have exactly the same prime factorisation, with the same number of 2s. Even cannot equal odd. Contradiction.","So √2 is not a fraction: it is **irrational**. The same argument works for √3, √5 and the square root of any number that is not a perfect square.",{"simplerExplanation":258},"Squares always contain each prime an even number of times. 2 × (a square) contains 2 an odd number of times. So a square can never equal 2 × another square.",{"id":260,"type":116,"itemId":261,"prompt":262,"check":263,"hints":265,"feedback":268},"prac-smallest-12","prime-and-composite.deepen-smallest-with-12-factors","What is the **smallest** number with exactly **12** factors?",{"kind":238,"answer":264,"tolerance":240},60,[266,267],"12 = 12, 6 × 2, 4 × 3 or 3 × 2 × 2. Each is a pattern of (power + 1)s.","Put the biggest powers on the smallest primes: try 2² × 3 × 5 and 2³ × 3².",{"correct":269,"incorrect":270},"60 = 2² × 3 × 5 has 3 × 2 × 2 = 12 factors, and no smaller number does. (2¹¹ = 2,048, 2⁵ × 3 = 96 and 2³ × 3² = 72 all have 12 factors but are bigger.)","The answer is 60 = 2² × 3 × 5. The other shapes give 2¹¹ = 2,048, 2⁵ × 3 = 96 and 2³ × 3² = 72, all bigger than 60.",{"id":272,"type":273,"conceptId":274,"relation":275,"explanation":276},"conn-hcf-d","connection","hcf-and-lcm","helps_understand","Prime factorisations give the HCF (lowest powers of shared primes) and the LCM (highest powers of all primes) in one step.",{"id":278,"type":53,"title":279,"eyebrow":280,"navLabel":281},"ch4","Why the square root is enough","Chapter 04","4 The square-root rule",{"id":283,"type":43,"markdown":284},"sqrt-proof","**Claim:** if n is composite, then n has a prime factor p with p × p ≤ n.\n\n**Why:** n composite means n = a × b, where a and b are both bigger than 1 and a ≤ b. Then a × a ≤ a × b = n. So a ≤ √n. The smallest prime factor of a is at most a, and it divides n too. So n has a prime factor no bigger than √n.\n\n**Turned round:** if **no** prime up to √n divides n, then n cannot be composite, so it is prime. That is the trial-division rule.\n\n**The same argument explains the sieve.** When you sieve with a prime p, every multiple k × p with k smaller than p has a prime factor smaller than p (a factor of k), so it was already crossed out. The first multiple p has to deal with is **p × p**. And once p × p passes the end of the grid, there is nothing left to do.",{"id":286,"type":287,"caption":288,"columns":289,"rows":294},"table-sieve-200","table","Sieving 1 to 200: each prime starts at its square and removes fewer new numbers",[290,291,292,293],"Prime","Starts at","New numbers crossed out","How many",[295,300,305,310,315,318,322],[296,297,298,299],"2","4","4, 6, 8, … 200","99",[301,302,303,304],"3","9","9, 15, 21, … 195 (odd multiples of 3)","32",[306,307,308,309],"5","25","25, 35, 55, 65, 85, 95, 115, 125, 145, 155, 175, 185","12",[311,312,313,314],"7","49","49, 77, 91, 119, 133, 161","6",[33,316,317,301],"121","121, 143, 187",[319,320,320,321],"13","169","1",[323,324,325,326],"17","289","nothing: 289 is past 200","0",{"id":328,"type":73,"component":329,"componentVersion":5,"config":330,"objective":335,"textAlternative":336,"help":337},"lab-sieve-200","prime-sieve",{"max":331,"columns":332,"modes":333,"rounds":332},200,10,[334],"sieve","Sieve all the way to 200 and check that the last prime needed is 13, because 17 × 17 is past 200.","The lab shows 1 to 200 in rows of 10. Sieve in order:\n\n- 2 crosses out 99 numbers; 3 crosses out 32 more; 5 crosses out 12 more; 7 crosses out 6 more (49, 77, 91, 119, 133, 161).\n- 11 crosses out only 121, 143 and 187. 13 crosses out only 169 = 13 × 13.\n- 17 crosses out nothing new, because 17 × 17 = 289 is more than 200.\n\nThat leaves 200 − 1 − 99 − 32 − 12 − 6 − 3 − 1 = **46 primes**. The last six primes up to 200 are 173, 179, 181, 191, 193 and 197, and the final one is 199.\n\nWhen you circle the survivors, notice the composites the sieve caught between 100 and 200 that look prime: 119 (7 × 17), 133 (7 × 19), 143 (11 × 13), 161 (7 × 23), 169 (13 × 13) and 187 (11 × 17).",{"hints":338},[339,340],"√200 is a little more than 14, so the last prime to sieve with is 13.","Before tapping 11 and 13, predict what they will cross out.",{"id":342,"type":343,"prompt":344,"options":345,"explanation":354},"pred-sqrt-1000","prediction","To test whether **997** is prime, what is the **largest** prime you need to try?",[346,348,350,352],{"id":123,"label":347},"29",{"id":126,"label":349},"31",{"id":129,"label":351},"37",{"id":132,"label":353},"499","**31.