[{"data":1,"prerenderedAt":1026},["ShallowReactive",2],{"layer:hcf-and-lcm:extend":3},{"layer":4,"contentHash":1001,"dependencyHashes":1002,"approval":1020,"releaseId":1025},{"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":996,"reviewStatus":997,"authoring":998},1,"hcf-and-lcm","en","extend","Cycles, gears and puzzles: HCF and LCM in the wild","Calendars, cicadas, tabla, bicycles, jugs, screens and olympiad problems","Expeditions beyond the textbook: cycles with head starts, calendars and planetary alignments (and why they are not LCMs), prime-cycle cicadas, gears and bicycle chains, tala rhythms, water jugs, ancient remainder puzzles, screen ratios, fractions, olympiad problems, careers and open questions.",[13,14,15,16,17],"Decide whether two cycles with different starting points can ever coincide, using the HCF.","Apply HCF and LCM to gears, rhythms, screens, calendars and jug puzzles, and explain where the models break down.","Solve remainder puzzles, including Sunzi’s, using the LCM and systematic listing.","Find the HCF and LCM of fractions and use them in problems.","Tackle olympiad-style problems that combine HCF, LCM and counting.",60,{"title":20,"rows":21},"Lesson plate",[22,25,28,31,34,37],{"label":23,"value":24},"Depth","Extend",{"label":26,"value":27},"Reading time","≈ 60 minutes",{"label":29,"value":30},"Prior knowledge","Understand and Investigate layers",{"label":32,"value":33},"Chapters","12",{"label":35,"value":36},"Contexts","Calendars, cicadas, gears, tala, screens",{"label":38,"value":39},"Labs","Cycle race, cicadas, screens, sort, match",[41,45,51,96,102,105,116,121,126,129,134,139,163,178,183,186,228,233,243,248,253,256,291,296,301,304,330,357,362,365,374,383,396,401,404,438,452,455,468,473,476,486,496,506,520,532,597,602,611,616,624,633,644,656,668,677,688,700,705,766,771,775,780,785,817,942,957,961,967,972,976,980],{"id":42,"type":43,"markdown":44},"intro-extend","prose","HCF and LCM are not just exam topics. They are the mathematics of **cycles that repeat** and **things that must fit exactly**, and the world is full of both: calendars and festivals, planets and insects, gears and bicycle chains, tabla rhythms, screen shapes, water jugs, and the codes that protect online payments.\n\nThis layer is a set of expeditions. Some are real-world contexts with honest caveats. Some are puzzles, including olympiad-style problems. Some end in questions nobody has answered yet. Pick the ones that interest you; you do not have to go in order.",{"id":46,"type":47,"variant":48,"title":49,"markdown":50},"how-to-extend","callout","observation","How to use this layer","Every chapter has something to **do**: a lab, a puzzle, a practice question or a project idea. The olympiad-style problems in Chapter 8 are meant to be hard. Spend at least ten minutes on one before reading the solution; getting stuck and unstuck is where the learning happens.",{"id":52,"type":53,"title":54,"terms":55},"glossary-extend","glossary","Words from the expeditions",[56,60,64,68,72,76,80,84,88,92],{"term":57,"meaning":58,"example":59},"cycle \u002F period","The length of time (or count) after which something repeats.","A 4-year Olympic cycle",{"term":61,"meaning":62,"example":63},"offset (head start)","How far apart two cycles start.","FIFA World Cup is offset 2 years from the Olympics.",{"term":65,"meaning":66,"example":67},"synodic period","How often two orbiting bodies line up again as seen from the Sun (or Earth).","Jupiter–Saturn: about 19.9 years",{"term":69,"meaning":70,"example":71},"Metonic cycle","19 years, almost exactly 235 lunar months: the basis of many lunisolar calendars.","19 × 365.24 ≈ 235 × 29.53 days",{"term":73,"meaning":74,"example":75},"hunting tooth","An extra gear tooth that makes the two tooth counts co-prime, so every tooth meets every other.","12 and 17 teeth",{"term":77,"meaning":78,"example":79},"tala \u002F sam","A rhythmic cycle in Indian classical music, and its first beat.","Teentaal has 16 beats.",{"term":81,"meaning":82,"example":83},"polyrhythm","Two or more different beat cycles played at the same time.","3 against 2 repeats every 6 pulses.",{"term":85,"meaning":86,"example":87},"aspect ratio","The shape of a rectangle as width : height in simplest form.","1920 × 1080 → 16 : 9",{"term":89,"meaning":90,"example":91},"Chinese Remainder Theorem","For pairwise co-prime divisors, any set of remainders has exactly one solution below their LCM.","Sunzi: 23",{"term":93,"meaning":94,"example":95},"inclusion–exclusion","Count \"a or b\" as count(a) + count(b) − count(both).","250 + 166 − 83 = 333",{"id":97,"type":98,"title":99,"eyebrow":100,"navLabel":101},"ch01","chapter","Calendars, festivals and cycles with a head start","Chapter 01","1 Calendars",{"id":103,"type":43,"markdown":104},"calendar-prose","Suppose a village has a weekly bazaar every **7 days** and a travelling cattle fair that returns every **5 days**. If both happen today, they coincide again in LCM(7, 5) = **35 days**. So far, so Discover.\n\nReal cycles, though, often **do not start together**. Here is a case that surprises most people. The Summer Olympics are held in years divisible by 4 (2024, 2028, 2032 …). The men's FIFA World Cup is also every 4 years, but in the years in between (2026, 2030, 2034 …). Both have a 4-year cycle, LCM(4, 4) = 4, yet they **never** fall in the same year. Their head starts differ by 2 years, and no amount of waiting fixes that.\n\nThe rule for cycles with a head start: two cycles of lengths a and b can only ever meet if the **difference in their starting points is a multiple of HCF(a, b)**. If they meet once, they then meet every LCM(a, b).",{"id":106,"type":107,"title":108,"problem":109,"steps":110},"we-offset","worked_example","Two festivals with a head start","A dance festival is held every **4 years** starting in 2024. A science fair is held every **6 years** starting in 2026. In which year do both first happen? How often after that?",[111,112,113,114,115],"Festival years: 2024, 2028, 2032, 2036, … Fair years: 2026, 2032, 2038, …","Difference in start years: 2026 − 2024 = 2. HCF(4, 6) = 2, and 2 is a multiple of 2, so they **can** meet.","Scanning the lists, the first shared year is **2032**.","After that they coincide every LCM(4, 6) = 12 years: 2032, 2044, 2056, …","If the fair had started in 2025 instead (a head start of 1, not a multiple of 2), they would never meet: festival years are all even, fair years all odd.",{"id":117,"type":47,"variant":118,"title":119,"markdown":120},"example-metonic","example","A calendar built on an almost-common multiple","Lunisolar calendars, such as Hindu calendars and the Hebrew calendar, have to keep months (set by the Moon) in step with years (set by the Sun). A month is about 29.53 days and a year about 365.24 days. They have no exact common multiple, but **19 years** ≈ 6,939.6 days and **235 months** ≈ 6,939.7 days, which agree to within about two hours. This near-match is the **Metonic cycle**. The Hebrew calendar uses it directly, adding 7 leap months every 19 years. Hindu lunisolar calendars also add an extra month, the **adhik maas**, but they work it out from the real positions of the Sun and Moon rather than from the 19-year cycle: it comes about once every 32 to 33 months, which works out at roughly seven extra months in 19 years. Real calendars like this are built on \"close enough\" LCMs.",{"id":122,"type":98,"title":123,"eyebrow":124,"navLabel":125},"ch02","Planets and cicadas: nature’s cycles","Chapter 02","2 Sky and insects",{"id":127,"type":43,"markdown":128},"planets-prose","Here is a curiosity that appears in many puzzle books: \"Three planets orbit a star in 3, 4 and 6 years. They are lined up today. When will they line up again?