PrepShorts · Study sheet · Class 7 Mathematics · Chapter 3, Finding Common Ground
Chapter 3 · Finding Common Ground
LCM: take the most occurrences of each prime
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Two paper chains, one of 6 cm links and one of 8 cm, whole links only, both finishing at the same length.
The idea
The LCM argument is the HCF argument run backwards, and the flip is worth seeing rather than being told. A common factor has to fit inside both factorisations, so each prime is capped by whichever number owns fewer copies. A common multiple has to hold both factorisations, so each prime is instead demanded by whichever number wants more — and supplying exactly that many, and not one more, is the cheapest way to satisfy both demands at once. The asymmetry in the names follows from the same picture: common factors cannot exceed the numbers themselves, so a highest one exists; common multiples run on without end, so there is no largest, and the interesting extreme is the least.
What you should be able to do
- Recognise a "when do two repeating things line up" problem and say why it is a common-multiple problem
- State what the lowest common multiple of two numbers is
- Explain why there is no largest common multiple, and why that is the reason the lowest one is the useful extreme
- State what the factorisation of a multiple must contain, and check it on an example
- Build a common multiple by supplying every prime that either number needs
- Compute the LCM directly by taking, for each prime present, the larger of its two counts
- Say why supplying extra copies of a prime still gives a common multiple, and why it stops being the lowest
- Extend the same procedure to three numbers
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| multiple | what you get by multiplying the number by a whole number | Class 6; used throughout §3.2, Part II, pp.55–58 |
| common multiple | a number both of the given numbers divide exactly | printed in §3.2, Part II, p.55 |
| Lowest Common Multiple (LCM) | the smallest of the common multiples | printed in bold in §3.2, Part II, p.56 |
| occurrences | the chapter's word for how many copies of a prime a factorisation holds | printed repeatedly in §3.2, Part II, pp.57–58 |
| maximum | the larger of the two counts of a prime, which is what the LCM must take | printed in §3.2, Part II, p.58 |
| subpart | a piece cut out of a factorisation and multiplied together | printed in §3.1, Part II, p.51, and reused in §3.2 |
| toran | the strip-of-cloth hanging the section opens with | printed in §3.2, Part II, p.55 |
| gajak | the sweet in the second problem, made of sesame, jaggery and ghee | printed in §3.2, Part II, p.55 |
| least | one of the three words the chapter offers for the same extreme, with lowest and smallest | all three printed together in §3.2, Part II, p.56, and Least stands in the section title |
| largest common multiple | the thing the section asks after, which no number can be | printed in §3.2, Part II, p.55, in the question that closes the toran problem |
Where people slip up
- "The LCM is the two numbers multiplied together." It is for 7 and 11, and it is not for 6 and 8, where the product is 48 and the answer is 24. The product is always a common multiple — that is why the LCM can never exceed it — but the shared primes get counted twice in it. Show 6 × 8 and 24 side by side.
- "Take the smaller count, as we did for the HCF." The constraint has reversed. For the HCF, a prime taken too often falls out of one of the numbers; for the LCM, a prime taken too rarely fails to hold one of them. Show the two failures next to each other.
- "More copies would be safer." Extra copies still give a common multiple — the book prints two such and poses a third as a question — they just stop it being the lowest. The word lowest is the whole specification.
- "There must be a largest common multiple as well as a lowest." There is not: double any common multiple and you have another one. The section asks after a largest one in those words and leaves the question open; an explanation that skips it wastes the chapter's best invitation.
- "A prime has to be in both numbers to enter the LCM." It does not. 5 appears only in 360 and still has to be in the LCM of 96 and 360, because without it the result is not a multiple of 360. Caution when explaining it: the book's sentence introducing this step on p.57 describes 2, 3 and 5 as appearing in both numbers, which is not so of the 5 — the working immediately below it makes clear that what is meant is the primes appearing across the two. Do not repeat the sentence as printed; state the correct condition.