** 31 × 31 = 961 is less than 997, but 37 × 37 = 1,369 is more (and there is no prime between 31 and 37). So you try 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31: eleven divisions. None of them divides 997, so **997 is prime**: the largest prime below 1,000.",{"id":356,"type":116,"itemId":357,"prompt":358,"check":359,"hints":361,"feedback":364},"prac-2021","prime-and-composite.deepen-sqrt-2021","To test **2,021** for being prime, you try primes up to √2,021. What is the **largest** prime you need to try? (Hint: 45 × 45 = 2,025.)",{"kind":238,"answer":360,"tolerance":240},43,[362,363],"√2,021 is just under 45.","Which prime is just below 45?",{"correct":365,"incorrect":366},"Yes: 43 × 43 = 1,849 is less than 2,021 but 47 × 47 = 2,209 is more. And in fact 2,021 = 43 × 47, so the very last trial finds the factor!","The largest prime not above √2,021 (just under 45) is 43. Try it: 2,021 = 43 × 47, so 2,021 is composite and you only discover it at the last step.",{"id":368,"type":53,"title":369,"eyebrow":370,"navLabel":371},"ch5","Why the divisibility rules work","Chapter 05","5 Why the rules work",{"id":373,"type":43,"markdown":374},"div-why-base","Every divisibility rule comes from two facts.\n\n1. **Place value.** 7,128 means 7 × 1,000 + 1 × 100 + 2 × 10 + 8.\n2. **Sharing a divisor.** If d divides two numbers, it divides their sum and their difference. So if d divides one part of a number, d divides the whole exactly when it divides the rest.\n\n**Rules for 2, 5 and 10.** 10 = 2 × 5, so every multiple of 10 is divisible by 2, 5 and 10. Write 7,128 = 7,120 + 8. The 7,120 part is a multiple of 10, so only the last digit, 8, decides. 8 is even, so 7,128 is divisible by 2 but not by 5 or 10.\n\n**Rule for 4.** 100 = 4 × 25, so every multiple of 100 is divisible by 4. Write 7,128 = 7,100 + 28. Only the last two digits decide: 28 = 4 × 7. ✓ (The same reasoning gives a rule for 25: look at the last two digits.)\n\n**Rule for 8.** 1,000 = 8 × 125. Write 15,128 = 15,000 + 128. Only the last three digits decide: 128 = 8 × 16. ✓",{"id":376,"type":43,"markdown":377},"div-why-39","**Rules for 3 and 9.** The key fact is that 10 is **one more than 9**. So\n\n- 10 = 9 + 1\n- 100 = 99 + 1\n- 1,000 = 999 + 1\n\nand 9, 99, 999, … are all multiples of 9 (and of 3). Now write 7,128 like this:\n\n7 × 1,000 + 1 × 100 + 2 × 10 + 8\n= 7 × (999 + 1) + 1 × (99 + 1) + 2 × (9 + 1) + 8\n= (7 × 999 + 1 × 99 + 2 × 9) + (7 + 1 + 2 + 8).\n\nThe first bracket is a multiple of 9. The second bracket is the **digit sum**, 18. So 7,128 and its digit sum leave the **same remainder** when divided by 9 (or by 3). Divisible by 9 exactly when the digit sum is. That is also why a number minus its reverse (same digits, same digit sum) is always a multiple of 9.",{"id":379,"type":43,"markdown":380},"div-why-11","**Rule for 11.** Here the key fact is that 10 is **one less than 11**. So\n\n- 10 = 11 − 1\n- 100 = 99 + 1, and 99 = 9 × 11\n- 1,000 = 1,001 − 1, and 1,001 = 91 × 11\n- 10,000 = 9,999 + 1, and 9,999 = 909 × 11\n\nThe powers of 10 are alternately 1 more and 1 less than a multiple of 11. So a number leaves the same remainder on division by 11 as its digits added with alternating signs, starting with + on the units digit. For 9,284: 4 − 8 + 2 − 9 = −11, a multiple of 11, so 9,284 = 11 × 844. ✓\n\n**Rule for 6.** 6 = 2 × 3 and 2, 3 are **co-prime**. If a number is divisible by 2 and by 3, its prime factorisation contains a 2 and a 3, so it contains 6. With non-co-prime pieces, like 2 and 4, the 2 in 4 might be the same 2, which is why 12 passes both tests for 2 and 4 but fails for 8.",{"id":382,"type":73,"component":383,"componentVersion":5,"config":384,"objective":440,"textAlternative":441,"help":442},"lab-why-sort","sort-game",{"prompt":385,"bins":386,"items":399,"seconds":240},"Which fact about 10, 100 or 1,000 explains each divisibility test?",[387,390,393,396],{"id":388,"label":389},"ten","10 = 2 × 5",{"id":391,"label":392},"nine","10 = 9 + 1",{"id":394,"label":395},"eleven","10 = 11 − 1",{"id":397,"label":398},"hund","100 = 4 × 25 or 1,000 = 8 × 125",[400,404,408,412,416,420,424,428,432,436],{"id":401,"label":402,"bin":388,"why":403},"r2","Test for 2: last digit even","Multiples of 10 are even, so only the units digit