[{"data":1,"prerenderedAt":983},["ShallowReactive",2],{"layer:hcf-and-lcm:deepen":3},{"layer":4,"contentHash":963,"dependencyHashes":964,"approval":977,"releaseId":982},{"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":958,"reviewStatus":959,"authoring":960},1,"hcf-and-lcm","en","deepen","Why it works: proofs, Euclid and the edges","Unique prime recipes, the product rule, Euclid’s algorithm and Bézout","Proofs in plain language: unique prime factorisation, why HCF takes smallest powers and LCM largest, why HCF × LCM = a × b (and why not for three numbers), why Euclid’s method works and how fast it is, Bézout’s identity, edge cases, harder problems and history.",[13,14,15,16,17],"Explain why the prime-factor rules for HCF and LCM follow from unique prime factorisation.","Prove HCF × LCM = a × b for two numbers and explain why it fails for three.","Prove that Euclid’s algorithm gives the HCF and always stops, and describe its worst case.","Write the HCF as a combination ax + by and use this to decide which amounts can be measured.","Solve harder problems: three remainders, pairs with given HCF and LCM, and edge cases with 0 and 1.",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","All four methods; index form",{"label":32,"value":33},"Chapters","13",{"label":35,"value":36},"Proofs","Product rule, Euclid, LCM multiples",{"label":38,"value":39},"Labs","Factor trees, Venn, Euclid, sort, match",[41,45,51,88,94,97,102,105,118,123,126,148,169,174,179,182,186,189,210,225,230,233,237,242,245,261,266,294,299,302,340,358,363,366,370,395,400,403,417,427,438,443,476,481,493,498,508,517,527,539,544,549,559,568,572,576,588,600,620,630,639,651,656,693,698,703,736,791,912,927,931,937,942,947],{"id":42,"type":43,"markdown":44},"intro-deepen","prose","You can now find an HCF or LCM four different ways, and you have discovered patterns by testing. This layer asks the questions a mathematician asks next: **Why do the methods work? Are the patterns true for every number, or just the ones we tried? What happens at the edges?**\n\nThe arguments here are short, but they are real proofs, the same ones found in university number theory books, written in plain language. Read them slowly, with a pencil. After each one, try it on your own example: a proof you have checked with numbers is a proof you understand.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"how-to-read-d","callout","observation","Notation used in this layer","- **a | b** means \"a divides b\": b ÷ a is a whole number. So 4 | 12 is true and 5 | 12 is false.\n- **min(x, y)** is the smaller of x and y; **max(x, y)** is the larger. min(2, 5) = 2 and max(2, 5) = 5.\n- Letters a, b, c stand for positive whole numbers unless we say otherwise.",{"id":52,"type":53,"title":54,"terms":55},"glossary-deepen","glossary","Words for reasoning and proof",[56,60,64,68,72,76,80,84],{"term":57,"meaning":58,"example":59},"proof","An argument that shows a statement is true in every case, not just the ones tested.","The min + max argument proves HCF × LCM = a × b.",{"term":61,"meaning":62,"example":63},"Fundamental Theorem of Arithmetic","Every whole number greater than 1 has exactly one prime factorisation, apart from order.","360 = 2³ × 3² × 5 and no other way.",{"term":65,"meaning":66,"example":67},"algorithm","A step-by-step method that always finishes and gives the answer.","Euclid’s algorithm for the HCF",{"term":69,"meaning":70,"example":71},"Euclid’s algorithm","Replace (a, b) by (b, remainder of a ÷ b) until the remainder is 0; the last divisor is the HCF.","(252, 198) → (198, 54) → (54, 36) → (36, 18) → 18",{"term":73,"meaning":74,"example":75},"Bézout’s identity","The HCF of a and b can be written as ax + by for some whole numbers x and y.","18 = 4 × 252 − 5 × 198",{"term":77,"meaning":78,"example":79},"kuttaka","Aryabhata’s \"pulveriser\" method (499 CE) for solving equations like ax − by = c by repeated division.","Used for astronomical cycle problems",{"term":81,"meaning":82,"example":83},"min \u002F max","The smaller \u002F larger of two numbers.","min(3, 5) = 3, max(3, 5) = 5",{"term":85,"meaning":86,"example":87},"∎ (end of proof)","A symbol marking that a proof is complete.","Every common multiple is a multiple of the LCM. ∎",{"id":89,"type":90,"title":91,"eyebrow":92,"navLabel":93},"ch01","chapter","The foundation: every number has one prime recipe","Chapter 01","1 Unique recipes",{"id":95,"type":43,"markdown":96},"fta-prose","The prime-factor method rests on one big fact:\n\n> **Fundamental Theorem of Arithmetic.** Every whole number greater than 1 can be written as a product of primes in **exactly one way**, apart from the order of the factors.\n\nThe first half (it *can* be done) is easy: keep splitting until you only have primes. The second half (only **one** way) is the surprising part. Why could 360 not be 2³ × 3² × 5 one way and something completely different another way?\n\nThe key step is a property of primes proved by Euclid: **if a prime divides a product, it must divide one of the factors.** For example, 7 | 21 × 10 = 210, and indeed 7 | 21. (A composite number does not have this property: 6 | 4 × 9 = 36, but 6 divides neither 4 nor 9.) With that property, if two prime recipes gave the same number, each prime in the first recipe would have to divide, and so equal, some prime in the second, and you could cancel them one pair at a time until nothing is left. The two recipes must be the same.",{"id":98,"type":47,"variant":99,"title":100,"markdown":101},"nuance-hilbert","nuance","A world where recipes are not unique","Uniqueness is not automatic. Imagine a world whose only numbers are 1, 5, 9, 13, 17, 21, … (every number that leaves remainder 1 on division by 4). Multiplying two of them always gives another one, so this world has its own \"primes\": numbers that cannot be split within the world. 9, 21 and 49 are all \"primes\" there, because 3 and 7 do not exist in that world.\n\nBut **441 = 9 × 49 = 21 × 21**: two different \"prime\" recipes for the same number! In such a world, HCF by prime factors would give nonsense. Our ordinary whole numbers are special, and that is what the Fundamental Theorem guarantees.",{"id":103,"type":43,"markdown":104},"divisors-prose","Uniqueness has a powerful consequence. Write 360 = 2³ × 3² × 5. **A number divides 360 exactly when its recipe uses only 2, 3 and 5, with at most three 2s, at most two 3s and at most one 5.** Nothing else can divide it, because any other prime, or an extra copy of one of these, would have to appear in 360's one and only recipe.