\" The textbook answer is LCM(3, 4, 6) = **12 years**, because then each planet has made a whole number of orbits (4, 3 and 2) and is back at its starting point.\n\nThat is correct for the puzzle, but it answers a slightly different question from real astronomy. In the puzzle, every planet returns to its **starting position**. Real alignments only need the planets to line up in the **same direction from the Sun**, anywhere around the orbit, and that happens much more often.",{"id":130,"type":47,"variant":131,"title":132,"markdown":133},"model-limit-planets","model_limit","Why real planetary alignments are not LCMs","Jupiter takes about 11.86 years to orbit the Sun and Saturn about 29.46 years. Their periods are not whole numbers, so there is no exact LCM at all. Yet Jupiter and Saturn line up (a \"great conjunction\") about every **19.9 years**, most recently in December 2020.\n\nThat number comes from **relative motion**, not from an LCM: Jupiter gains one full lap on Saturn every 1 ÷ (1⁄11.86 − 1⁄29.46) ≈ 19.9 years, like a fast runner lapping a slow one on a track. The LCM picture is a helpful toy model for whole-number cycles, but astronomers use the relative-motion formula.",{"id":135,"type":47,"variant":136,"title":137,"markdown":138},"nuance-cicadas","nuance","Why do some cicadas count in primes?","In eastern North America, periodical cicadas spend **13 or 17 years** underground, then emerge together in their millions. Both are prime numbers. One well-known hypothesis says a prime cycle helps them avoid predators that boom on shorter cycles.\n\nThe LCM shows the idea. A 12-year cicada would meet a predator with a 2-, 3-, 4- or 6-year cycle every **12 years**, at every emergence. A 17-year cicada meets a 2-year predator only every 34 years, a 3-year one every 51 years, and a 6-year one every 102 years.\n\nThis is a hypothesis, not a settled result, and biologists still debate it. A second leading idea is that prime cycles stop broods of different cycle lengths from emerging together and interbreeding, which would blur the cycle. It is a great example of mathematics suggesting an explanation that scientists must then test.",{"id":140,"type":141,"component":142,"componentVersion":5,"config":143,"objective":161,"textAlternative":162},"lab-frogs-cycles","interactive","hcf-lcm",{"sets":144,"views":157,"challenge":160},[145,149,151,152,155],[146,147,148],3,4,6,[150,147],12,[150,148],[153,154],16,10,[150,156],7,[158,159],"frogs","lists",true,"Race cycles of different lengths, from toy planets to tala rhythms, and predict when they first coincide.","This lab races two or three cycles, shown as frogs hopping along a number line. The first shared stone is when the cycles coincide.\n\n- **Toy planets:** cycles 3 and 4 and 6 → first coincide at **12**.\n- **12-year cicada, 4-year predator:** cycles 12 and 4 → first coincide at **12**.\n- **12-year cicada, 6-year predator:** cycles 12 and 6 → first coincide at **12**.\n- **Teentaal (16 beats) and Jhaptaal (10 beats):** cycles 16 and 10 → first coincide at **80**.\n- **Ektaal (12 beats) and Rupak (7 beats):** cycles 12 and 7 → first coincide at **84**.\n\nA cycle with many factors (12) meets short cycles at every step; co-prime cycles such as 12 and 7 take the full product to meet.",{"id":164,"type":141,"component":142,"componentVersion":5,"config":165,"objective":176,"textAlternative":177},"lab-cicada-lists",{"sets":166,"views":174,"challenge":160},[167,169,171,172,173],[168,147],17,[170,148],13,[168,148],[150,147],[150,148],[159,175],"venn","Compare prime and non-prime cicada cycles with predator cycles and see how rarely prime cycles coincide.","This lab lists multiples and shows the prime factors of a cicada cycle and a predator cycle.\n\n- **17-year cicada, 4-year predator:** HCF 1, they coincide every **68 years**, which is every 4 cicada emergences.\n- **13-year cicada, 6-year predator:** HCF 1, they coincide every **78 years**, which is every 6 cicada emergences.\n- **17-year cicada, 6-year predator:** HCF 1, they coincide every **102 years**, which is every 6 cicada emergences.\n- **12-year cicada, 4-year predator:** HCF 4, they coincide every **12 years**, which is every 1 cicada emergence.\n- **12-year cicada, 6-year predator:** HCF 6, they coincide every **12 years**, which is every 1 cicada emergence.\n\nA prime cycle shares no factor with any shorter cycle, so the LCM is the full product and meetings are rare. The 12-year cycle meets 4- and 6-year predators at every single emergence.",{"id":179,"type":98,"title":180,"eyebrow":181,"navLabel":182},"ch03","Gears, chains and the hunting tooth","Chapter 03","3 Gears",{"id":184,"type":43,"markdown":185},"gears-prose","When two gears mesh, their teeth take turns. Paint one tooth on each gear red, and start with the two red teeth touching. When do they touch again?\n\nEvery time a tooth passes the meeting point, both gears move on by one tooth. The small gear's red tooth is back after every 12 teeth (for a 12-tooth gear); the big gear's after every 18. They are both back together after LCM(12, 18) = **36 teeth**, which is **3 turns** of the small gear and **2 turns** of the big one.