matters.",{"id":405,"label":406,"bin":388,"why":407},"r5","Test for 5: last digit 0 or 5","Multiples of 10 are multiples of 5.",{"id":409,"label":410,"bin":388,"why":411},"r10","Test for 10: last digit 0","Everything except the units digit is already a multiple of 10.",{"id":413,"label":414,"bin":391,"why":415},"r3","Test for 3: digit sum","Each place value is (a multiple of 9) + 1, so the number and its digit sum share a remainder on division by 3.",{"id":417,"label":418,"bin":391,"why":419},"r9","Test for 9: digit sum","Each place value is 9, 99, 999, … plus 1.",{"id":421,"label":422,"bin":391,"why":423},"rev","A number minus its reverse is a multiple of 9","Same digits, same digit sum, same remainder on dividing by 9.",{"id":425,"label":426,"bin":394,"why":427},"r11","Test for 11: alternating digit sum","Place values alternate between 1 more and 1 less than a multiple of 11.",{"id":429,"label":430,"bin":397,"why":431},"r4","Test for 4: last two digits","100 is a multiple of 4, so the hundreds and above never matter.",{"id":433,"label":434,"bin":397,"why":435},"r25","Test for 25: last two digits 00, 25, 50 or 75","100 is a multiple of 25 too.",{"id":437,"label":438,"bin":397,"why":439},"r8","Test for 8: last three digits","1,000 = 8 × 125, so thousands and above are multiples of 8.","Match each divisibility test to the place-value fact that makes it work.","This game has 10 test cards and 4 bins, each a fact about powers of 10.\n\n- **10 = 2 × 5:** the tests for 2, 5 and 10 (look at the last digit only).\n- **10 = 9 + 1:** the digit-sum tests for 3 and 9, and the fact that a number minus its reverse is a multiple of 9.\n- **10 = 11 − 1:** the alternating-sum test for 11.\n- **100 = 4 × 25 or 1,000 = 8 × 125:** the last-two-digits tests for 4 and 25, and the last-three-digits test for 8.\n\nIn every case the idea is the same: split the number into a part that is certainly a multiple of the divisor and a small part that decides the answer.",{"hints":443},[444,445],"Tests that only look at the end of the number use a power of 10 that the divisor goes into.","Tests that use all the digits use 10 = 9 + 1 or 10 = 11 − 1.",{"id":447,"type":53,"title":448,"eyebrow":449,"navLabel":450},"ch6","A test for 7 (and 13)","Chapter 06","6 Testing for 7",{"id":452,"type":43,"markdown":453},"rule-7","There is no digit-sum test for 7, because 10 is neither 1 more nor 1 less than a multiple of 7. But there are two good tricks.\n\n**Chop and double.** Remove the last digit, double it, and subtract it from what is left. Repeat until the number is small. The original is divisible by 7 exactly when the result is.\n\nWhy it works: write the number as 10a + b (b is the last digit). The trick gives a − 2b. Now 2 × (10a + b) = 20a + 2b, and 20a + 2b + (a − 2b) = 21a, a multiple of 7. So 10a + b and a − 2b are either both multiples of 7 or neither (7 does not divide 2).\n\n**Use 1,001.** 1,001 = 7 × 11 × 13. Split the number into groups of three digits from the right and add and subtract them alternately (because 1,000 is one less than 1,001). The result is divisible by 7, 11 or 13 exactly when the original is. For 7,028,011: 11 − 28 + 7 = −10, so it is not divisible by 7, 11 or 13.",{"id":455,"type":62,"title":456,"problem":457,"steps":458},"we-rule7","Is 3,794 divisible by 7?","Use chop and double.",[459,460,461,462],"3,794 → chop 4, double it to 8, and 379 − 8 = 371.","371 → chop 1, double it to 2, and 37 − 2 = 35.","35 = 7 × 5, so 35 is divisible by 7.","So **3,794 is divisible by 7**. Check: 3,794 = 7 × 542. ✓",{"id":464,"type":53,"title":39,"eyebrow":465,"navLabel":466},"ch7","Chapter 07","7 Infinitely many primes",{"id":468,"type":43,"markdown":469},"euclid","Around **300 BCE**, the Greek mathematician **Euclid** of Alexandria wrote the *Elements*, a set of thirteen books that was used to teach mathematics for over 2,000 years. In Book IX he proved that there is no largest prime. His argument is short enough to fit on a postcard.\n\n**Claim:** there are more primes than any list you can make.\n\n**Why:** suppose someone hands you a finite list of primes, say 2, 3, 5 and 7. Multiply them all together and **add 1**:\n\nN = 2 × 3 × 5 × 7 + 1 = 211.\n\nNow look at N.\n\n- Dividing N by 2 leaves remainder 1 (N is one more than a multiple of 2).\n- Dividing N by 3 leaves remainder 1. The same happens for 5 and for 7.\n- So **none of the primes on the list divides N**.