\n\nSo each divisor is a choice: 0, 1, 2 or 3 twos (4 choices), 0, 1 or 2 threes (3 choices), 0 or 1 five (2 choices). That gives 4 × 3 × 2 = **24 divisors**, which you can check by listing them.",{"id":106,"type":107,"items":108},"formulas-divisors","formulas",[109,112,115],{"expression":110,"caption":111},"n = p^a × q^b × r^c","Prime recipe of n, with different primes p, q, r (written ^ for \"to the power\").",{"expression":113,"caption":114},"d | n ⇔ d = p^x × q^y × r^z, x ≤ a…","A divisor uses the same primes, each at most as many times as n does.",{"expression":116,"caption":117},"number of divisors = (a+1)(b+1)(c+1)","Each index can be chosen from 0 up to its maximum. 360: (3+1)(2+1)(1+1) = 24.",{"id":119,"type":90,"title":120,"eyebrow":121,"navLabel":122},"ch02","Why smallest powers for HCF, highest for LCM","Chapter 02","2 Why the rules work",{"id":124,"type":43,"markdown":125},"why-rules-prose","Take a = 360 = 2³ × 3² × 5 and b = 588 = 2² × 3 × 7².\n\n**The HCF.** A common divisor d must divide 360, so it has at most three 2s; and divide 588, so at most two 2s. It must satisfy **both**, so at most min(3, 2) = 2 twos. Similarly at most min(2, 1) = 1 three. It cannot contain 5 (588 has none) or 7 (360 has none). The **biggest** number obeying all these limits takes the maximum allowed of each: 2² × 3 = **12**. That is the HCF, and it is the product of the common primes with the **smallest** powers.\n\n**The LCM.** A common multiple m must contain 360's recipe, so at least three 2s, two 3s and one 5; and contain 588's recipe, so at least two 2s, one 3 and two 7s. It needs at least max(3, 2) = 3 twos, max(2, 1) = 2 threes, one 5 and two 7s. The **smallest** such number has exactly those: 2³ × 3² × 5 × 7² = **17,640**.",{"id":127,"type":128,"caption":129,"columns":130,"rows":137},"table-min-max","table","Prime by prime for 360 and 588: min goes to the HCF, max goes to the LCM",[131,132,133,134,135,136],"Prime","Power in 360","Power in 588","min → HCF","max → LCM","min + max",[138,142,144,146],[139,140,139,139,140,141],"2","3","5",[140,139,143,143,139,140],"1",[141,143,145,145,143,143],"0",[147,145,139,145,139,139],"7",{"id":149,"type":150,"component":151,"componentVersion":5,"config":152,"objective":163,"textAlternative":164,"help":165},"lab-factor-tree-d","interactive","factor-tree",{"numbers":153,"showIndexForm":162},[154,155,156,157,158,159,160,161],360,588,1260,2520,5040,9240,1147,1591,true,"Factorise larger numbers into index form and use the powers to predict HCFs, LCMs and numbers of divisors.","This lab builds factor trees for larger numbers and writes them in index form.\n\n- 360 = **2³ × 3² × 5** (24 divisors)\n- 588 = **2² × 3 × 7²** (18 divisors)\n- 1,260 = **2² × 3² × 5 × 7** (36 divisors)\n- 2,520 = **2³ × 3² × 5 × 7** (48 divisors; the smallest number divisible by every number from 1 to 10)\n- 5,040 = **2⁴ × 3² × 5 × 7** (60 divisors)\n- 9,240 = **2³ × 3 × 5 × 7 × 11** (64 divisors)\n- 1,147 = **31 × 37** and 1,591 = **37 × 43**: products of two primes, hard to spot by divisibility tests. Their HCF is **37**.\n\nFrom the powers: HCF(360, 588) = 2² × 3 = 12, and LCM(2,520, 9,240) = 2³ × 3² × 5 × 7 × 11 = 27,720.",{"hints":166},[167,168],"For 1,147 try dividing by primes near its square root (about 34).","Use (a+1)(b+1)(c+1)… to count divisors from the index form.",{"id":170,"type":47,"variant":171,"title":172,"markdown":173},"aha-common-divisors","aha","Every common divisor divides the HCF","The argument above shows more than you might notice. **Every** common divisor of 360 and 588 has at most two 2s and at most one 3, so every common divisor divides 2² × 3 = 12. The common divisors of 360 and 588 are exactly the divisors of 12: 1, 2, 3, 4, 6, 12.\n\nThis is why, in Discover, the common factors of 12 and 18 turned out to be the factors of 6. It also explains the name GCD in advanced maths: the HCF is \"greatest\" not just in size, but in the stronger sense that **every** other common divisor divides it.",{"id":175,"type":90,"title":176,"eyebrow":177,"navLabel":178},"ch03","Proof: HCF × LCM = a × b","Chapter 03","3 Product proof",{"id":180,"type":43,"markdown":181},"proof-product","Look at the last column of the table above: for every prime, **min + max = the sum of the two powers**. That is not a coincidence. For any two numbers x and y, one of them is the smaller and the other the larger, so min(x, y) + max(x, y) = x + y. (If x = 3 and y = 2: 2 + 3 = 3 + 2.)\n\nNow multiply powers of the same prime by **adding indices**: 2² × 2³ = 2⁵.\n\n- In HCF × LCM, the power of each prime p is min + max.\n- In a × b, the power of each prime p is (power in a) + (power in b).\n\nThese are equal for **every** prime, so by the uniqueness of prime recipes, **HCF(a, b) × LCM(a, b) = a × b**. For 360 and 588: 12 × 17,640 = 2,11,680 = 360 × 588. ✓\n\nThat is a complete proof, for every pair of positive whole numbers, not just the ones we tested.",{"id":183,"type":47,"variant":99,"title":184,"markdown":185},"nuance-proof-three","Why the same proof fails for three numbers","For three numbers the powers of a prime are three values, say x, y, z. HCF × LCM gets min + max, but a × b × c gets x + y + z. These are equal only if the **middle** value is 0.\n\nFor 2, 4, 8 the powers of 2 are 1, 2, 3: min + max = 4 but the sum is 6, so the product is 2² = 4 times bigger than HCF × LCM. Check: 2 × 4 × 8 = 64 and HCF × LCM = 2 × 8 = 16. ✓\n\nThis explains the Investigate result: the rule holds when, for every prime, at most two of the numbers contain it, and in practice that happens when the numbers are **pairwise co-prime** (then each prime is in only one of them).",{"id":187,"type":43,"markdown":188},"three-formula","For strong learners, here is a formula that **does** work for three numbers. It corrects for the primes shared by exactly two of them:\n\n**LCM(a, b, c) = a × b × c × HCF(a, b, c) ÷ [HCF(a, b) × HCF(b, c) × HCF(c, a)]**\n\nTry it on 6, 10, 15: the product is 900, HCF(6, 10, 15) = 1, and the pairwise HCFs are 2, 5 and 3. So LCM = 900 × 1 ÷ 30 = **30**. ✓ On 12, 18, 30: 6,480 × 6 ÷ (6 × 6 × 6) = **180**. ✓ It can be proved prime by prime, like the two-number rule, using this fact about any three powers x, y, z: **max(x, y, z) = x + y + z − min(x, y) − min(y, z) − min(z, x) + min(x, y, z)**. Try it with 1, 2, 3: 6 − 1 − 2 − 1 + 1 = 3. ✓",{"id":190,"type":150,"component":191,"componentVersion":5,"config":192,"objective":208,"textAlternative":209},"lab-venn-three-d","hcf-lcm",{"sets":193,"views":206,"challenge":162},[194,198,202,204],[195,196,197],6,10,15,[199,200,201],12,18,30,[203,195,196],4,[200,201,205],42,[207],"venn","Use a three-circle prime Venn diagram to check the three-number LCM formula and see which primes are double-counted.","This lab places the prime factors of three numbers in three overlapping circles. The centre is the HCF of all three; each two-circle overlap is the extra part shared by just that pair.