\n\nEngineers care about this. If the same pairs of teeth meet over and over, any tiny flaw on one tooth wears the same partner teeth again and again. Choosing tooth counts that are **co-prime**, for example 12 and 17, makes every tooth on one gear meet every tooth on the other before the pattern repeats: LCM(12, 17) = **204** teeth, 17 turns of the small gear. The extra tooth that makes the counts co-prime is traditionally called a **hunting tooth**.",{"id":187,"type":188,"caption":189,"columns":190,"rows":197},"table-gears","table","Two meshing gears: when do the same two teeth meet again? (all computed)",[191,192,193,194,195,196],"Teeth","HCF","LCM (teeth passed)","Turns of small gear","Turns of big gear","Different partner teeth for each tooth",[198,204,209,213,218,223],[199,200,201,202,203,202],"12 and 18","6","36","3","2",[205,206,207,208,33,208],"12 and 17","1","204","17",[210,211,212,202,203,202],"20 and 30","10","60",[214,206,215,216,217,216],"20 and 31","620","31","20",[219,220,221,222,202,222],"15 and 40","5","120","8",[224,206,225,226,227,226],"16 and 41","656","41","16",{"id":229,"type":47,"variant":230,"title":231,"markdown":232},"aha-hunting","aha","Each tooth meets LCM ÷ small teeth partners","In the last column: a tooth on the small gear meets a **different** tooth of the big gear on each turn until the pattern repeats. With 12 and 18, each small-gear tooth only ever meets 3 big-gear teeth. With 12 and 17 it meets all 17. The number of partners is (big teeth) ÷ HCF: 18 ÷ 6 = 3, and 17 ÷ 1 = 17.",{"id":234,"type":107,"title":235,"problem":236,"steps":237},"we-bicycle","The bicycle chain","A bicycle chain has **116 links** and runs over a front chainring with **48 teeth**. A particular chain link sits on a particular tooth. After how many links have passed will that same link sit on that same tooth again? How many pedal turns is that?",[238,239,240,241,242],"Each link that passes moves the chain on by one link and the chainring on by one tooth.","The link returns to its starting place every 116 links; the tooth every 48.","Both return together after LCM(116, 48) links. 116 = 2² × 29 and 48 = 2⁴ × 3, so LCM = 2⁴ × 3 × 29 = **1,392 links**.","That is 29 turns of the chainring (so **29 pedal turns**) and 12 trips of the chain all the way round.","Because HCF(116, 48) = 4, each chain link only ever sits on 48 ÷ 4 = 12 of the 48 teeth. Wear is spread over a quarter of the teeth for each link.",{"id":244,"type":47,"variant":245,"title":246,"markdown":247},"careful-gear-ratio","careful","Gear ratio is a different question","Do not mix up \"when do the same teeth meet?\" (an LCM question) with the **gear ratio**, which says how fast the wheels turn. With a 48-tooth chainring and a 16-tooth rear sprocket, one pedal turn pulls 48 links, which turns the rear sprocket 48 ÷ 16 = 3 times. The ratio 48 : 16 simplifies, using HCF 16, to **3 : 1**. So both HCF and LCM appear on a bicycle, answering different questions.",{"id":249,"type":98,"title":250,"eyebrow":251,"navLabel":252},"ch04","Rhythm: tabla cycles and polyrhythms","Chapter 04","4 Rhythm",{"id":254,"type":43,"markdown":255},"rhythm-prose","Indian classical music is organised in **tala** cycles. **Teentaal** has 16 beats, **Jhaptaal** 10, **Ektaal** 12 and **Rupak** 7. The first beat of each cycle is the **sam**, where musicians often land together with a flourish.\n\nImagine a duet where one drummer plays in Teentaal (16) and another in Jhaptaal (10), starting together on sam. Their sams coincide again after LCM(16, 10) = **80 beats**: 5 cycles of Teentaal and 8 of Jhaptaal. With Teentaal and Rupak, it takes LCM(16, 7) = **112 beats**, because 16 and 7 are co-prime.\n\nIn a **tihai**, a phrase is played three times so that it ends exactly on sam; composers use the arithmetic of cycles to make the landing come out right. And in West African, Latin American and Western music, **polyrhythms** like \"3 against 2\" repeat every LCM(3, 2) = 6 pulses, and \"4 against 3\" every 12.",{"id":257,"type":188,"caption":258,"columns":259,"rows":264},"table-tala","When do two tala cycles land on sam together? (all computed)",[260,261,192,262,263],"Tala pair","Beats per cycle","Sams coincide every","Cycles of each",[265,270,276,281,286],[266,267,203,268,269],"Teentaal + Jhaptaal","16 and 10","80 beats","5 and 8",[271,272,273,274,275],"Teentaal + Ektaal","16 and 12","4","48 beats","3 and 4",[277,278,206,279,280],"Teentaal + Rupak","16 and 7","112 beats","7 and 16",[282,283,203,284,285],"Jhaptaal + Ektaal","10 and 12","60 beats","6 and 5",[287,288,206,289,290],"Ektaal + Rupak","12 and 7","84 beats","7 and 12",{"id":292,"type":47,"variant":293,"title":294,"markdown":295},"try-it-rhythm","try_it","Clap a polyrhythm","Count steadily from 1 to 6, over and over. Clap with your **left** hand on 1, 3 and 5 (every 2 counts) and with your **right** hand on 1 and 4 (every 3 counts). Both hands clap together only on count 1, every LCM(2, 3) = 6 counts. Once it feels smooth, try 3 against 4: count to 12, left hand every 3, right hand every 4.",{"id":297,"type":98,"title":298,"eyebrow":299,"navLabel":300},"ch05","Water jugs and the power of the HCF","Chapter 05","5 Jug puzzles",{"id":302,"type":43,"markdown":303},"jugs-prose","A classic puzzle: you have an unmarked **3-litre** jug and an unmarked **5-litre** jug, and a tap. You may fill a jug, empty a jug, or pour from one into the other until one is full or the other empty. **Can you measure exactly 4 litres?**\n\nYes, and the Deepen layer explains why: HCF(3, 5) = 1, and any amount that is a multiple of the HCF (up to the bigger jug) can be made. With a **4-litre** and a **6-litre** jug you can make 2, 4 and 6 litres, but never 1, 3 or 5, because every combination of 4s and 6s is even: HCF(4, 6) = 2.",{"id":305,"type":306,"title":307,"items":308},"steps-jugs","steps","Measuring 4 litres with 3 L and 5 L jugs",[309,313,317,321,324,327],{"title":310,"tag":311,"text":312},"Fill the 5 L jug","(0, 5)","Small jug empty, big jug 5 litres.",{"title":314,"tag":315,"text":316},"Pour big into small","(3, 2)","The 3 L jug fills up; 2 litres stay in the big jug.",{"title":318,"tag":319,"text":320},"Empty the small jug","(0, 2)","Pour the 3 litres away.",{"title":314,"tag":322,"text":323},"(2, 0)","The 2 litres move into the small jug.",{"title":310,"tag":325,"text":326},"(2, 5)","Small jug has 2 litres, big jug full.",{"title":314,"tag":328,"text":329},"(3, 4)","Only 1 litre fits into the small jug, so exactly **4 litres** remain in the big jug.",{"id":331,"type":332,"itemId":333,"prompt":334,"check":335,"hints":351,"feedback":354},"practice-jugs","practice","hcf-and-lcm.extend-jugs-which","With a **6-litre** and a **9-litre** jug (fill, empty, pour), which amount can you measure in the 9-litre jug?",{"kind":336,"options":337,"correct":350},"choice",[338,341,344,347],{"id":339,"label":340},"a","1 litre",{"id":342,"label":343},"b","3 litres",{"id":345,"label":346},"c","4 litres",{"id":348,"label":349},"d","5 litres",[342],[352,353],"Every amount is a multiple of HCF(6, 9).","HCF(6, 9) = 3.",{"correct":355,"incorrect":356},"Yes: fill the 9 L jug and pour into the 6 L jug; 3 litres remain. Only multiples of HCF(6, 9) = 3 are possible.","All reachable amounts are multiples of HCF(6, 9) = 3: 0, 3, 6, 9. So 3 litres is the only possible choice here.",{"id":358,"type":98,"title":359,"eyebrow":360,"navLabel":361},"ch06","Ancient remainder puzzles: the Chinese Remainder Theorem","Chapter 06","6 Remainder puzzles",{"id":363,"type":43,"markdown":364},"crt-prose","Around the 3rd to 5th century CE, the Chinese book *Sunzi Suanjing* posed this puzzle:\n\n> There are some things whose number is unknown. Counted in threes, 2 are left over; counted in fives, 3 are left over; counted in sevens, 2 are left over. How many things are there?