\n- But N is bigger than 1, so it has at least one prime factor (Chapter 1).\n- That prime factor is therefore **not on the list**.\n\nThe list was missing a prime. This works for **any** finite list, so no finite list can contain every prime. There are **infinitely many primes**.",{"id":471,"type":47,"variant":472,"title":473,"markdown":474},"careful-euclid","careful","N itself need not be prime","A common mistake is to say \"N is a new prime\". Sometimes it is: 2 × 3 × 5 × 7 + 1 = 211 and 2 × 3 × 5 × 7 × 11 + 1 = 2,311 are prime. But\n\n2 × 3 × 5 × 7 × 11 × 13 + 1 = 30,031 = 59 × 509.\n\n30,031 is composite, but its prime factors 59 and 509 are both missing from the list 2, 3, 5, 7, 11, 13. The proof only claims that N has **a prime factor not on the list**, which is always true.",{"id":476,"type":43,"markdown":477},"saidak","**A second proof, found in 2006.** The mathematician Filip Saidak noticed a proof that uses only one fact from this topic: **consecutive numbers are co-prime**.\n\nStart with any number, say 2. Multiply it by the next number: 2 × 3 = 6. Since 2 and 3 are co-prime, 6 has at least **two** different prime factors.\n\nNow do it again with 6: 6 × 7 = 42. Since 6 and 7 are co-prime, 7 brings at least one prime that is not in 6. So 42 has at least **three** different prime factors (2, 3, 7).\n\nAgain: 42 × 43 = 1,806 = 2 × 3 × 7 × 43. At least **four** primes. Then 1,806 × 1,807 has at least five, because 1,807 = 13 × 139 shares no prime with 1,806.\n\nEach step adds at least one new prime, and you can take as many steps as you like. So there must be infinitely many primes.",{"id":479,"type":62,"title":480,"problem":481,"steps":482},"we-euclid-list","Running Euclid’s machine","Start with the list {3, 7}. Use Euclid’s method to find a prime not on it.",[483,484,485,486],"Multiply and add 1: N = 3 × 7 + 1 = 22.","N = 22 = 2 × 11. Its prime factors are 2 and 11.","Neither 2 nor 11 is on the list {3, 7}. Euclid’s argument promised at least one new prime; here we got two.","Check the reason: 22 ÷ 3 leaves remainder 1, and 22 ÷ 7 leaves remainder 1, so 3 and 7 cannot divide it.",{"id":488,"type":343,"prompt":489,"options":490,"explanation":499},"pred-euclid","Your list is {2, 3, 5, 7, 11, 13}, and N = 30,031. Which of these statements is **definitely** true?",[491,493,495,497],{"id":123,"label":492},"N is prime",{"id":126,"label":494},"N has a prime factor bigger than 13",{"id":129,"label":496},"N is divisible by 13",{"id":132,"label":498},"N is even","**N has a prime factor bigger than 13.** N leaves remainder 1 when divided by each of 2, 3, 5, 7, 11 and 13, so none of them divides it. Every prime factor of N must therefore be a prime **other than** these six, and all other primes are bigger than 13. (In fact 30,031 = 59 × 509, so (a) is false.)",{"id":501,"type":53,"title":502,"eyebrow":503,"navLabel":504},"ch8","Co-primes: why they behave so well","Chapter 08","8 Co-prime reasons",{"id":506,"type":43,"markdown":507},"coprime-deep","**Co-prime means no shared prime.** By unique factorisation, the common factors of two numbers are built from the primes they share. So HCF = 1 exactly when their factorisations have **no prime in common**. 8 = 2³ and 15 = 3 × 5 share no prime, so they are co-prime. That is the fastest test.\n\n**Consecutive numbers are co-prime.** If d divides both n and n + 1, it divides their difference, 1. So d = 1.\n\n**The product rule.** If a and b are co-prime and both divide n, then **a × b divides n**. Reason: a's primes appear in n, and b's primes appear in n; since they share no primes, they are different primes and all of them appear in n together. This is exactly why the test for 6 (2 and 3), 12 (3 and 4), 15 (3 and 5), 18 (2 and 9) and 24 (3 and 8) works, and why it fails for non-co-prime pieces.\n\n**Fractions in lowest terms.