\n\n- **6, 10, 15:** product 900; HCF(a, b, c) = 1; pairwise HCFs 2, 5, 3; formula gives 900 × 1 ÷ 30 = **30** = LCM.\n- **12, 18, 30:** product 6,480; HCF(a, b, c) = 6; pairwise HCFs 6, 6, 6; formula gives 6,480 × 6 ÷ 216 = **180** = LCM.\n- **4, 6, 10:** product 240; HCF(a, b, c) = 2; pairwise HCFs 2, 2, 2; formula gives 240 × 2 ÷ 8 = **60** = LCM.\n- **18, 30, 42:** product 22,680; HCF(a, b, c) = 6; pairwise HCFs 6, 6, 6; formula gives 22,680 × 6 ÷ 216 = **630** = LCM.\n\nThe pairwise HCFs remove the primes counted twice in the product; multiplying back by the HCF of all three restores the primes that were removed once too often.",{"id":211,"type":212,"itemId":213,"prompt":214,"check":215,"hints":219,"feedback":222},"practice-product-proof","practice","hcf-and-lcm.deepen-missing-large","The HCF of two numbers is **18** and their LCM is **1,512**. One number is **216**. What is the other?",{"kind":216,"answer":217,"tolerance":218},"number",126,0,[220,221],"HCF × LCM = product.","18 × 1,512 = 27,216.",{"correct":223,"incorrect":224},"Yes: 27,216 ÷ 216 = 126. Check: HCF(216, 126) = 18 and LCM = 1,512.","18 × 1,512 = 27,216 = 216 × other, so other = 126.",{"id":226,"type":90,"title":227,"eyebrow":228,"navLabel":229},"ch04","Every common multiple is a multiple of the LCM","Chapter 04","4 Multiples of LCM",{"id":231,"type":43,"markdown":232},"cm-proof","In Discover you noticed that the common multiples of 4 and 6 are 12, 24, 36, … : all multiples of the LCM. Here is a proof that avoids prime factors entirely and uses only division with remainder.\n\nLet L = LCM(a, b) and let M be **any** common multiple of a and b. Divide M by L:\n\n**M = q × L + r**, where the remainder r is 0, 1, 2, …, up to L − 1.\n\n- a divides M and a divides L (so a divides q × L). Therefore a divides M − q × L = r.\n- In the same way, b divides r.\n- So r is a common multiple of a and b (or 0). But r is **smaller than L**, and L is the *lowest* positive common multiple. So r cannot be a positive common multiple: r must be **0**.\n\nTherefore M = q × L: every common multiple is a multiple of the LCM. ∎",{"id":234,"type":47,"variant":48,"title":235,"markdown":236},"observation-difference-trick","The trick used here","The argument used one simple principle twice: **if d divides x and d divides y, then d divides x − y** (and x + y, and x − q × y for any whole q). Write x = d × s and y = d × t; then x − y = d × (s − t).\n\nThe same principle proves that the HCF divides the difference of two numbers (Investigate), and it is the whole engine of Euclid’s algorithm in the next chapter.",{"id":238,"type":90,"title":239,"eyebrow":240,"navLabel":241},"ch05","Why Euclid’s long division method works","Chapter 05","5 Euclid’s proof",{"id":243,"type":43,"markdown":244},"euclid-proof","Euclid’s method replaces the pair (a, b) by the pair (b, r), where a = q × b + r. Here is why the HCF does not change.\n\n**Claim.** The pairs (a, b) and (b, r) have **exactly the same common divisors**.\n\n- If d divides a and b, then d divides a − q × b = r. So d is a common divisor of b and r.\n- If d divides b and r, then d divides q × b + r = a. So d is a common divisor of a and b.\n\nSame common divisors means same **greatest** common divisor: HCF(a, b) = HCF(b, r).\n\n**Why it stops.** Each remainder is smaller than the divisor before it, so the numbers keep getting smaller. Whole numbers cannot decrease for ever, so eventually a remainder is 0. At that point we have a pair (g, 0), and HCF(g, 0) = g, because every number divides 0. So the last non-zero remainder is the HCF. ∎",{"id":246,"type":247,"title":248,"problem":249,"steps":250,"help":259},"we-euclid-big","worked_example","HCF(1,591, 1,147) without factorising","Find HCF(1,591, 1,147) by Euclid’s method, and use the result to factorise both numbers.",[251,252,253,254,255,256,257,258],"1,591 = 1,147 × 1 + 444. HCF(1,591, 1,147) = HCF(1,147, 444).","1,147 = 444 × 2 + 259. HCF(1,147, 444) = HCF(444, 259).","444 = 259 × 1 + 185. HCF(444, 259) = HCF(259, 185).","259 = 185 × 1 + 74. HCF(259, 185) = HCF(185, 74).","185 = 74 × 2 + 37. HCF(185, 74) = HCF(74, 37).","74 = 37 × 2 + 0. Remainder 0, so the HCF is **37**.","So HCF = 37. Now divide: 1,591 ÷ 37 = 43 and 1,147 ÷ 37 = 31. Both 43 and 31 are prime.","So 1,591 = 37 × 43 and 1,147 = 31 × 37, and LCM = 31 × 37 × 43 = 49,321.",{"simplerExplanation":260},"Each division keeps the same common factors but makes the numbers smaller, until the answer is obvious.",{"id":262,"type":47,"variant":263,"title":264,"markdown":265},"example-subtraction","example","The subtraction version: ancient Chinese \"mutual subtraction\"","You do not even need division. Since HCF(a, b) = HCF(b, a − b), you can just keep subtracting the smaller number from the larger:\n\n(98, 63) → (63, 35) → (35, 28) → (28, 7) → (21, 7) → (14, 7) → (7, 7) → HCF = **7**.\n\nThe Chinese classic *The Nine Chapters on the Mathematical Art* uses exactly this \"mutual subtraction\" (更相减损) to simplify fractions. Division is just a faster way of doing many subtractions at once: 28 − 7 − 7 − 7 = 7 is the same as 28 = 7 × 3 + 7.",{"id":267,"type":150,"component":191,"componentVersion":5,"config":268,"objective":288,"textAlternative":289,"help":290},"lab-euclid",{"sets":269,"views":286,"challenge":162},[270,272,275,278,281,283],[271,18],89,[273,274],377,233,[276,277],987,610,[279,280],1000,625,[279,282],999,[284,285],598,403,[287],"division","Run Euclid’s algorithm on pairs chosen to be fast and slow, count the steps, and find which pairs make it work hardest.","This lab runs long division for the HCF step by step and counts the divisions.\n\n- **(89, 55):** 9 divisions, HCF = 1. Quotients: 1, 1, 1, 1, 1, 1, 1, 1, 2.\n- **(377, 233):** 12 divisions, HCF = 1. Quotients: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2.\n- **(987, 610):** 14 divisions, HCF = 1. Quotients: 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2.\n- **(1000, 625):** 4 divisions, HCF = 125. Quotients: 1, 1, 1, 2.\n- **(1000, 999):** 2 divisions, HCF = 1.\n- **(598, 403):** 3 divisions, HCF = 13.\n\nThe slowest pairs are neighbouring Fibonacci numbers (55, 89, 233, 377, 610, 987 are all Fibonacci numbers): every quotient is 1, so each step shrinks the numbers as little as possible. (1,000, 999) looks big but finishes in 2 steps.",{"hints":291},[292,293],"Fibonacci numbers: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, …","Count a step every time you divide.",{"id":295,"type":90,"title":296,"eyebrow":297,"navLabel":298},"ch06","How fast is Euclid? The Fibonacci worst case","Chapter 06","6 Speed",{"id":300,"type":43,"markdown":301},"speed-prose","Try Euclid on two neighbouring **Fibonacci numbers** (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …, each the sum of the two before). Every quotient is 1 and every remainder is the previous Fibonacci number:\n\n89 = 55 × 1 + 34; 55 = 34 × 1 + 21; 34 = 21 × 1 + 13; … all the way down to 2 = 1 × 2 + 0.