\n\nThe smallest answer is **23**. And here is where the LCM comes in: the answers repeat every LCM(3, 5, 7) = **105**, giving 23, 128, 233, 338, …\n\nThe general result is called the **Chinese Remainder Theorem**: if the divisors are pairwise co-prime, then for **any** choice of remainders there is exactly one answer between 0 and the LCM, and the rest are that answer plus multiples of the LCM. Indian mathematicians, from Aryabhata’s kuttaka to Brahmagupta and Bhaskara, developed methods for the same kind of problem, often to work out when astronomical cycles would line up.",{"id":366,"type":107,"title":367,"problem":368,"steps":369},"we-sieve-crt","Solving a remainder puzzle by sieving","Find the smallest number that leaves remainder **1** when divided by **3** and remainder **2** when divided by **4**.",[370,371,372,373],"List numbers that leave remainder 2 on division by 4 (start with the bigger divisor): 2, 6, 10, 14, 18, 22, …","Test each on division by 3: 2 → 2 ✗; 6 → 0 ✗; 10 → 1 ✓.","**10** works. Check: 10 = 3 × 3 + 1 and 10 = 4 × 2 + 2. ✓","All answers: 10, 22, 34, …, stepping by LCM(3, 4) = 12.",{"id":375,"type":107,"title":376,"problem":377,"steps":378},"we-eggs","The egg-basket puzzle","A woman carrying a basket of eggs is bumped and the eggs break. She cannot remember how many there were, but when she took them out 2, 3, 4, 5 or 6 at a time, one egg was always left over; taken out 7 at a time, none were left. What is the smallest number of eggs she could have had?",[379,380,381,382],"One left over for 2, 3, 4, 5 and 6 means (number − 1) is a common multiple of 2, 3, 4, 5, 6.","LCM(2, 3, 4, 5, 6) = 60. So the number is 61, 121, 181, 241, 301, … (60k + 1).","Now it must also be divisible by 7. Test: 61 ✗, 121 ✗, 181 ✗, 241 ✗, 301 = 7 × 43 ✓.","**301 eggs.** (The next answer is 301 + LCM(60, 7) = 301 + 420 = 721.)",{"id":384,"type":332,"itemId":385,"prompt":386,"check":387,"hints":390,"feedback":393},"practice-sunzi-style","hcf-and-lcm.extend-crt-practice","Find the smallest positive number that leaves remainder **2** when divided by **5**, and remainder **3** when divided by **7**.",{"kind":388,"answer":168,"tolerance":389},"number",0,[391,392],"List numbers with remainder 3 on division by 7: 3, 10, 17, 24, …","Which is the first with remainder 2 on division by 5?",{"correct":394,"incorrect":395},"17 works: 17 = 5 × 3 + 2 and 17 = 7 × 2 + 3. The next is 17 + 35 = 52.","From 3, 10, 17, 24, … the first with remainder 2 on division by 5 is 17.",{"id":397,"type":98,"title":398,"eyebrow":399,"navLabel":400},"ch07","Screens, ratios and HCF of fractions","Chapter 07","7 Screens and fractions",{"id":402,"type":43,"markdown":403},"screens-prose","Every screen size is a ratio in disguise. A Full HD screen is **1920 × 1080** pixels. HCF(1920, 1080) = 120, and 1920 ÷ 120 = 16 while 1080 ÷ 120 = 9, so its shape is **16 : 9**. The screen is exactly a 16 × 9 grid of 120-pixel squares, the biggest square \"tile\" that fits, just like the courtyard in Discover.",{"id":405,"type":188,"caption":406,"columns":407,"rows":411},"table-screens","Screen resolutions simplified with the HCF (all computed)",[408,192,409,410],"Resolution","Aspect ratio","Where you meet it",[412,416,420,424,428,433],[413,221,414,415],"1920 × 1080","16 : 9","Full HD TVs, laptops",[417,418,414,419],"1280 × 720","80","HD video",[421,422,414,423],"3840 × 2160","240","4K TVs",[425,221,426,427],"2400 × 1080","20 : 9","many phones held sideways",[429,430,431,432],"1024 × 768","256","4 : 3","older tablets and monitors",[434,435,436,437],"2560 × 1600","320","8 : 5","some laptops",{"id":439,"type":141,"component":440,"componentVersion":5,"config":441,"objective":450,"textAlternative":451},"lab-factor-tree-screens","factor-tree",{"numbers":442,"showIndexForm":160},[443,444,445,446,447,448,449],1920,1080,1280,720,2400,3840,2160,"Factorise screen dimensions and use the shared primes to find each screen’s aspect ratio.","This lab builds factor trees for screen widths and heights.\n\n- 1,920 = **2⁷ × 3 × 5**\n- 1,080 = **2³ × 3³ × 5**\n- 1,280 = **2⁸ × 5**\n- 720 = **2⁴ × 3² × 5**\n- 2,400 = **2⁵ × 3 × 5²**\n- 3,840 = **2⁸ × 3 × 5**\n- 2,160 = **2⁴ × 3³ × 5**\n\nShared part of 1,920 and 1,080: 2³ × 3 × 5 = 120, leaving 2⁴ = 16 and 3² = 9: the ratio 16 : 9. For 2,400 and 1,080 the shared part is also 120, leaving 20 and 9: the ratio 20 : 9. 3,840 × 2,160 is exactly twice 1,920 × 1,080, so the ratio stays 16 : 9.",{"id":453,"type":43,"markdown":454},"fractions-hcf-prose","Can fractions have an HCF and LCM? Yes, if you ask the right question. Two runners take **¾ minute** and **⅚ minute** per lap. When are they next together at the start? We need the smallest time that is a whole number of laps for both: a common multiple of ¾ and ⅚.\n\nThe rule is: **LCM of fractions = LCM of the numerators ÷ HCF of the denominators** (with each fraction in simplest form). Here LCM(3, 5) ÷ HCF(4, 6) = 15 ÷ 2 = **7½ minutes**. Check: 7½ ÷ ¾ = 10 laps and 7½ ÷ ⅚ = 9 laps, both whole. ✓\n\nSimilarly, **HCF of fractions = HCF of the numerators ÷ LCM of the denominators**: the largest length that fits a whole number of times into both ¾ m and ⅚ m is HCF(3, 5) ÷ LCM(4, 6) = **1⁄12 m**, which fits 9 and 10 times.",{"id":456,"type":332,"itemId":457,"prompt":458,"check":459,"hints":462,"feedback":465},"practice-frac-lcm","hcf-and-lcm.extend-lcm-fractions","Two lights blink every **⅔ second** and every **¾ second**. They blink together now. After how many seconds do they next blink together? (Give a fraction.)",{"kind":460,"numerator":148,"denominator":5,"acceptEquivalent":461},"fraction",false,[463,464],"LCM of fractions = LCM of numerators ÷ HCF of denominators.","LCM(2, 3) = 6 and HCF(3, 4) = 1.",{"correct":466,"incorrect":467},"Yes: 6 ÷ 1 = 6 seconds. That is 9 blinks of the first light and 8 of the second.","LCM(2, 3) ÷ HCF(3, 4) = 6 ÷ 1 = 6 seconds. Check: 6 ÷ ⅔ = 9 and 6 ÷ ¾ = 8, both whole numbers.",{"id":469,"type":98,"title":470,"eyebrow":471,"navLabel":472},"ch08","Olympiad corner","Chapter 08","8 Olympiad corner",{"id":474,"type":43,"markdown":475},"olympiad-intro","These problems use nothing beyond this topic, but they need you to combine ideas. Try each for ten minutes before opening the solution.",{"id":477,"type":107,"title":478,"problem":479,"steps":480},"we-oly-sum","Sum and HCF","Two numbers add up to **528** and their HCF is **33**. How many such pairs are there?",[481,482,483,484,485],"Write the numbers as 33m and 33n with HCF(m, n) = 1.","Then 33m + 33n = 528, so m + n = 16.","Pairs with m smaller than n adding to 16: (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9). (8, 8) would make the numbers equal with HCF 264, so it is out.","Keep only co-prime pairs: (1, 15), (3, 13), (5, 11), (7, 9). The even ones share a 2.","**Four pairs:** (33, 495), (99, 429), (165, 