** A fraction like 8⁄15 is in its simplest form exactly when the top and bottom are co-prime.",{"id":509,"type":62,"title":510,"problem":511,"steps":512},"we-2376","Is 2,376 divisible by 24?","Use the product rule with co-prime pieces.",[513,514,515,516,517],"24 = 3 × 8, and 3 and 8 are co-prime.","Test for 3: digit sum 2 + 3 + 7 + 6 = 18. ✓","Test for 8: last three digits 376 = 8 × 47. ✓","Both pass and 3, 8 are co-prime, so 2,376 is divisible by 24. Check: 2,376 = 24 × 99. ✓","Using 4 and 6 instead would be unsafe: they share the factor 2, so passing both only proves divisibility by their LCM, 12.",{"id":519,"type":287,"caption":520,"columns":521,"rows":525},"table-phi","How many numbers from 1 to n are co-prime with n? (Euler’s totient, written φ(n))",[522,523,293,524],"n","Co-prime numbers from 1 to n","Note",[526,528,531,534,536,539,541,544,546],[321,321,321,527],"1 is co-prime to itself",[306,529,297,530],"1, 2, 3, 4","prime: n − 1",[314,532,296,533],"1, 5","—",[311,535,314,530],"1, 2, 3, 4, 5, 6",[537,538,297,533],"8","1, 3, 5, 7",[302,540,314,533],"1, 2, 4, 5, 7, 8",[542,543,297,533],"10","1, 3, 7, 9",[33,545,542,530],"1, 2, 3, 4, 5, 6, 7, 8, 9, 10",[309,547,297,533],"1, 5, 7, 11",{"id":549,"type":47,"variant":48,"title":550,"markdown":551},"obs-phi","Primes are the most co-prime","For a prime p, every number from 1 to p − 1 is co-prime with it, so φ(p) = p − 1. A number with many small prime factors has few co-prime partners: φ(12) = 4 and φ(30) = 8. Euler’s function φ turns up again in the Extend layer, at the heart of the RSA code that protects online banking.",{"id":553,"type":53,"title":554,"eyebrow":555,"navLabel":556},"ch9","Twin prime facts you can prove","Chapter 09","9 Twin prime proofs",{"id":558,"type":43,"markdown":559},"twin-proofs","Nobody can prove that twin primes go on forever, but several facts about them are easy to prove.\n\n**1. After (3, 5), twin primes are 6k − 1 and 6k + 1.** The middle number of twins p and p + 2 is even (it lies between two odd numbers). One of p, p + 1, p + 2 is a multiple of 3, and it cannot be p or p + 2 (they are primes bigger than 3). So the middle is a multiple of 2 and of 3, and since 2 and 3 are co-prime, a multiple of 6.\n\n**2. The sum of twin primes (after 3, 5) is a multiple of 12.** p + (p + 2) = 2 × (p + 1) = 2 × 6k = 12k. Check: 17 + 19 = 36 = 12 × 3; 71 + 73 = 144 = 12 × 12.\n\n**3. The product of twin primes, plus 1, is a perfect square.** p × (p + 2) + 1 = p × p + 2 × p + 1 = (p + 1) × (p + 1). Check: 11 × 13 + 1 = 144 = 12 × 12; 29 × 31 + 1 = 900 = 30 × 30.\n\n**4. The only prime triplet is 3, 5, 7.** One of n, n + 2, n + 4 is always a multiple of 3 (check the three possible remainders), so for all three to be prime, that one must be 3.",{"id":561,"type":116,"itemId":562,"prompt":563,"check":564,"hints":566,"feedback":568},"prac-twin-product","prime-and-composite.deepen-twin-product","Without a calculator: 41 × 43 + 1 is a perfect square. **Which number squared?**",{"kind":238,"answer":565,"tolerance":240},42,[567],"Use p × (p + 2) + 1 = (p + 1) × (p + 1).",{"correct":569,"incorrect":570},"Yes: 41 × 43 + 1 = 42 × 42 = 1,764.","p × (p + 2) + 1 = (p + 1)², so the answer is 42: 41 × 43 + 1 = 1,764 = 42 × 42.",{"id":572,"type":53,"title":573,"eyebrow":574,"navLabel":575},"ch10","The long story of primes","Chapter 10","10 History",{"id":577,"type":578,"title":579,"items":580},"timeline-primes","timeline","Two thousand years of prime numbers",[581,585,589,593,597,601,605,609,613,617,621,625],{"time":582,"title":583,"text":584},"c. 300 BCE","Euclid’s Elements","Euclid proves there are infinitely many primes and lays the groundwork for unique factorisation (Euclid’s lemma).",{"time":586,"title":587,"text":588},"c. 240 BCE","Eratosthenes’ sieve","Eratosthenes of Cyrene, librarian at Alexandria, describes the sieve for listing primes.",{"time":590,"title":591,"text":592},"1640","Fermat","Pierre de Fermat states his \"little theorem\" about primes and the two-squares result for 4k + 1 primes.",{"time":594,"title":595,"text":596},"1644","Mersenne","Marin Mersenne, a French friar, publishes a list of exponents p for which he claimed 2ᵖ − 1 is prime; it contained errors.",{"time":598,"title":599,"text":600},"1737","Euler","Leonhard Euler gives a new proof that primes never end and links primes to the sum 1 + ½ + ⅓ + …",{"time":602,"title":603,"text":604},"1742","Goldbach","In letters between Christian Goldbach and Euler appears the guess that every even number above 2 is a sum of two primes. Still unproved.",{"time":606,"title":607,"text":608},"1801","Gauss","Carl Friedrich Gauss gives the first complete proof that prime factorisation is unique.",{"time":610,"title":611,"text":612},"1896","Prime Number Theorem","Hadamard and de la Vallée Poussin prove that the primes up to n number roughly n divided by its natural logarithm.",{"time":614,"title":615,"text":616},"1915","Ramanujan","Srinivasa Ramanujan publishes his long paper