- "The LCM is bigger than both numbers, always." It is at least as big as each of them, and it equals the larger one whenever the smaller divides it — which is the observation What HCF and LCM do for consecutive, even, and co-prime numbers opens on.
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Worked answers to this chapter’s exercises · this video explains Figure it Out · 4 Q1, Figure it Out · 6 Q1, Figure it Out · 6 Q2, Figure it Out · 6 Q5, Figure it Out · 6 Q8, Figure it Out · 6 Q9, Figure it Out · 6 Q12, Figure it Out · 6 Q13
Transcript1,447 words
Two paper chains, hanging side by side. One is built from links six centimetres long, the other from links eight. Whole links only. You cannot cut one in half. And the two chains have to finish at exactly the same length. So that length has to be something six can reach, laying links end to end. And the very same length has to be something eight can reach. It has to be a multiple of both — a common multiple problem, before anybody says the words.
And of the lengths that work, we want the shortest, because a shorter chain is less work. So climb the two ladders and see where they meet. Six, twelve, eighteen, twenty-four, thirty, thirty-six, forty-two, forty-eight, fifty-four. Eight, sixteen, twenty-four, thirty-two, forty, forty-eight, fifty-six, sixty-four, seventy-two. Twenty-four is on both ladders. That is four of the six-centimetre links, and three of the eight. Keep climbing and forty-eight is on both as well.
So there is more than one answer, and the shortest is twenty-four. Notice what forty-eight is, though. Six times eight. Multiplying the two lengths always gives a length both chains can reach, just not the shortest. There is a question hiding here, better than it looks. We asked for the shortest length. Is there a longest? Let the ladders keep climbing. Twenty-four, forty-eight, seventy-two, ninety-six, a hundred and twenty, a hundred and forty-four.
Every twenty-four, going up forever. And here is why it never stops. Take any length that works, and double it. If six fits into that length a whole number of times, it fits into twice it too. So does eight. So whatever answer you name, twice that is another. There is no largest. Which is why the shortest is the interesting end of the list — it is the only end there is.
The same problem turns up wearing different clothes. A bakery gives away a free loaf every Monday. That is every seven days. A friend of yours passes through town every ten days. Today is a Monday, she is here, and has just had a loaf. When does that happen again? Loaves on days seven, fourteen, twenty-one, twenty-eight, thirty-five, forty-two, forty-nine, fifty-six, sixty-three, seventy. Visits on days ten, twenty, thirty, forty, fifty, sixty, seventy.
Day seventy. Ten loaves later, seven visits later. And seven times ten is seventy — here the product really is the answer. That shortest chain, and that seventieth day, have a name. The smallest number both of your numbers divide into is called the lowest common multiple. You will hear it called the least common multiple too. Same quantity. And the word lowest is doing real work. Hunting for common factors we wanted the highest, and a highest exists because a factor is never bigger than what it divides. That list has a top.
Common multiples run the other way. They climb forever, so there is no top at all. The only end that list has is the bottom. The two ideas are mirror images, and even their names mirror. Highest for one. Lowest for the other. Here it is once more, as a game. You count around a circle. One, two, three, and so on. Land on a multiple of four and you say apple instead of the number. Land on a multiple of six and you say banana.
The first person saying both words stands on the first number that is a multiple of four and of six. That is twelve, which is the lowest common multiple. Not a coincidence — it is the same question asked out loud. Seven and eleven give seventy-seven. You count a long way round for that one. Fourteen and thirty give two hundred and ten. Fifteen and fifty-five give a hundred and sixty-five.
Which is where listing hurts. Reaching two hundred and ten takes fifteen multiples of fourteen and seven of thirty, and one slip gives the wrong answer. So do to multiples what we did to factors. Stop listing, start building. The key is one observation about what a multiple is made of. Take thirty-six, which is two, two, three, three. And take six hundred and forty-eight, which is thirty-six times eighteen.