363), (231, 297).",{"id":487,"type":107,"title":488,"problem":489,"steps":490},"we-oly-or","Divisible by 4 or 6","How many numbers from 1 to 1,000 are divisible by **4 or 6** (or both)?",[491,492,493,494,495],"Divisible by 4: 1,000 ÷ 4 = **250**.","Divisible by 6: 1,000 ÷ 6 = 166 remainder 4, so **166**.","Adding counts the numbers divisible by both twice. Divisible by both = divisible by LCM(4, 6) = 12: 1,000 ÷ 12 = 83 remainder 4, so **83**.","Total = 250 + 166 − 83 = **333**.","A common error is to subtract the multiples of 24 (4 × 6) instead of 12; that gives the wrong answer.",{"id":497,"type":107,"title":498,"problem":499,"steps":500},"we-oly-lcm-1-10","The smallest number every digit divides","What is the smallest positive number that is divisible by every whole number from 1 to 10? And from 1 to 12?",[501,502,503,504,505],"We need LCM(1, 2, 3, …, 10). Take the highest power of each prime that is at most 10.","Primes up to 10: 2, 3, 5, 7. Highest powers ≤ 10: 2³ = 8, 3² = 9, 5, 7.","LCM = 8 × 9 × 5 × 7 = **2,520**.","Up to 12: 11 is a new prime, and 12 = 2² × 3 needs nothing new. So LCM(1, …, 12) = 2,520 × 11 = **27,720**.","Notice 1 to 12 needed only one more factor, 11, even though we added two numbers.",{"id":507,"type":508,"prompt":509,"options":510,"explanation":519},"predict-lcm-growth","prediction","LCM(1, 2, …, n) is 2,520 for n = 10. For which of these n does the LCM **stay the same** when you go from n − 1 to n?",[511,513,515,517],{"id":339,"label":512},"n = 11",{"id":342,"label":514},"n = 13",{"id":345,"label":516},"n = 14",{"id":348,"label":518},"n = 16","**n = 14.** 14 = 2 × 7 and both 2 and 7 are already in the LCM with high enough powers, so LCM(1, …, 14) = LCM(1, …, 13) = 3,60,360. The LCM only grows when n is a **power of a prime**: 11 and 13 are primes, and 16 = 2⁴ needs a fourth 2 (LCM(1, …, 16) = 7,20,720).",{"id":521,"type":332,"itemId":522,"prompt":523,"check":524,"hints":526,"feedback":529},"practice-oly","hcf-and-lcm.extend-oly-hcf-sum","Two numbers have HCF **12** and LCM **144**, and neither number equals 12 or 144. What is their sum?",{"kind":388,"answer":525,"tolerance":389},84,[527,528],"Write them as 12m and 12n with m × n = 144 ÷ 12 = 12 and m, n co-prime.","Co-prime splits of 12: 1 × 12 and 3 × 4.",{"correct":530,"incorrect":531},"Yes: the split 3 × 4 gives 36 and 48, and 36 + 48 = 84.","m × n = 12 with HCF(m, n) = 1: (1, 12) gives 12 and 144 (not allowed); (3, 4) gives 36 and 48. Sum = 84.",{"id":533,"type":141,"component":534,"componentVersion":5,"config":535,"objective":595,"textAlternative":596},"lab-sort-extend","sort-game",{"prompt":536,"bins":537,"items":546,"seconds":389},"What does each problem need: the HCF, the LCM, or both?",[538,540,543],{"id":539,"label":192},"hcf",{"id":541,"label":542},"lcm","LCM",{"id":544,"label":545},"both","Both",[547,551,555,559,563,567,571,575,579,583,587,591],{"id":548,"label":549,"bin":539,"why":550},"e1","Simplify a 1920 × 1080 screen to its aspect ratio","Divide both by HCF 120 to get 16 : 9.",{"id":552,"label":553,"bin":541,"why":554},"e2","When do Teentaal (16) and Jhaptaal (10) land on sam together?","Cycles meeting: LCM = 80 beats.",{"id":556,"label":557,"bin":539,"why":558},"e3","Which amounts can a 4 L and a 6 L jug measure?","Only multiples of HCF(4, 6) = 2.",{"id":560,"label":561,"bin":544,"why":562},"e4","Two numbers with HCF 12 and LCM 144: find all pairs","Split LCM ÷ HCF = 12 into co-prime parts, then multiply by the HCF.",{"id":564,"label":565,"bin":541,"why":566},"e5","Sunzi’s puzzle: remainders 2, 3, 2 on ÷ 3, 5, 7","Answers repeat every LCM(3, 5, 7) = 105; the smallest is 23.",{"id":568,"label":569,"bin":544,"why":570},"e6","Can two 4-year and 6-year cycles starting 1 year apart ever meet?","They can meet only if 1 is a multiple of HCF 2 (it is not); if they did, it would repeat every LCM 12.",{"id":572,"label":573,"bin":541,"why":574},"e7","Gears of 12 and 18 teeth: when do the same teeth meet again?","After LCM(12, 18) = 36 teeth pass.",{"id":576,"label":577,"bin":539,"why":578},"e8","Gear ratio of a 48-tooth chainring and 16-tooth sprocket in simplest form","Divide by HCF 16 to get 3 : 1.",{"id":580,"label":581,"bin":544,"why":582},"e9","Runners with ¾ min and ⅚ min laps meet at the start","LCM of fractions = LCM(3, 5) ÷ HCF(4, 6) = 7½ minutes.",{"id":584,"label":585,"bin":541,"why":586},"e10","Numbers 1 to 1,000 divisible by 4 or 6","Subtract the double-counted multiples of LCM(4, 6) = 12: 333.",{"id":588,"label":589,"bin":539,"why":590},"e11","Squares crossed by the diagonal of a 12 × 18 grid","m + n − HCF(m, n) = 12 + 18 − 6 = 24.",{"id":592,"label":593,"bin":541,"why":594},"e12","How often do a 17-year cicada and 6-year predator emerge together?","LCM(17, 6) = 102 years.","Classify real-world and puzzle problems by whether they need the HCF, the LCM, or both.","This game has twelve cards and three bins: HCF, LCM and Both.\n\nHCF: simplifying 1920 × 1080 to 16 : 9 (HCF 120); which amounts 4 L and 6 L jugs can measure (multiples of 2); simplifying the 48 : 16 gear ratio to 3 : 1; squares crossed by the diagonal of a 12 × 18 grid (12 + 18 − 6 = 24).\n\nLCM: Teentaal and Jhaptaal landing on sam together (80 beats); Sunzi’s remainder puzzle (answers repeat every 105; smallest 23); when 12- and 18-tooth gears repeat (36 teeth); numbers up to 1,000 divisible by 4 or 6 (333, subtracting multiples of 12); a 17-year cicada and a 6-year predator (every 102 years).\n\nBoth: pairs with HCF 12 and LCM 144 (split 12 into co-prime parts); whether offset 4- and 6-year cycles can meet (HCF decides if, LCM decides how often); runners with ¾ and ⅚ minute laps (LCM of numerators ÷ HCF of denominators = 7½ minutes).",{"id":598,"type":98,"title":599,"eyebrow":600,"navLabel":601},"ch09","More puzzles from daily life in India","Chapter 09","9 Daily-life puzzles",{"id":603,"type":107,"title":604,"problem":605,"steps":606},"we-temple-bells","Temple bells from morning to evening","Three temple bells toll at intervals of **9, 12 and 15 minutes**. They toll together at **9:00 a.m.** How many times do they toll together from 9:00 a.m. up to and including 6:00 p.m.?",[607,608,609,610],"Together every LCM(9, 12, 15) minutes. 9 = 3², 12 = 2² × 3, 15 = 3 × 5, so LCM = 2² × 3² × 5 = **180 minutes** = 3 hours.","Times: 9:00 a.m., 12:00 noon, 3:00 p.m., 6:00 p.m.","Including the start at 9:00 a.m., that is **4 times**.","If the question says \"after 9:00 a.m.