\n\nThat is 9 divisions for numbers under 100, the most for any pair that small. In 1844 the French mathematician **Gabriel Lamé** proved that this is the worst case: **the number of division steps is never more than 5 times the number of digits in the smaller number.** So even for numbers with 100 digits, Euclid needs at most 500 steps, which a computer does in a blink. Factorising two 100-digit numbers, by contrast, can be impossibly slow. That is why computers find HCFs by Euclid, not by prime factors.",{"id":303,"type":128,"caption":304,"columns":305,"rows":311},"table-euclid-steps","Division steps for some pairs (counted by computer)",[306,307,308,309,310],"Pair","Kind of pair","Steps","HCF","Lamé limit: 5 × digits of smaller",[312,317,321,325,328,333,337],[313,314,315,143,316],"89, 55","Fibonacci neighbours","9","10",[318,314,319,143,320],"987, 610","14","15",[322,314,323,143,324],"10,946, 6,765","19","20",[326,327,139,143,320],"1,000, 999","consecutive",[329,330,331,332,324],"1,591, 1,147","products of primes","6","37",[334,335,143,336,320],"10,000, 625","one divides the other","625",[338,339,147,331,324],"1,23,456, 7,890","random-looking",{"id":341,"type":342,"title":343,"note":344,"scale":345,"rungs":346},"ladder-speed","ladder","Numbers up to… and the worst-case number of Euclid steps","Checked by computer over every pair below each limit: the slowest pair is always a pair of neighbouring Fibonacci numbers.","linear",[347,350,354],{"label":348,"value":203,"display":349},"Below 10","4 steps (8, 5)",{"label":351,"value":352,"display":353},"Below 100",9,"9 steps (89, 55)",{"label":355,"value":356,"display":357},"Below 1,000",14,"14 steps (987, 610)",{"id":359,"type":90,"title":360,"eyebrow":361,"navLabel":362},"ch07","The HCF as a combination: Bézout’s identity","Chapter 07","7 Bézout",{"id":364,"type":43,"markdown":365},"bezout-prose","Run Euclid’s steps **backwards** and something remarkable appears. For 252 and 198 the steps were:\n\n252 = 198 × 1 + 54; 198 = 54 × 3 + 36; 54 = 36 × 1 + 18.\n\nNow rewrite, starting from the last useful line and substituting upwards:\n\n- 18 = 54 − 36\n- 36 = 198 − 3 × 54, so 18 = 54 − (198 − 3 × 54) = 4 × 54 − 198\n- 54 = 252 − 198, so 18 = 4 × (252 − 198) − 198 = **4 × 252 − 5 × 198**\n\nCheck: 4 × 252 = 1,008 and 5 × 198 = 990, and 1,008 − 990 = **18**. ✓\n\nThis always works: **the HCF of a and b can be written as a × x + b × y for some whole numbers x and y (one of them negative).** It is called **Bézout’s identity**, after Étienne Bézout (1730–1783) — although for whole numbers it was stated much earlier, by Claude Gaspard Bachet de Méziriac in 1624, and Bézout’s own proof was for polynomials. Earlier still, the Indian mathematician **Aryabhata** used the same back-substitution around 499 CE in his *kuttaka* (\"pulveriser\") method for solving equations like a × x − b × y = c.",{"id":367,"type":47,"variant":171,"title":368,"markdown":369},"aha-bezout","What Bézout means for jugs and stamps","Bézout says that with a 3-litre jug and a 5-litre jug you can measure out exactly 1 litre, because HCF(3, 5) = 1 and 1 = 2 × 3 − 5 (fill the 3-litre jug twice, pour out one full 5-litre jug). And with a 4-litre and a 6-litre jug you can **never** measure an odd amount, because every amount you can make is a combination 4x + 6y, which is always a multiple of HCF(4, 6) = 2.\n\nThe amounts you can make with two jugs are exactly the multiples of their HCF (up to the size of the jugs). The Extend layer turns this into puzzles.",{"id":371,"type":212,"itemId":372,"prompt":373,"check":374,"hints":389,"feedback":392},"practice-bezout","hcf-and-lcm.deepen-bezout-which","Which of these amounts can be written as **6x + 15y** for some whole numbers x and y (positive, negative or zero)?",{"kind":375,"options":376,"correct":388},"choice",[377,379,382,385],{"id":378,"label":316},"a",{"id":380,"label":381},"b","21",{"id":383,"label":384},"c","25",{"id":386,"label":387},"d","32",[380],[390,391],"Every combination 6x + 15y is a multiple of HCF(6, 15).","HCF(6, 15) = 3.",{"correct":393,"incorrect":394},"Right. HCF(6, 15) = 3 and 21 is a multiple of 3: for example 21 = 6 × 1 + 15 × 1. 10, 25 and 32 are not multiples of 3.","Any 6x + 15y is divisible by 3, since both 6 and 15 are. Only 21 is a multiple of 3 (21 = 6 + 15).",{"id":396,"type":90,"title":397,"eyebrow":398,"navLabel":399},"ch08","Reasoning with letters: HCF for every n","Chapter 08","8 Letters",{"id":401,"type":43,"markdown":402},"letters-prose","The \"divides the difference\" principle lets you prove facts about **infinitely many** pairs at once. Use letters for the numbers, and subtract cleverly, exactly as Euclid would.\n\n**Example.** Show that n and 2n + 1 are co-prime for every whole number n.\n\nAny common divisor d of n and 2n + 1 also divides 2 × n, and so divides (2n + 1) − 2n = **1**. The only divisor of 1 is 1. So HCF(n, 2n + 1) = 1, whatever n is. (Try n = 7: HCF(7, 15) = 1. n = 100: HCF(100, 201) = 1.)",{"id":404,"type":405,"prompt":406,"options":407,"explanation":416},"predict-letters","prediction","What is HCF(3n + 1, 2n + 1) for every whole number n?",[408,410,412,414],{"id":378,"label":409},"It depends on n",{"id":380,"label":411},"Always 1",{"id":383,"label":413},"Always n",{"id":386,"label":415},"Always 2","**Always 1.** Run Euclid with letters: (3n + 1) − (2n + 1) = n, so HCF(3n + 1, 2n + 1) = HCF(2n + 1, n). Then (2n + 1) − 2 × n = 1, so it equals HCF(n, 1) = 1. Checked by computer for every n up to 3,000.",{"id":418,"type":247,"title":419,"problem":420,"steps":421},"we-letters","HCF(n² + 1, n + 1) can only be 1 or 2","Show that HCF(n² + 1, n + 1) is always 1 or 2, and find when it is 2.",[422,423,424,425,426],"Let d be a common divisor. d divides n + 1, so it divides (n + 1) × (n − 1) = n² − 1.","d also divides n² + 1. So d divides (n² + 1) − (n² − 1) = **2**.","The only divisors of 2 are 1 and 2, so the HCF is 1 or 2.","It is 2 exactly when both numbers are even, which happens when n is odd (then n + 1 and n² + 1 are both even).","Check: n = 3 gives HCF(10, 4) = 2; n = 4 gives HCF(17, 5) = 1. ✓",{"id":428,"type":212,"itemId":429,"prompt":430,"check":431,"hints":432,"feedback":435},"practice-letters","hcf-and-lcm.deepen-letters-5n","For every whole number n, HCF(5n + 3, 3n + 2) is the same number. What is it?",{"kind":216,"answer":5,"tolerance":218},[433,434],"Subtract: (5n + 3) − (3n + 2) = 2n + 1.","Then (3n + 2) − (2n + 1) = n + 1, and keep going.",{"correct":436,"incorrect":437},"Yes: the Euclid steps with letters end at 1, so the numbers are always co-prime.","(5n + 3) − (3n + 2) = 2n + 1; (3n + 2) − (2n + 1) = n + 1; (2n + 