Write eighteen as its primes, two, three, three, and put the rows together. Six hundred and forty-eight is two, two, two, three, three, three, three. Look at what is sitting inside that row. Two, two, three, three. The whole of thirty-six, with a two and two threes left over. That is what being a multiple means. A multiple's row of primes holds the other number's entire row, and then some.
Now turn that into a method. Fourteen is two times seven. Thirty-five is five times seven. We want the smallest number whose row holds both of those rows at once. It needs a two, because fourteen has one. A five, because thirty-five has one. And a seven. Two, five, seven. That is seventy. Does it hold fourteen? A two and a seven — yes. Thirty-five? A five and a seven — yes.
And nothing in it is spare. Take the five out and you have fourteen, which cannot hold thirty-five. Take the two out and you have thirty-five, which cannot hold fourteen. Every prime in seventy is there because one of the two demanded it. That is what makes it the lowest. Now a pair you could not do by eye. Ninety-six is two, two, two, two, two, three. Three hundred and sixty is two, two, two, three, three, five.
Go prime by prime and count, but ask the opposite question. Twos: ninety-six has five, three hundred and sixty has three. We need five — anything with fewer cannot hold ninety-six. Threes: ninety-six has one, three hundred and sixty has two. We need two. Fives: ninety-six has none at all, three hundred and sixty has one. We still need one. That last line is the one to watch. Five is in only one of the two numbers and still has to be in the answer — without it, the result is not a multiple of three hundred and sixty.
Five twos, two threes and one five. That is one thousand four hundred and forty. Why the larger count, and not the smaller? For common factors we took the smaller, and taking more made the answer stop fitting inside one of them. Here the failure runs the other way. Take four twos instead of five and you get seven hundred and twenty. That is a fine multiple of three hundred and sixty. But ninety-six does not divide it — you get seven and a half.
There are five twos inside ninety-six, and four cannot hold five. So a prime here is not capped by the smaller count. It is demanded by the larger one. Too many copies and you fall out of one of the numbers. Too few and you fail to hold one. The whole difference is one word. Smaller becomes larger. The other direction is far more forgiving. Go back to fourteen and thirty-five, where the answer was seventy.
Throw an extra three in. Two, three, five, seven, which is two hundred and ten. That still holds fourteen and still holds thirty-five. Still a common multiple, simply bigger than it needs to be. Or two, seven, five, seven and three, which is one thousand four hundred and seventy. Both are correct answers to what is a common multiple. Neither answers what we asked. Which brings back the shortcut everybody reaches for. Just multiply the two numbers.
Never wrong — the product is always a common multiple. It just counts the shared primes twice. Six times eight is forty-eight, and the two they both own got counted in both, which is why the answer is half of that. The rule, then, in one line. Write your numbers as primes. For every prime appearing in any of them, take the largest count you see. Multiply. And it does not care how many numbers you hand it.
A hundred and five is three, five, seven. A hundred and ninety-five, three, five, thirteen. Sixty-five, five and thirteen. One three, one five, one seven, one thirteen. That is one thousand three hundred and sixty-five. Seven is in only one of the three and still has to be there. A prime only has to be demanded by somebody. One to finish on. A herd goes through three gates in equal numbers, then five, then seven.
So the herd is a multiple of three, of five and of seven. One of each, a hundred and five. The next one up is two hundred and ten, so under two hundred that is the only herd there is.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Breaking a number down to primes, and why the result is uniqueClass 7 · Ch 3, Finding Common Ground
- Reading every factor of a number off its prime factorisationClass 7 · Ch 3, Finding Common Ground
- HCF: take the fewest occurrences of each primeClass 7 · Ch 3, Finding Common Ground
Comes up again in
- What HCF and LCM do for consecutive, even, and co-prime numbersClass 7 · Ch 3, Finding Common Ground
- Getting the HCF and the LCM out of one division ladderClass 7 · Ch 3, Finding Common Ground
- Conjecture and generalisation: what mathematicians mean by those wordsClass 7 · Ch 3, Finding Common Ground