on highly composite numbers, the opposite of primes.",{"time":618,"title":619,"text":620},"2002","AKS test","Manindra Agrawal, Neeraj Kayal and Nitin Saxena at IIT Kanpur find the first fast, always-correct primality test.",{"time":622,"title":623,"text":624},"2013","Bounded gaps","Yitang Zhang proves infinitely many prime pairs differ by less than 70 million; others soon cut this to 246.",{"time":626,"title":627,"text":628},"2024","Largest known prime","A GIMPS volunteer finds 2¹³⁶²⁷⁹⁸⁴¹ − 1, a prime with over 41 million digits.",{"id":630,"type":47,"variant":631,"title":632,"markdown":633},"ex-ramanujan","example","Ramanujan and 1,729","When the English mathematician G. H. Hardy visited Srinivasa Ramanujan in hospital, he mentioned that his taxi had the \"rather dull\" number **1,729**. Ramanujan replied at once that it was very interesting: it is the smallest number that can be written as the sum of two cubes in two different ways, **1³ + 12³ = 9³ + 10³ = 1,729**.\n\nIts prime factorisation is also neat: 1,729 = 7 × 13 × 19, three primes that are each 6 apart. Ramanujan, who grew up in Kumbakonam in Tamil Nadu and taught himself from borrowed books, had an extraordinary feel for the personality of numbers.",{"id":635,"type":53,"title":636,"eyebrow":637,"navLabel":638},"ch11","Edge cases and careful thinking","Chapter 11","11 Edge cases",{"id":640,"type":47,"variant":111,"title":641,"markdown":642},"nuance-edges","Where the definitions need care","- **0** is divisible by every number (0 = n × 0), but it is neither prime nor composite, and nothing is divisible by 0.\n- **Negative numbers** such as −7 are not called prime in school mathematics. (Advanced mathematics allows \"−7 is a prime element\", but the counting-number definition is what we use.)\n- **Units.** 1 is the only counting number that divides every number. It is called a *unit*, which is another reason it is kept apart from primes.\n- **Composite does not mean \"has small factors\".** 10,403 = 101 × 103 has no factor below 101. Trial division must go all the way to √n.\n- **\"Exactly two factors\"** already excludes 1: its two \"factors\" 1 and 1 are the same number.",{"id":644,"type":47,"variant":645,"title":646,"markdown":647},"misc-deepen","misconception","“Euclid’s product-plus-one is always prime”","30,031 = 59 × 509 is the first counterexample. The proof never needed N to be prime: it only needed N’s prime factors to be missing from the list.",{"id":649,"type":650,"title":651,"terms":652},"gloss-deepen","glossary","Words for proofs",[653,657,661,665,668,672,675,679,683,687,691,694],{"term":654,"meaning":655,"example":656},"theorem","A statement that has been proved true.","There are infinitely many primes.",{"term":658,"meaning":659,"example":660},"lemma","A smaller result used as a stepping stone to prove a theorem.","Euclid’s lemma",{"term":662,"meaning":663,"example":664},"Fundamental Theorem of Arithmetic","Every number bigger than 1 is a product of primes in exactly one way, apart from order.","60 = 2² × 3 × 5 only",{"term":660,"meaning":666,"example":667},"If a prime divides a product, it divides at least one of the numbers multiplied.","7 divides 42 × 5, and 7 divides 42.",{"term":669,"meaning":670,"example":671},"proof by contradiction","Showing something is true by assuming it is false and reaching an impossibility.","Assume there is a last prime…",{"term":673,"meaning":674},"infinitely many","More than any number you can name; the list never ends.",{"term":676,"meaning":677,"example":678},"perfect square","A number that is a whole number multiplied by itself. All powers in its factorisation are even.","3,600 = 2⁴ × 3² × 5²",{"term":680,"meaning":681,"example":682},"perfect cube","A whole number multiplied by itself three times.","1,728 = 12³",{"term":684,"meaning":685,"example":686},"Euler’s totient φ(n)","How many numbers from 1 to n are co-prime with n.","φ(10) = 4: 1, 3, 7, 9",{"term":688,"meaning":689,"example":690},"remainder","What is left after exact division. Two numbers with the same remainder on division by d differ by a multiple of d.","17 and 2 both leave 2 on dividing by 5.",{"term":692,"meaning":693},"unit","A number that divides every number. Among counting numbers, only 1.",{"term":695,"meaning":696,"example":697},"primality test","A method for deciding whether a number is prime.","Trial