1) − (n + 1) = n; (n + 1) − n = 1. So the HCF is 1.",{"id":439,"type":90,"title":440,"eyebrow":441,"navLabel":442},"ch09","Edge cases: 0, 1, equal numbers and negatives","Chapter 09","9 Edge cases",{"id":444,"type":128,"caption":445,"columns":446,"rows":450},"table-edge","What happens at the edges",[447,309,448,449],"Case","LCM","Why",[451,454,457,461,465,469,472],[452,143,378,453],"a and 1","1 divides everything, and nothing bigger than 1 divides 1.",[455,378,378,456],"a and a","A number is its own biggest factor and its own smallest positive multiple.",[458,378,459,460],"a and a multiple of a, say ka","ka","a divides both, and ka is already a multiple of both.",[462,143,463,464],"two different primes p, q","p × q","They share no prime factor.",[466,378,467,468],"a and 0","not defined (or 0)","Every number divides 0, so the common divisors of a and 0 are the divisors of a. The only common multiple of a and 0 is 0, which the positive LCM excludes.",[470,467,467,471],"0 and 0","Every number divides 0, so there is no greatest one. Computer languages usually return 0 by convention.",[473,331,474,475],"negative numbers, e.g. −12 and 18","36","Signs do not affect divisibility, so HCF and LCM are taken as positive.",{"id":477,"type":47,"variant":478,"title":479,"markdown":480},"careful-zero","careful","Careful with zero","School problems never ask for the HCF or LCM with 0, but Euclid’s algorithm itself ends at a pair like (18, 0), and says HCF(18, 0) = 18. This is not a trick: 18 divides 18, and 18 divides 0 (0 = 18 × 0), and no bigger number divides 18. The **L**CM is different: 0 is a multiple of everything, so we insist the LCM be **positive**, and then there is no LCM with 0 at all.",{"id":482,"type":212,"itemId":483,"prompt":484,"check":485,"hints":487,"feedback":490},"practice-edge","hcf-and-lcm.deepen-edge-hcf-lcm","What is LCM(HCF(84, 0), 1)?",{"kind":216,"answer":486,"tolerance":218},84,[488,489],"HCF(84, 0) = 84, because every number divides 0.","LCM(a, 1) = a.",{"correct":491,"incorrect":492},"Yes: HCF(84, 0) = 84, and LCM(84, 1) = 84.","Work from the inside: HCF(84, 0) = 84. Then LCM(84, 1) = 84.",{"id":494,"type":90,"title":495,"eyebrow":496,"navLabel":497},"ch10","Harder calculations","Chapter 10","10 Harder problems",{"id":499,"type":247,"title":500,"problem":501,"steps":502},"we-ncert-remainders","Three remainders, one divisor","Find the largest number that divides **1,251**, **9,377** and **15,628** leaving remainders **1**, **2** and **3** respectively.",[503,504,505,506,507],"Subtract each remainder: the number must divide 1,250, 9,375 and 15,625 exactly.","Prime factors: 1,250 = 2 × 5⁴; 9,375 = 3 × 5⁵; 15,625 = 5⁶.","Common primes: only 5, and the smallest power is 5⁴. HCF = 5⁴ = **625**.","Check: 1,251 = 625 × 2 + 1 ✓; 9,377 = 625 × 15 + 2 ✓; 15,628 = 625 × 25 + 3 ✓.","Also check the divisor (625) is bigger than every remainder, which it is.",{"id":509,"type":247,"title":510,"problem":511,"steps":512},"we-lcm-range","The largest 4-digit number with a given remainder","Find the **largest 4-digit number** that leaves remainder **5** when divided by **12**, **18** and **30**.",[513,514,515,516],"LCM(12, 18, 30) = 180. Numbers with remainder 5 are 180k + 5.","The largest multiple of 180 up to 9,999: 9,999 ÷ 180 = 55 remainder 99, so 180 × 55 = 9,900.","Add 5: 9,900 + 5 = **9,905**, which is still 4 digits.","Check: 9,905 ÷ 12, ÷ 18 and ÷ 30 each leave remainder 5.",{"id":518,"type":247,"title":519,"problem":520,"steps":521},"we-pairs-hcf-lcm","How many pairs have a given HCF and LCM?","Find all pairs of whole numbers (a, b) with a ≤ b, **HCF = 12** and **LCM = 360**.",[522,523,524,525,526],"Write a = 12m and b = 12n with HCF(m, n) = 1 (the 12 takes all the shared part).","Then LCM = 12 × m × n, so m × n = 360 ÷ 12 = 30.","Split 30 into two co-prime factors with m ≤ n: 1 × 30, 2 × 15, 3 × 10, 5 × 6. (Each prime 2, 3, 5 must go wholly to m or to n.)","Pairs: (12, 360), (24, 180), (36, 120), (60, 72). **Four pairs.**","Check one: HCF(60, 72) = 12 and LCM(60, 72) = 360. ✓",{"id":528,"type":212,"itemId":529,"prompt":530,"check":531,"hints":533,"feedback":536},"practice-pairs","hcf-and-lcm.deepen-count-pairs","How many pairs (a, b) with a ≤ b have **HCF 6** and **LCM 210**?",{"kind":216,"answer":532,"tolerance":218},2,[534,535],"Write a = 6m, b = 6n with m × n = 210 ÷ 6 = 35 and HCF(m, n) = 1.","35 = 5 × 7. How many ways can you split it into co-prime m ≤ n?",{"correct":537,"incorrect":538},"Yes: m × n = 35 gives (1, 35) and (5, 7), so the pairs are (6, 210) and (30, 42).","m × n = 35 with m, n co-prime: (1, 35) and (5, 7). Pairs: (6, 210) and (30, 42). So 2.",{"id":540,"type":47,"variant":541,"title":542,"markdown":543},"misconception-pairs","misconception","\"Any factor pair of LCM ÷ HCF will do\"","The split m × n must use **co-prime** m and n. Try HCF 4 and LCM 48: 48 ÷ 4 = 12, and 12 = 2 × 6. That gives (8, 24), but HCF(8, 24) is 8, not 4, because 2 and 6 share a 2. The valid splits of 12 are only 1 × 12 and 3 × 4, giving (4, 48) and (12, 16).",{"id":545,"type":90,"title":546,"eyebrow":547,"navLabel":548},"ch11","More proofs and olympiad ideas","Chapter 11","11 More proofs",{"id":550,"type":247,"title":551,"problem":552,"steps":553},"we-proof-even","Proof: consecutive even numbers have HCF 2","Prove that HCF(2k, 2k + 2) = 2 for every whole number k ≥ 1.",[554,555,556,557,558],"Both 2k and 2k + 2 are even, so 2 is a common divisor. The HCF is at least 2.","Any common divisor d divides the difference (2k + 2) − 2k = 2.","So d is 1 or 2. The greatest common divisor is therefore at most 2.","At least 2 and at most 2: the HCF is exactly **2**. ∎","Check: HCF(14, 16) = 2, HCF(98, 100) = 2.",{"id":560,"type":247,"title":561,"problem":562,"steps":563},"we-mersenne","Olympiad classic: HCF(2⁶ − 1, 2⁸ − 1)","Find HCF(63, 255), and notice that 63 = 2⁶ − 1 and 255 = 2⁸ − 1.",[564,565,566,567],"Euclid: 255 = 63 × 4 + 3; 63 = 3 × 21 + 0. HCF = **3**.","Now 3 = 2² − 1, and 2 = HCF(6, 8). So HCF(2⁶ − 1, 2⁸ − 1) = 2^HCF(6, 8) − 1.","This is true in general: **HCF(2^m − 1, 2^n − 1) = 2^HCF(m, n) − 1**. The reason is that Euclid’s steps on the exponents (8, 6) → (6, 2) → (2, 0) are mirrored by the numbers: 2⁸ − 1 = 2² × (2⁶ − 1) + (2² − 1).","Try: HCF(2¹² − 1, 2¹⁸ − 1) = HCF(4,095, 2,62,143) = 63 = 2⁶ − 1, since HCF(12, 18) = 6.",{"id":569,"type":47,"variant":541,"title":570,"markdown":571},"misconception-sum-divides","\"If d divides a + b, then d divides a and b\"","The difference trick only works in one direction. If d divides **both** a and b, then it divides a + b and a − b. But d dividing the **sum** tells you nothing about each part: 3 divides 1 + 2 = 3, yet 3 divides neither 1 nor 2. Students who reverse the rule \"prove\" false results, so always check