\", the answer is 3. Read carefully whether the starting moment counts.",{"id":612,"type":47,"variant":613,"title":614,"markdown":615},"misconception-count-start","misconception","Counting the start (or forgetting it)","In \"how many times together\" problems, many students divide the total time by the LCM and stop: 540 minutes ÷ 180 = 3. That counts the meetings **after** the start. If the start is included, add 1. Draw a quick timeline with the start marked as 0; it makes the answer obvious.",{"id":617,"type":107,"title":618,"problem":619,"steps":620},"we-trains","Two trains at a platform","At a junction, a Vande Bharat train passes every **25 minutes** and a goods train every **40 minutes** (in this made-up timetable). Both pass at noon. When do they next pass at the same time?",[621,622,623],"25 = 5² and 40 = 2³ × 5. LCM = 2³ × 5² = **200 minutes**.","200 minutes = 3 hours 20 minutes, so at **3:20 p.m.**","HCF(25, 40) = 5: sharing a 5 is why they meet after 200 minutes rather than 25 × 40 = 1,000.",{"id":625,"type":107,"title":626,"problem":627,"steps":628},"we-rangoli-grid","A rangoli on a dot grid","Priya’s rangoli board is **84 cm × 60 cm**. She wants to mark dots in a square grid so that there are dots on all four edges and the corners, with the dots as **far apart** as possible. How far apart should they be, and how many dots will there be?",[629,630,631,632],"The spacing must divide both 84 and 60 so dots land on every edge: an **HCF** problem.","HCF(84, 60) = **12 cm**.","Along 84 cm: 84 ÷ 12 = 7 gaps, so 8 dots. Along 60 cm: 60 ÷ 12 = 5 gaps, so 6 dots.","Total dots = 8 × 6 = **48**.",{"id":634,"type":332,"itemId":635,"prompt":636,"check":637,"hints":639,"feedback":641},"practice-school-bells","hcf-and-lcm.extend-school-events","A school holds a fire drill every **15 days**, a parent meeting every **20 days** and a science quiz every **30 days**. All three happen on 1 July. After how many days do all three next happen on the same day?",{"kind":388,"answer":18,"tolerance":389,"unit":638},"days",[640],"LCM of 15, 20 and 30.",{"correct":642,"incorrect":643},"Yes: LCM(15, 20, 30) = 60 days, so 30 August.","15 = 3 × 5, 20 = 2² × 5, 30 = 2 × 3 × 5. LCM = 2² × 3 × 5 = 60.",{"id":645,"type":332,"itemId":646,"prompt":647,"check":648,"hints":650,"feedback":653},"practice-cloth","hcf-and-lcm.extend-cloth-pieces","A handloom weaver in Kanchipuram has silk lengths of **72 m**, **108 m** and **180 m**. She cuts all of them into equal pieces, as long as possible. How many pieces does she get altogether?",{"kind":388,"answer":154,"tolerance":389,"unit":649},"pieces",[651,652],"Piece length = HCF(72, 108, 180).","Then divide each length by it and add.",{"correct":654,"incorrect":655},"Yes: HCF = 36 m, giving 2 + 3 + 5 = 10 pieces.","HCF(72, 108, 180) = 36. Pieces: 72 ÷ 36 = 2, 108 ÷ 36 = 3, 180 ÷ 36 = 5. Total 10.",{"id":657,"type":332,"itemId":658,"prompt":659,"check":660,"hints":662,"feedback":665},"practice-count-together","hcf-and-lcm.extend-count-together","Two lighthouses flash every **8 s** and **14 s**, together at the start. How many times do they flash together in **10 minutes**, counting the start?",{"kind":388,"answer":661,"tolerance":389},11,[663,664],"Find the LCM, then how many multiples fit in 600 seconds.","Remember to add 1 for the start.",{"correct":666,"incorrect":667},"Yes: LCM(8, 14) = 56 s; 600 ÷ 56 = 10 remainder 40, so 10 times after the start, 11 including it.","LCM(8, 14) = 56 s. In 600 s: 600 ÷ 56 = 10 (remainder 40), plus the start makes 11.",{"id":669,"type":107,"title":670,"problem":671,"steps":672},"we-ribbon-fractions","HCF of fractions: cutting ribbons","A craft teacher has ribbons of **2½ m** and **3¾ m**. She wants to cut both into equal pieces, as long as possible, with nothing left. How long is each piece?",[673,674,675,676],"Write the lengths as fractions in simplest form: 2½ = 5⁄2 and 3¾ = 15⁄4.","HCF of fractions = HCF of numerators ÷ LCM of denominators.","HCF(5, 15) = 5 and LCM(2, 4) = 4, so the piece is **5⁄4 m = 1¼ m**.","Check: 5⁄2 ÷ 5⁄4 = 2 pieces and 15⁄4 ÷ 5⁄4 = 3 pieces. Both whole ✓.",{"id":678,"type":508,"prompt":679,"options":680,"explanation":687},"predict-olympiad-lcm","Which is bigger: LCM(1, 2, …, 20) ÷ LCM(1, 2, …, 19), or LCM(1, 2, …, 19) ÷ LCM(1, 2, …, 18)?",[681,683,685],{"id":339,"label":682},"The first",{"id":342,"label":684},"The second",{"id":345,"label":686},"They are equal","**The second.** Going from 19 to 20 adds nothing new (20 = 2² × 5, and 16 and 5 are already in), so the first ratio is 1. Going from 18 to 19 adds the new prime 19, so the second ratio is 19.",{"id":689,"type":332,"itemId":690,"prompt":691,"check":692,"hints":694,"feedback":697},"practice-oly-2","hcf-and-lcm.extend-oly-hcf-lcm-sum","The HCF of two numbers is **6**, their LCM is **90**, and both are bigger than 6. What is their sum?",{"kind":388,"answer":693,"tolerance":389},48,[695,696],"Write them as 6m and 6n with m × n = 15 and HCF(m, n) = 1.","m = 1 would give 6 itself, which is not allowed.",{"correct":698,"incorrect":699},"Yes: m = 3, n = 5, so the numbers are 18 and 30, sum 48.","90 ÷ 6 = 15 = 3 × 5. Excluding 1 × 15 (which gives 6), the numbers are 18 and 30. Sum 48.",{"id":701,"type":98,"title":702,"eyebrow":703,"navLabel":704},"ch10","Who uses HCF and LCM?","Chapter 10","10 Careers",{"id":706,"type":707,"title":708,"prompt":709,"options":710},"explorer-careers","explorer","HCF and LCM at work","Pick a job to see where the ideas turn up.",[711,723,735,746,756],{"id":712,"label":713,"chain":714,"badge":719,"note":722},"coder","Software engineer",[715,716,717,718],"Fractions in code","Simplify with gcd","Euclid in a loop","Fast for huge numbers",{"text":720,"tone":721},"Uses Euclid daily","yes","Programming languages such as Python include gcd and lcm functions, built on Euclid’s algorithm. They are used to keep fractions in simplest form, to resize images without distortion, and to schedule repeating tasks. Every time your phone shows a photo in a 4 : 3 or 16 : 9 frame, a gcd was probably computed.",{"id":724,"label":725,"chain":726,"badge":732,"note":734},"crypto","Cryptographer",[727,728,729,730,731],"Two huge primes","Public key","Extended Euclid","Private key","Secure payment",{"text":733,"tone":721},"Uses Bézout","The RSA system that protects many online payments uses the extended Euclidean algorithm (Bézout’s identity) to build its secret key, and relies on the fact that multiplying two huge primes is easy but factorising the result is extremely hard. Secure websites and online payments depend on number theory like this.",{"id":736,"label":737,"chain":738,"badge":743,"note":745},"engineer","Mechanical engineer",[739,740,741,742],"Choose tooth counts","Check the LCM","Co-prime = even wear","Longer-lasting gears",{"text":744,"tone":721},"Uses LCM","Gearbox designers choose tooth counts to get the right speed ratio (an HCF-simplified ratio) while making counts co-prime where possible, so wear is spread evenly. Clockmakers and bicycle designers think the same way.",{"id":747,"label":748,"chain":749,"badge":754,"note":755},"planner","Transport planner",[750,751,752,753],"Bus and train headways","Find the LCM","Plan connections","Shorter waits",{"text":744,"tone":721},"Metro and bus planners set how often services run (the \"headway\"). Services with headways that share big factors, like 10 and 15 minutes, line up every 30 minutes, making reliable connections possible. Traffic-signal engineers coordinate cycle lengths along a road for the same reason.",{"id":757,"label":758,"chain":759,"badge":764,"note":765},"musician","Musician",[760,761,762,763],"Tala or metre","Cycle lengths","LCM of cycles","Land on sam together",{"text":744,"tone":721},"Tabla players, drummers and composers work with cycles of beats. Knowing when different cycles line up lets them build tihais, polyrhythms and crossing patterns that resolve on the first beat exactly when the audience expects it.",{"id":767,"type":98,"title":768,"eyebrow":769,"navLabel":770},"ch11","Projects and open questions","Chapter 11","11 Projects",{"id":772,"type":47,"variant":293,"title":773,"markdown":774},"try-it-projects","Project ideas","1. **School timetable audit.