division; AKS",{"id":699,"type":700,"title":701,"questions":702},"quiz-deepen","quiz","Reasons, not just rules",[703,716,729,740,752,765,774,786,799,812,825],{"itemId":704,"prompt":705,"options":706,"correct":126,"why":715},"prime-and-composite.deepen-q-smallest-factor","Why is the smallest factor (bigger than 1) of any number always prime?",[707,709,711,713],{"id":123,"label":708},"Because small numbers are prime",{"id":126,"label":710},"If it were composite, it would have an even smaller factor that also divides the number",{"id":129,"label":712},"Because 2 is prime",{"id":132,"label":714},"It is not always prime","A composite smallest factor would have a smaller factor, which would also divide the original number, contradicting \"smallest\".",{"itemId":717,"prompt":718,"options":719,"correct":123,"why":728},"prime-and-composite.deepen-q-even-world","In the \"even world\", why does 60 fail to have a unique factorisation?",[720,722,724,726],{"id":123,"label":721},"60 = 2 × 30 = 6 × 10, with all four being E-primes",{"id":126,"label":723},"60 is odd",{"id":129,"label":725},"Even numbers have no factors",{"id":132,"label":727},"It does have a unique factorisation","2, 6, 10 and 30 all leave remainder 2 on division by 4, so none is a product of two even numbers. Two different E-prime recipes give 60.",{"itemId":730,"prompt":731,"options":732,"correct":129,"why":739},"prime-and-composite.deepen-q-count","How many factors does 2³ × 3² × 7 have?",[733,734,735,737],{"id":123,"label":314},{"id":126,"label":309},{"id":129,"label":736},"24",{"id":132,"label":738},"42","(3 + 1)(2 + 1)(1 + 1) = 4 × 3 × 2 = 24.",{"itemId":741,"prompt":742,"options":743,"correct":129,"why":751},"prime-and-composite.deepen-q-square","Which of these is a perfect square?",[744,746,748,750],{"id":123,"label":745},"2³ × 3²",{"id":126,"label":747},"2² × 3² × 5",{"id":129,"label":749},"2⁴ × 5²",{"id":132,"label":207},"Every power must be even. 2⁴ × 5² = 16 × 25 = 400 = 20².",{"itemId":753,"prompt":754,"options":755,"correct":126,"why":764},"prime-and-composite.deepen-q-nine-why","The digit-sum test for 9 works because…",[756,758,760,762],{"id":123,"label":757},"9 is the biggest digit",{"id":126,"label":759},"10, 100, 1,000, … are each one more than a multiple of 9",{"id":129,"label":761},"9 is a square",{"id":132,"label":763},"9 = 3 × 3","10 = 9 + 1, 100 = 99 + 1 and so on, so each digit contributes itself plus a multiple of 9.",{"itemId":766,"prompt":767,"options":768,"correct":123,"why":773},"prime-and-composite.deepen-q-eleven","Is 918,082 divisible by 11?",[769,771],{"id":123,"label":770},"Yes",{"id":126,"label":772},"No","From the right: 2 − 8 + 0 − 8 + 1 − 9 = −22, a multiple of 11. So yes: 918,082 = 11 × 83,462.",{"itemId":775,"prompt":776,"options":777,"correct":126,"why":785},"prime-and-composite.deepen-q-euclid","In Euclid’s proof, why can none of the listed primes divide N = (product of the list) + 1?",[778,780,782,783],{"id":123,"label":779},"N is too big",{"id":126,"label":781},"Each leaves remainder 1",{"id":129,"label":492},{"id":132,"label":784},"N is odd","N is one more than a multiple of each listed prime, so dividing leaves remainder 1.",{"itemId":787,"prompt":788,"options":789,"correct":126,"why":798},"prime-and-composite.deepen-q-30031","2 × 3 × 5 × 7 × 11 × 13 + 1 = 30,031. What does this example show?",[790,792,794,796],{"id":123,"label":791},"Euclid was wrong",{"id":126,"label":793},"N need not be prime, but its prime factors are new",{"id":129,"label":795},"30,031 is the largest prime",{"id":132,"label":797},"13 is the last prime","30,031 = 59 × 509, and 59 and 509 are not on the list.",{"itemId":800,"prompt":801,"options":802,"correct":126,"why":811},"prime-and-composite.deepen-q-sqrt","Why is it enough to try primes up to √n when testing n?",[803,805,807,809],{"id":123,"label":804},"Big primes never divide anything",{"id":126,"label":806},"If n = a × b with a ≤ b, then a × a ≤ n, so a ≤ √n",{"id":129,"label":808},"Because √n is always prime",{"id":132,"label":810},"It is only a rule of thumb","The smaller factor of any factor pair is at most √n, and its smallest prime factor divides n too.",{"itemId":813,"prompt":814,"options":815,"correct":123,"why":824},"prime-and-composite.deepen-q-twin-sum","The sum of twin primes 59 and 61 is 120. Which is guaranteed for every twin pair after (3, 5)?",[816,818,820,822],{"id":123,"label":817},"The sum is a multiple of 12",{"id":126,"label":819},"The