which facts you already know and which you are trying to show.",{"id":573,"type":47,"variant":541,"title":574,"markdown":575},"misconception-hcf-product","\"HCF(a, bc) = HCF(a, b) × HCF(a, c)\"","It looks natural, but it fails when b and c share a factor with a in overlapping ways. HCF(4, 2 × 2) = 4, and HCF(4, 2) × HCF(4, 2) = 4: works here. But HCF(2, 2 × 2) = 2 while HCF(2, 2) × HCF(2, 2) = 4. Wrong. It **is** true when b and c are co-prime. When a formula \"looks right\", test an edge case with small repeated primes before trusting it.",{"id":577,"type":212,"itemId":578,"prompt":579,"check":580,"hints":582,"feedback":585},"practice-mersenne","hcf-and-lcm.deepen-mersenne-hcf","Use the pattern HCF(2^m − 1, 2^n − 1) = 2^HCF(m, n) − 1 to find **HCF(2¹⁰ − 1, 2¹⁵ − 1)**, that is HCF(1,023, 32,767).",{"kind":216,"answer":581,"tolerance":218},31,[583,584],"HCF(10, 15) = 5.","So the answer is 2⁵ − 1.",{"correct":586,"incorrect":587},"Yes: 2⁵ − 1 = 31.","HCF(10, 15) = 5, so the HCF is 2⁵ − 1 = 31. (Check: 1,023 = 31 × 33 and 32,767 = 31 × 1,057.)",{"id":589,"type":212,"itemId":590,"prompt":591,"check":592,"hints":594,"feedback":597},"practice-bezout-2","hcf-and-lcm.deepen-bezout-17-5","Find the smallest positive whole number x for which **17x − 5y = 1** has a whole-number solution y.",{"kind":216,"answer":593,"tolerance":218},3,[595,596],"Try x = 1, 2, 3, … and check whether 17x − 1 is a multiple of 5.","HCF(17, 5) = 1 guarantees a solution exists.",{"correct":598,"incorrect":599},"17 × 3 = 51 and 51 − 1 = 50 = 5 × 10. So x = 3, y = 10.","x = 1: 16 ✗; x = 2: 33 ✗; x = 3: 50 = 5 × 10 ✓. So x = 3. This is the kind of equation Aryabhata’s kuttaka solves.",{"id":601,"type":212,"itemId":602,"prompt":603,"check":604,"hints":615,"feedback":617},"practice-proof-step","hcf-and-lcm.deepen-proof-step","In the proof that HCF(n, n + 1) = 1, which step is the key one?",{"kind":375,"options":605,"correct":614},[606,608,610,612],{"id":378,"label":607},"n and n + 1 are both even",{"id":380,"label":609},"Any common divisor divides (n + 1) − n = 1",{"id":383,"label":611},"n + 1 is always prime",{"id":386,"label":613},"The LCM is n + 1",[380],[616],"What does a common divisor of two numbers always divide?",{"correct":618,"incorrect":619},"Right. A common divisor divides the difference, which is 1, so it must be 1.","The key step is that a common divisor divides the difference, (n + 1) − n = 1. Options a and c are false, and d is irrelevant.",{"id":621,"type":247,"title":622,"problem":623,"steps":624},"we-same-remainder-proof","Why \"same remainder\" problems use differences","Explain why the largest number that leaves the **same remainder** when dividing 62, 132 and 237 is HCF(132 − 62, 237 − 132), and find it.",[625,626,627,628,629],"Suppose d leaves remainder r each time: 62 = d × q₁ + r, 132 = d × q₂ + r, 237 = d × q₃ + r.","Subtract: 132 − 62 = d × (q₂ − q₁), so d divides 70. Likewise 237 − 132 = d × (q₃ − q₂), so d divides 105.","So d is a common divisor of 70 and 105. Conversely any common divisor of 70 and 105 leaves equal remainders on 62, 132, 237 (their differences are multiples of it).","The largest is HCF(70, 105) = **35**. Check: 62, 132 and 237 all leave remainder 27 on division by 35.","The unknown remainder r cancels out when you subtract. That is the whole trick.",{"id":631,"type":247,"title":632,"problem":633,"steps":634},"we-lcm-proof-division","Proof: LCM = a × b ÷ HCF without prime factors","Show that a × b ÷ HCF(a, b) is a common multiple of a and b, and check it with 18 and 24.",[635,636,637,638],"Let h = HCF(a, b), and write a = h × m, b = h × n. Here m and n are whole numbers with HCF(m, n) = 1.","Then a × b ÷ h = h × m × n = a × n = b × m. So it is a multiple of a (a × n) and of b (b × m).","Showing it is the **lowest** common multiple needs the fact that m and n share no factor, which is where co-primality enters.","Check: 18 = 6 × 3 and 24 = 6 × 4, so a × b ÷ h = 6 × 3 × 4 = **72** = LCM(18, 24) = 72 ✓.",{"id":640,"type":212,"itemId":641,"prompt":642,"check":643,"hints":645,"feedback":648},"practice-three-rem","hcf-and-lcm.deepen-same-remainder","Find the largest number that leaves the **same remainder** when it divides **1,305**, **4,665** and **6,905**.",{"kind":216,"answer":644,"tolerance":218},1120,[646,647],"Take the differences: 4,665 − 1,305 and 6,905 − 4,665.","The answer is the HCF of the differences.",{"correct":649,"incorrect":650},"Differences 3,360 and 2,240; HCF = 1120. Each number leaves remainder 185 on division by 1,120.","4,665 − 1,305 = 3,360 and 6,905 − 4,665 = 2,240. HCF(3,360, 2,240) = 1120.",{"id":652,"type":90,"title":653,"eyebrow":654,"navLabel":655},"ch12","A short history of the HCF","Chapter 12","12 History",{"id":657,"type":658,"title":659,"items":660},"timeline-history","timeline","From Alexandria to your phone",[661,665,669,673,677,681,685,689],{"time":662,"title":663,"text":664},"c. 300 BCE","Euclid’s Elements","Book VII, Propositions 1–2 give the \"anthyphairesis\" (mutual subtraction) method for the greatest common measure of two numbers: the oldest non-trivial algorithm still in daily use.",{"time":666,"title":667,"text":668},"c. 100 CE","The Nine Chapters","The Chinese classic uses \"mutual subtraction\" to simplify fractions, finding the HCF of numerator and denominator. Its date is uncertain: historians place the text anywhere from about 200 BCE to the 1st century CE.",{"time":670,"title":671,"text":672},"499 CE","Aryabhata’s kuttaka","In the Aryabhatiya, Aryabhata solves equations like a × x − b × y = c using repeated division, the same steps as Euclid run backwards.",{"time":674,"title":675,"text":676},"628 CE","Brahmagupta","The Brahmasphutasiddhanta extends the kuttaka to problems in astronomy, where planetary cycles must line up.",{"time":678,"title":679,"text":680},"1624","Bachet","Claude Gaspard Bachet de Méziriac states the identity for whole numbers in the second edition of his Problèmes plaisants et délectables; Étienne Bézout (1730–1783), whose name it now carries, later proved the version for polynomials.",{"time":682,"title":683,"text":684},"1801","Gauss","Article 16 of Carl Friedrich Gauss’s Disquisitiones Arithmeticae appears to be the first proof that a number’s prime factorisation is unique.",{"time":686,"title":687,"text":688},"1844","Lamé","Gabriel Lamé proves the Fibonacci worst case: Euclid needs at most 5 steps per digit, one of the first results in the study of how fast algorithms run.",{"time":690,"title":691,"text":692},"1970s–today","Computers","Euclid and its extended version run inside computer algebra, fraction arithmetic and the RSA encryption that protects online payments.",{"id":694,"type":47,"variant":695,"title":696,"markdown":697},"model-limit-history","model_limit","History is simplified