** Find two things at school that repeat on different cycles (bells, sports days, library periods). Work out when they coincide over a term, then check against the real calendar.\n2. **Gear lab.** Using LEGO or cardboard gears, mark one tooth on each of two meshing gears. Count turns until the marks meet again. Compare with the LCM. Then change one gear by one tooth.\n3. **Rhythm composer.** Write a short piece for two clappers in cycles of 3 and 4 (or 5 and 7). Mark where they land together, and perform it.\n4. **Screen survey.** Collect the resolutions of phones, laptops and TVs at home. Use the HCF to find each aspect ratio and make a table. Which ratio is most common?\n5. **Euclid speed test.** Write down 10 pairs of numbers and count the division steps for each. Can you find a pair under 1,000 that beats the Fibonacci pair (987, 610)?",{"id":776,"type":47,"variant":777,"title":778,"markdown":779},"open-questions","question","Open questions for curious learners","Nobody has fully solved these yet.\n\n- **The lonely runner conjecture.** Runners with different constant speeds start together on a circular track. Is every runner, at some moment, far from all the others (at least 1⁄(number of runners) of a lap away)? It has been proved for up to 13 runners — by hand up to 7, and for 8 to 13 only recently, with computer-assisted searches — but nobody has proved it for every number of runners.\n- **Why prime cycles?** Do periodical cicadas really have 13- and 17-year cycles because of predators, because prime cycles keep broods from interbreeding, or for some other reason? Biologists still debate it.\n- **Fast factorising.** Euclid finds an HCF in a blink, but nobody knows a fast way to factorise huge numbers on ordinary computers. If someone found one, much of today’s internet security would need to change.\n- **Fairer schedules.** How should a city choose bus and metro frequencies so that connections are frequent but no route waits too long? This is an active research area mixing LCMs with data.",{"id":781,"type":98,"title":782,"eyebrow":783,"navLabel":784},"ch12","Check yourself","Chapter 12","12 Wrap-up",{"id":786,"type":141,"component":787,"componentVersion":5,"config":788,"objective":815,"textAlternative":816},"lab-match-extend","match-pairs",{"prompt":789,"mode":790,"pairs":791},"Match each situation to its answer.","memory",[792,794,797,800,803,806,809,812],{"a":793,"b":414},"1920 × 1080 screen ratio",{"a":795,"b":796},"Teentaal and Jhaptaal sams meet","every 80 beats",{"a":798,"b":799},"Sunzi’s puzzle (2, 3, 2 on ÷ 3, 5, 7)","23",{"a":801,"b":802},"Egg basket: 1 left for 2–6, none for 7","301 eggs",{"a":804,"b":805},"LCM(1, 2, …, 10)","2,520",{"a":807,"b":808},"Gears of 12 and 17 teeth repeat after","204 teeth",{"a":810,"b":811},"Runners with ¾ and ⅚ minute laps meet","after 7½ minutes",{"a":813,"b":814},"Olympics and FIFA World Cup in same year","never","Match real-world cycles and puzzles from this layer to their answers.","This memory game has sixteen cards forming eight pairs:\n\n- 1920 × 1080 screen → 16 : 9 (divide by HCF 120).\n- Teentaal (16) and Jhaptaal (10) → sams meet every 80 beats.\n- Sunzi’s puzzle → 23.\n- Egg basket → 301 eggs.\n- LCM of 1 to 10 → 2,520.\n- Gears of 12 and 17 teeth → the same teeth meet again after 204 teeth.\n- Runners with ¾ and ⅚ minute laps → together after 7½ minutes.\n- Olympics (years divisible by 4) and FIFA World Cup (2 years later) → never in the same year.",{"id":818,"type":819,"title":820,"questions":821},"quiz-extend","quiz","Extend check",[822,835,848,861,873,886,898,908,921,931],{"itemId":823,"prompt":824,"options":825,"correct":345,"why":834},"hcf-and-lcm.extend-q-offset","Cycle A repeats every 6 days starting on day 0; cycle B every 9 days starting on day 4. Can they ever fall on the same day?",[826,828,830,832],{"id":339,"label":827},"Yes, on day 18",{"id":342,"label":829},"Yes, every 54 days",{"id":345,"label":831},"No, because 4 is not a multiple of HCF(6, 9) = 3",{"id":348,"label":833},"No, because 6 and 9 are not co-prime","Two cycles can meet only if the difference in their starts is a multiple of the HCF. 4 is not a multiple of 3.",{"itemId":836,"prompt":837,"options":838,"correct":339,"why":847},"hcf-and-lcm.extend-q-planets","Why is the Jupiter–Saturn conjunction period (about 19.9 years) not an LCM?",[839,841,843,845],{"id":339,"label":840},"Their periods are not whole numbers, and alignment only needs the same direction, not the starting point",{"id":342,"label":842},"Because Jupiter is bigger",{"id":345,"label":844},"Because LCMs only work for primes",{"id":348,"label":846},"It is the LCM of 11.86 and 29.46","Alignments come from relative motion: 1 ÷ (1⁄11.86 − 1⁄29.46) ≈ 19.9 years.",{"itemId":849,"prompt":850,"options":851,"correct":345,"why":860},"hcf-and-lcm.extend-q-hunting","Which pair of gears spreads wear most evenly (every tooth meets every tooth)?",[852,854,856,858],{"id":339,"label":853},"20 and 30 teeth",{"id":342,"label":855},"15 and 45 teeth",{"id":345,"label":857},"20 and 31 teeth",{"id":348,"label":859},"16 and 24 teeth","20 and 31 are co-prime (HCF 1), so the pattern repeats only after 620 teeth and every pairing occurs.",{"itemId":862,"prompt":863,"options":864,"correct":345,"why":872},"hcf-and-lcm.extend-q-tala","Ektaal (12 beats) and Rupak (7 beats) start together on sam. When do they next land on sam together?",[865,867,869,870],{"id":339,"label":866},"19 beats",{"id":342,"label":868},"42 beats",{"id":345,"label":289},{"id":348,"label":871},"168 beats","12 and 7 are co-prime, so LCM = 12 × 7 = 84 beats.",{"itemId":874,"prompt":875,"options":876,"correct":342,"why":885},"hcf-and-lcm.extend-q-jugs","With 8 L and 12 L jugs, which amount can you measure?",[877,879,881,883],{"id":339,"label":878},"2 L",{"id":342,"label":880},"4 L",{"id":345,"label":882},"6 L",{"id":348,"label":884},"10 L","Only multiples of HCF(8, 12) = 4: 4, 8, 12.",{"itemId":887,"prompt":888,"options":889,"correct":345,"why":897},"hcf-and-lcm.extend-q-crt","Smallest number (after 2) leaving remainder 2 when divided by 3, 5 and 7?",[890,891,893,895],{"id":339,"label":208},{"id":342,"label":892},"37",{"id":345,"label":894},"107",{"id":348,"label":896},"212","Same remainder: LCM(3, 5, 7) + 2 = 107.",{"itemId":899,"prompt":900,"options":901,"correct":348,"why":907},"hcf-and-lcm.extend-q-screen","A monitor is 2560 × 1600. What is its aspect ratio?",[902,903,905,906],{"id":339,"label":414},{"id":342,"label":904},"16 : 10",{"id":345,"label":431},{"id":348,"label":436},"HCF(2560, 1600) = 320; 2560 ÷ 320 = 8 and 1600 ÷ 320 = 5. (Marketing often calls this 16 : 10, which is the same ratio unsimplified.)",{"itemId":909,"prompt":910,"options":911,"correct":339,"why":920},"hcf-and-lcm.extend-q-frac","LCM of 2⁄3 and 4⁄5 (as lap times in minutes) is…",[912,914,916,918],{"id":339,"label":913},"4 minutes",{"id":342,"label":915},"8⁄15 minute",{"id":345,"label":917},"4⁄15 minute",{"id":348,"label":919},"12 minutes","LCM(2, 4) ÷ HCF(3, 5) = 4 ÷ 1 = 4 minutes: 6 laps of ⅔ and 5 laps of ⅘.",{"itemId":922,"prompt":923,"options":924,"correct":342,"why":930},"hcf-and-lcm.extend-q-growth","LCM(1, …, 16) is how many times LCM(1, …, 15)?",[925,926,927,928],{"id":339,"label":206},{"id":342,"label":203},{"id":345,"label":227},{"id":348,"label":929},"15","16 = 2⁴ needs one more 2 than 8 = 2³, so the LCM doubles. No other prime changes.",{"itemId":932,"prompt":933,"options":934,"correct":342,"why":941},"hcf-and-lcm.extend-q-or","How many numbers from 1 to 100 are divisible by 6 or 10?",[935,937,938,939],{"id":339,"label":936},"26",{"id":342,"label":799},{"id":345,"label":227},{"id":348,"label":940},"30","100 ÷ 6 gives 16 multiples of 6, 100 ÷ 10 gives 10 multiples of 10, and 3 numbers (multiples of LCM 30) were counted twice: 16 + 10 − 3 = 23.",{"id":943,"type":944,"title":945,"points":946},"cheat-sheet-extend","summary","Cheat sheet",[947,948,949,950,951,952,953,954,955,956],"**Cycles with a head start** can meet only if the difference in starts is a multiple of the HCF; then they meet every LCM.","**Real sky cycles** are not whole numbers: alignments come from relative motion (Jupiter–Saturn ≈ 19.9 years). Calendars use near-LCMs like the 19-year Metonic cycle.","**Cicadas** with 13- and 17-year cycles meet short predator cycles rarely (big LCMs): a debated hypothesis.","**Gears:** the same teeth meet again after LCM(teeth) teeth; co-prime counts (a hunting tooth) spread wear evenly. Gear ratios are simplified with the HCF.","**Rhythm:** tala cycles land on sam together every LCM of their beats: Teentaal and Jhaptaal every 80.","**Jugs:** two jugs can measure exactly the multiples of their HCF (up to the bigger jug).","**Remainder puzzles:** solutions repeat every LCM of the divisors (Chinese Remainder Theorem).","**Screens:** divide width and height by their HCF to get the aspect ratio: 1920 × 1080 → 16 : 9.","**Fractions:** LCM = LCM(numerators) ÷ HCF(denominators); HCF = HCF(numerators) ÷ LCM(denominators).","**Counting \"a or b\":** multiples of a + multiples of b − multiples of LCM(a, b).",{"id":958,"type":959,"prompt":960},"reflect-extend","reflection","Choose one expedition from this layer (calendars, cicadas, gears, rhythm, jugs, remainders or screens). Explain it to someone at home using your own numbers, and write down one question it raised that you could not answer.",{"id":962,"type":963,"conceptId":964,"relation":965,"explanation":966},"conn-patterns-e","connection","patterns","related_to","Cycles, rhythms and repeating remainders are patterns whose period is an LCM.",{"id":968,"type":963,"conceptId":969,"relation":970,"explanation":971},"conn-data-e","data-handling","applied_in","Screen-size surveys and timetable audits turn HCF and LCM into real data projects.",{"id":973,"type":963,"conceptId":974,"relation":970,"explanation":975},"conn-shape-e","shape-and-space","Tiling rectangles, grid diagonals and gear circles bring HCF and LCM into geometry.",{"id":977,"type":963,"conceptId":978,"relation":965,"explanation":979},"conn-electricity-e","electricity","Blinking lights, generator poles and AC cycles are repeating signals whose coincidences follow the LCM.",{"id":981,"type":982,"sourceIds":983},"sources-extend","sources",[984,985,986,987,988,989,990,991,992,993,994,995],"hcf-and-lcm-wiki-lcm","hcf-and-lcm-wiki-gcd","hcf-and-lcm-wiki-gear-train","hcf-and-lcm-wiki-chinese-remainder-theorem","hcf-and-lcm-wiki-euclidean-algorithm","hcf-and-lcm-mathsisfun-lcm","hcf-and-lcm-wiki-metonic-cycle","hcf-and-lcm-wiki-hindu-calendar","hcf-and-lcm-wiki-great-conjunction","hcf-and-lcm-wiki-periodical-cicadas","hcf-and-lcm-wiki-lonely-runner","hcf-and-lcm-wiki-tala",[984,985,986,987,988,989,990,991,992,993,994,995],"needs_review",{"generatedBy":999,"notes":1000},"claude-code","Draft generated with Python-checked arithmetic; pending owner review.","af03cd82735cf04b5cfcb68e181ad73f50678f12e555fb12902bf1bf8aefb725",{"component:hcf-lcm@1":1003,"logic:practice":1004,"component:factor-tree@1":1005,"component:sort-game@1":1006,"component:match-pairs@1":1007,"source:hcf-and-lcm-mathsisfun-lcm":1008,"source:hcf-and-lcm-wiki-chinese-remainder-theorem":1009,"source:hcf-and-lcm-wiki-euclidean-algorithm":1010,"source:hcf-and-lcm-wiki-gcd":1011,"source:hcf-and-lcm-wiki-gear-train":1012,"source:hcf-and-lcm-wiki-great-conjunction":1013,"source:hcf-and-lcm-wiki-hindu-calendar":1014,"source:hcf-and-lcm-wiki-lcm":1015,"source:hcf-and-lcm-wiki-lonely-runner":1016,"source:hcf-and-lcm-wiki-metonic-cycle":1017,"source:hcf-and-lcm-wiki-periodical-cicadas":1018,"source:hcf-and-lcm-wiki-tala":1019},"83f068508b17b70184e45fbcf95d356790bfa2b0a98dcdb859f8e9dc105e5d33","3d6b0fe1b15255975a32b0fcd94e8019bc959ad45cbf12e136e86149549c6878","dfb598a254f20e70550d586121be496f7a546ba0254060fd97b95cb919dfd934","b164f45a2c8ca08f26c450768ff0231e113e9fe45381eddb34dc6d0548596c38","2a8ee4ac87460b4e1175a4bb13c96b03d577db06dde95670eb7fcfe4ad787899","722e7f7db3c09e9e29bf5c0f363173f52eb56ef08d1bd84ad3f51890d1375d07","708278c7e46fb9cf937f0d73f700bd73f1b2cbef3d4ece9927812f71b14b2bc9","c37b91030f0a162c4968726c72618dce986659b01f396306c685ac9597ae61df","857640343ebbf479db694438bf4af134fd5baee47f017b64110ded8b40bb2dc6","ebf3f51e6b37254a0ad24e1c5c7e27c7848e236b3642abe0dd941c72cd20f31f","ecd41a4be00e639977cbe6eafa8a6b954f18905ffc67545c776125bbf5aaa361","9c4cd23eceaa430088b74bc71d60516784270365cbdf4a19ceb895af522ef57e","12412937f9bd209912735d662496fbe90c125f89d22844b5a753d934cc84ef2c","98fd3c3b31b5af36148e4a2ea2da0f1bad838bc61b14e2169497eee919ccf746","87d38c0a7307d002cc75e25823f23e93bd89ee1a72d53b4d17d3e767d90a1ca1","7c2930121994ad645888a544a9da889942165b6da79b4078aafbf604b25cef05","91bf8d71899e46e723ad38d746990146b47548be24c66de02f02f35254832bd7",{"state":1021,"reviewer":1022,"selfReview":160,"reviewedAt":1023,"method":1024},"approved","The library owner","2026-09-20T10:18:37.581Z","owner_bulk","preview-7e1cbbcc4f",1789899597912]