sum is a multiple of 24",{"id":129,"label":821},"The sum is a square",{"id":132,"label":823},"The sum is prime","p + (p + 2) = 2 × (6k) = 12k. 120 = 12 × 10. It is not always a multiple of 24: 17 + 19 = 36.",{"itemId":826,"prompt":827,"options":828,"correct":123,"why":835},"prime-and-composite.deepen-q-rule-7","Using chop and double, 1,869 → 186 − 18 = 168 → 16 − 16 = 0. What do you conclude?",[829,831,833],{"id":123,"label":830},"1,869 is divisible by 7",{"id":126,"label":832},"1,869 is not divisible by 7",{"id":129,"label":834},"The test failed","0 is divisible by 7, so 1,869 is too: 1,869 = 7 × 267.",{"id":837,"type":838,"prompt":839},"reflect-deepen","reflection","Retell Euclid’s proof that there are infinitely many primes in your own words, as if to a friend who has never seen it. Then explain why the example 30,031 = 59 × 509 does not break the proof.",{"id":841,"type":842,"title":843,"points":844},"cheat-deepen","summary","Cheat sheet: the reasons",[845,846,847,848,849,850,851,852,853,854,855],"**Existence:** the smallest factor above 1 is always prime; keep dividing and the numbers shrink, so every n above 1 is a product of primes.","**Euclid’s lemma:** a prime dividing a × b divides a or b. False for composites (6 divides 4 × 9).","**Fundamental Theorem of Arithmetic:** each n above 1 has exactly one prime factorisation. In the \"even world\" this fails: 60 = 2 × 30 = 6 × 10.","**1 is not prime** so that the theorem stays simple.","**Number of factors:** add 1 to each power and multiply. 360 = 2³ × 3² × 5 → 24 factors. Squares have all powers even.","**√n rule:** in any factor pair a × b = n with a ≤ b, a ≤ √n. The sieve for p starts at p × p.","**Rules from place value:** 10 = 2 × 5; 100 = 4 × 25; 1,000 = 8 × 125; 10 = 9 + 1 (digit sum); 10 = 11 − 1 (alternating sum).","**Test for 7:** 10a + b is divisible by 7 exactly when a − 2b is. Or use 1,001 = 7 × 11 × 13.","**Euclid (c. 300 BCE):** product of any list of primes + 1 has a prime factor not on the list. Infinitely many primes.","**Co-prime** = no shared prime. If co-prime a and b both divide n, then a × b divides n.","**Twin facts:** after (3, 5), twins are 6k ± 1, their sum is a multiple of 12, and their product + 1 is a square.",{"id":857,"type":858,"sourceIds":859},"sources-deepen","sources",[860,861,862,863,864,865,866,867],"prime-and-composite-britannica-prime-number","prime-and-composite-britannica-eratosthenes-sieve","prime-and-composite-mathisfun-divisibility","prime-and-composite-wiki-twin-prime","prime-and-composite-gimps-mersenne","prime-and-composite-ncert-class6-playing-numbers","prime-and-composite-wiki-euclid-theorem","prime-and-composite-wiki-aks",[860,861,862,863,864,865,866,867],"needs_review",{"generatedBy":871,"notes":872},"claude-code","Draft generated locally with Python-checked numbers; pending owner review.","e907a1a59d3aaa41de7ec3df9a96fcb1cc2112cbead3be3fdfeee300bb444705",{"component:factor-tree@1":875,"logic:practice":876,"component:match-pairs@1":877,"component:prime-sieve@1":878,"component:sort-game@1":879,"source:prime-and-composite-britannica-eratosthenes-sieve":880,"source:prime-and-composite-britannica-prime-number":881,"source:prime-and-composite-gimps-mersenne":882,"source:prime-and-composite-mathisfun-divisibility":883,"source:prime-and-composite-ncert-class6-playing-numbers":884,"source:prime-and-composite-wiki-aks":885,"source:prime-and-composite-wiki-euclid-theorem":886,"source:prime-and-composite-wiki-twin-prime":887},"dfb598a254f20e70550d586121be496f7a546ba0254060fd97b95cb919dfd934","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","76baccc44f3b2ac333537045fea84801ea3c46f2d75f3db8a02355ef0f267041","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","af8e7bbdca2e2859b452689299eeef54805896728cef8a361ea1ec0dd63d12d4","8985127013a5a393bbde77d9c7d1e0757013c9dd047aa324fe51fa4706c2edfa","be41b20420652d001b4454ec3ac0d159a0139fb79e77363b0afc6f376ecc405c","d26ef4a1f28e097000ad535bd8d0bb52463c4eb61d815cd4dd383a8db4bac076","d766224fdc533b260af62a8660196dabe3a5e506c901ab99ee8eba647c9a7616","b817c8cf9495d0a940a4d3d0df7582557ce0c189585178e0abe4ad9236fd3c6d","7f8377346cb21696b3902171a5daab30d77946fe52e91f17fd0f533c31d3a29f","b0b179308ab5f683db390f67f15856137356f04cbd838b8129c6b89a3c2b8265",{"state":889,"reviewer":890,"selfReview":86,"reviewedAt":891,"method":892},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899599098]