here","Dates for ancient texts are approximate, and credit for ideas is often shared across many people over centuries. Euclid collected and organised earlier Greek work; the Nine Chapters grew over several centuries; and Indian mathematicians after Aryabhata, such as Bhaskara I and Bhaskara II, refined the kuttaka. Treat the timeline as a map of landmarks, not a complete story.",{"id":699,"type":90,"title":700,"eyebrow":701,"navLabel":702},"ch13","Check your reasoning","Chapter 13","13 Wrap-up",{"id":704,"type":150,"component":705,"componentVersion":5,"config":706,"objective":734,"textAlternative":735},"lab-match-deepen","match-pairs",{"prompt":707,"mode":708,"pairs":709},"Connect each fact to the reason it is true.","connect",[710,713,716,719,722,725,728,731],{"a":711,"b":712},"HCF × LCM = a × b","min + max of two powers = their sum",{"a":714,"b":715},"HCF(a, b) = HCF(b, r)","d | a and d | b ⇔ d | b and d | r",{"a":717,"b":718},"Euclid’s method always stops","Remainders keep getting smaller",{"a":720,"b":721},"Common multiples are multiples of LCM","Remainder on ÷ LCM is a smaller common multiple",{"a":723,"b":724},"Common divisors divide the HCF","Their powers are at most the min powers",{"a":726,"b":727},"HCF of consecutive numbers is 1","A common divisor divides the difference, 1",{"a":729,"b":730},"Product rule fails for 2, 4, 8","min + max ≠ sum of three powers",{"a":732,"b":733},"Prime-factor method is valid","Prime recipes are unique","Match each HCF\u002FLCM fact with the key idea in its proof.","This game pairs eight facts with the reasons behind them:\n\n- HCF × LCM = a × b ← for each prime, min + max of two powers equals their sum.\n- HCF(a, b) = HCF(b, r) ← a number divides a and b exactly when it divides b and r, since r = a − q × b.\n- Euclid’s method always stops ← the remainders keep getting smaller, and whole numbers cannot shrink for ever.\n- Every common multiple is a multiple of the LCM ← its remainder on dividing by the LCM would be a smaller common multiple, so it must be 0.\n- Every common divisor divides the HCF ← its prime powers are at most the minimum powers.\n- The HCF of consecutive numbers is 1 ← a common divisor divides their difference, 1.\n- The product rule fails for 2, 4, 8 ← for three powers, min + max is not the sum.\n- The prime-factor method is valid ← every number has exactly one prime recipe.",{"id":737,"type":150,"component":738,"componentVersion":5,"config":739,"objective":789,"textAlternative":790},"lab-sort-deepen","sort-game",{"prompt":740,"bins":741,"items":748,"seconds":218},"Is each piece of reasoning valid, or is there a flaw?",[742,745],{"id":743,"label":744},"valid","Valid reasoning",{"id":746,"label":747},"flawed","Flawed reasoning",[749,753,757,761,765,769,773,777,781,785],{"id":750,"label":751,"bin":743,"why":752},"d1","d divides a and b, so d divides a − 3b.","If a = ds and b = dt then a − 3b = d(s − 3t).",{"id":754,"label":755,"bin":746,"why":756},"d2","HCF × LCM = product worked for 20 pairs, so it is proved.","Testing examples is evidence, not proof. The min + max argument is the proof.",{"id":758,"label":759,"bin":743,"why":760},"d3","HCF(2, 4, 8) × LCM = 16 ≠ 64, so the product rule is false for three numbers in general.","One counterexample is enough to show a general rule is false.",{"id":762,"label":763,"bin":746,"why":764},"d4","6 divides 4 × 9, so 6 divides 4 or 9.","That property holds only for primes. 6 = 2 × 3 splits between 4 and 9.",{"id":766,"label":767,"bin":743,"why":768},"d5","Remainders in Euclid keep shrinking, so the method must stop.","A strictly decreasing list of whole numbers cannot go on for ever.",{"id":770,"label":771,"bin":746,"why":772},"d6","m × n = 12 with m = 2, n = 6, so (8, 24) has HCF 4 and LCM 48.","2 and 6 are not co-prime; HCF(8, 24) is actually 8.",{"id":774,"label":775,"bin":743,"why":776},"d7","Any common divisor of n and n + 1 divides 1, so they are co-prime.","It divides the difference, which is 1.",{"id":778,"label":779,"bin":746,"why":780},"d8","LCM(a, b, c) = LCM(LCM(a, b), c), so HCF × LCM = abc for three numbers too.","The first fact is true, but it does not imply the second; 2, 4, 8 is a counterexample.",{"id":782,"label":783,"bin":743,"why":784},"d9","Every number divides 0, so HCF(15, 0) = 15.","The common divisors of 15 and 0 are the divisors of 15; the greatest is 15.",{"id":786,"label":787,"bin":746,"why":788},"d10","441 = 21 × 21 and 21 = 3 × 7, so 441 has two prime recipes.","In ordinary whole numbers 441 = 3² × 7² only. Two recipes happen only in the 1, 5, 9, 13… world.","Judge whether short arguments about HCF and LCM are valid proofs or contain a flaw.","This game has ten short arguments to sort into \"Valid reasoning\" and \"Flawed reasoning\".\n\nValid: if d divides a and b it divides a − 3b; one counterexample (2, 4, 8) shows the product rule fails for three numbers; shrinking remainders force Euclid to stop; a common divisor of n and n + 1 divides 1; HCF(15, 0) = 15 because every number divides 0.\n\nFlawed: 20 examples do not prove a rule; \"6 divides 4 × 9 so 6 divides 4 or 9\" wrongly uses a property that only primes have; (8, 24) has HCF 8, not 4, because the split 2 × 6 is not co-prime; LCM(a, b, c) = LCM(LCM(a, b), c) does not imply the three-number product rule; 441 has only one prime recipe, 3² × 7², among ordinary whole numbers.",{"id":792,"type":793,"title":794,"questions":795},"quiz-deepen","quiz","Reasoning check",[796,805,818,831,843,853,866,877,889,899],{"itemId":797,"prompt":798,"options":799,"correct":383,"why":804},"hcf-and-lcm.deepen-q-minmax","a = 2⁵ × 3 × 11 and b = 2² × 3⁴. What is the power of 2 in HCF(a, b) × LCM(a, b)?",[800,801,802,803],{"id":378,"label":139},{"id":380,"label":141},{"id":383,"label":147},{"id":386,"label":316},"min(5, 2) + max(5, 2) = 2 + 5 = 7, the same as in a × b.",{"itemId":806,"prompt":807,"options":808,"correct":380,"why":817},"hcf-and-lcm.deepen-q-euclid-step","Euclid’s method replaces (a, b) by (b, r). Which fact makes this safe?",[809,811,813,815],{"id":378,"label":810},"r is always prime",{"id":380,"label":812},"Common divisors of a and b are exactly the common divisors of b and r",{"id":383,"label":814},"r is always even",{"id":386,"label":816},"a and b are co-prime","Since r = a − q × b and a = q × b + r, any common divisor of one pair divides the other pair too.",{"itemId":819,"prompt":820,"options":821,"correct":383,"why":830},"hcf-and-lcm.deepen-q-fib","Which pair makes Euclid’s algorithm take the most steps?",[822,824,826,828],{"id":378,"label":823},"(100, 99)",{"id":380,"label":825},"(96, 48)",{"id":383,"label":827},"(89, 55)",{"id":386,"label":829},"(97, 2)","89 and 55 are neighbouring Fibonacci numbers: every quotient is 1. It takes 9 steps; (100, 99) takes 2.",{"itemId":832,"prompt":833,"options":834,"correct":386,"why":842},"hcf-and-lcm.deepen-q-divisors","How many divisors does 2⁴ × 3² × 7 have?",[835,836,838,840],{"id":378,"label":147},{"id":380,"label":837},"8",{"id":383,"label":839},"24",{"id":386,"label":841},"30","(4 + 1)(2 + 1)(1 + 1) = 30. Computed: 30.",{"itemId":844,"prompt":845,"options":846,"correct":380,"why":852},"hcf-and-lcm.deepen-q-common-count","How many common factors do 360 and 588 have?",[847,848,849,851],{"id":378,"label":139},{"id":380,"label":331},{"id":383,"label":850},"12",{"id":386,"label":839},"The common factors are the factors of HCF = 12: 1, 2, 3, 4, 6, 12. That is 6.",{"itemId":854,"prompt":855,"options":856,"correct":383,"why":865},"hcf-and-lcm.deepen-q-bezout","Using a 6-litre and a 10-litre jug (fill, empty, pour), which amount can you **never** measure?",[857,859,861,863],{"id":378,"label":858},"2 L",{"id":380,"label":860},"4 L",{"id":383,"label":862},"7 L",{"id":386,"label":864},"8 L","Every amount is a combination 6x + 10y, a multiple of HCF(6, 10) = 2. 7 is odd.",{"itemId":867,"prompt":868,"options":869,"correct":383,"why":876},"hcf-and-lcm.deepen-q-zero","HCF(45, 0) = ?",[870,871,872,874],{"id":378,"label":145},{"id":380,"label":143},{"id":383,"label":873},"45",{"id":386,"label":875},"not defined","Every number divides 0, so the common divisors are the divisors of 45; the greatest is 45.",{"itemId":878,"prompt":879,"options":880,"correct":380,"why":888},"hcf-and-lcm.deepen-q-three-formula","For 4, 6, 10: product 240, HCF(4, 6, 10) = 2, pairwise HCFs 2, 2, 2. What does the three-number formula give for the LCM?",[881,882,884,886],{"id":378,"label":841},{"id":380,"label":883},"60",{"id":383,"label":885},"120",{"id":386,"label":887},"240","240 × 2 ÷ (2 × 2 × 2) = 480 ÷ 8 = 60, and LCM(4, 6, 10) = 60. ✓",{"itemId":890,"prompt":891,"options":892,"correct":383,"why":898},"hcf-and-lcm.deepen-q-pairs","How many pairs a ≤ b have HCF 5 and LCM 150?",[893,894,895,897],{"id":378,"label":143},{"id":380,"label":139},{"id":383,"label":896},"4",{"id":386,"label":331},"150 ÷ 5 = 30 = 2 × 3 × 5 splits into co-prime m ≤ n as 1 × 30, 2 × 15, 3 × 10, 5 × 6: four pairs (5, 150), (10, 75), (15, 50), (25, 30).",{"itemId":900,"prompt":901,"options":902,"correct":380,"why":911},"hcf-and-lcm.deepen-q-history","Aryabhata’s kuttaka method (499 CE) is closely related to…",[903,905,907,909],{"id":378,"label":904},"the sieve of Eratosthenes",{"id":380,"label":906},"Euclid’s algorithm run backwards",{"id":383,"label":908},"the ladder method for LCM",{"id":386,"label":910},"counting divisors","The kuttaka solves a × x − b × y = c by repeated division and back-substitution, the same steps as extended Euclid.",{"id":913,"type":914,"title":915,"points":916},"cheat-sheet-deepen","summary","Cheat sheet",[917,918,919,920,921,922,923,924,925,926],"**Fundamental Theorem of Arithmetic:** every number > 1 has exactly one prime recipe. It rests on: if a prime divides a product, it divides a factor.","**Divisors:** d | n exactly when d’s prime powers are at most n’s. Number of divisors = product of (index + 1).","**HCF** = min powers of each prime; **LCM** = max powers. Every common divisor divides the HCF; every common multiple is a multiple of the LCM.","**Proof of HCF × LCM = a × b:** for each prime, min + max = sum of the two powers.","**Three numbers:** min + max ≠ sum of three powers, so the rule fails unless pairwise co-prime. Correct: LCM(a, b, c) = abc × HCF(a, b, c) ÷ [HCF(a, b) HCF(b, c) HCF(c, a)].","**Euclid:** HCF(a, b) = HCF(b, r) because the two pairs have the same common divisors. It stops because remainders shrink.","**Speed:** Fibonacci neighbours are the worst case; at most 5 steps per digit of the smaller number (Lamé, 1844).","**Bézout:** HCF(a, b) = ax + by for some whole x, y. Every combination ax + by is a multiple of the HCF.","**Edges:** HCF(a, 0) = a; LCM with 0 is not defined; HCF(a, 1) = 1; signs are ignored.","**Pairs with HCF h and LCM l:** a = hm, b = hn with m × n = l ÷ h and m, n co-prime.",{"id":928,"type":929,"prompt":930},"reflect-deepen","reflection","Explain to a friend, in your own words and without prime factors, why Euclid’s long division method gives the HCF. Then explain why it must eventually stop. Which step of your explanation would you find hardest to convince a sceptic about?",{"id":932,"type":933,"conceptId":934,"relation":935,"explanation":936},"conn-prime-d","connection","prime-and-composite","helps_understand","Unique prime factorisation, the foundation of this layer, is the deepest fact about primes.",{"id":938,"type":933,"conceptId":939,"relation":940,"explanation":941},"conn-four-ops-d","four-operations","applied_in","Division with remainder, a = q × b + r, drives every proof in Euclid’s method.",{"id":943,"type":933,"conceptId":944,"relation":945,"explanation":946},"conn-patterns-d","patterns","related_to","Fibonacci numbers are the slowest inputs for Euclid’s algorithm.",{"id":948,"type":949,"sourceIds":950},"sources-deepen","sources",[951,952,953,954,955,956,957],"hcf-and-lcm-ncert-class10-real-numbers","hcf-and-lcm-wiki-gcd","hcf-and-lcm-wiki-lcm","hcf-and-lcm-wiki-euclidean-algorithm","hcf-and-lcm-wiki-bezout-identity","hcf-and-lcm-wiki-fundamental-theorem-arithmetic","hcf-and-lcm-mactutor-nine-chapters",[951,952,953,954,955,956,957],"needs_review",{"generatedBy":961,"notes":962},"claude-code","Draft generated with Python-checked arithmetic; pending owner review.","4bbe91e1d951456a85ffa5e483775facd4b73b0c83d8815b9594f643df43324d",{"component:factor-tree@1":965,"component:hcf-lcm@1":966,"logic:practice":967,"component:match-pairs@1":968,"component:sort-game@1":969,"source:hcf-and-lcm-mactutor-nine-chapters":970,"source:hcf-and-lcm-ncert-class10-real-numbers":971,"source:hcf-and-lcm-wiki-bezout-identity":972,"source:hcf-and-lcm-wiki-euclidean-algorithm":973,"source:hcf-and-lcm-wiki-fundamental-theorem-arithmetic":974,"source:hcf-and-lcm-wiki-gcd":975,"source:hcf-and-lcm-wiki-lcm":976},"dfb598a254f20e70550d586121be496f7a546ba0254060fd97b95cb919dfd934","83f068508b17b70184e45fbcf95d356790bfa2b0a98dcdb859f8e9dc105e5d33","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","5178e268c9ef0180a9ef1b16753a988c0175b4429d5e94e8e7cd7b4ea6297ce5","48d1fd4c018916e9edeb2f76b15a8eb3cb83c72b08a9d3ab332a3c167fe9be0b","88847a983d29f0ee4313233f6f22ecc4ce290b238b4d3d9f0f79bfbdd0bcf646","c37b91030f0a162c4968726c72618dce986659b01f396306c685ac9597ae61df","50100e58ebbaddcff1524ff76d0091b4f415d0e612aea2fbce479c1b0babca86","857640343ebbf479db694438bf4af134fd5baee47f017b64110ded8b40bb2dc6","12412937f9bd209912735d662496fbe90c125f89d22844b5a753d934cc84ef2c",{"state":978,"reviewer":979,"selfReview":162,"reviewedAt":980,"method":981},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597289]