Least Common Multiple (L.C.M)

Two school bells start ring­ing together at nine o'clock. One rings every $8$ min­utes. The other rings every $12$ min­utes. When will they next ring at the same moment? The answer is the least com­mon mul­ti­ple of $8$ and $12$. The same idea tells you when two buses meet again at a stop, how many bis­cuits to buy so they pack evenly into two sizes of box, and, later on, how to add frac­tions such as $\displaystyle \frac{1}{8} + \frac{1}{12}$.

Take a num­ber and mul­ti­ply it by $1$. Then by $2$. Then by $3$. Keep going as long as you like. Each answer you get is a mul­ti­ple of that num­ber. So the mul­ti­ples of $3$ are $3, 6, 9, 12, 15$ and on with­out end.

Every num­ber has a list like that. You can carry the list on as far as you like.

The les­son before this one looked down from a num­ber. It looked at the fac­tors that divide into it. It asked for the biggest fac­tor two num­bers share.

This les­son looks the other way. It looks up at the num­bers a num­ber divides into. It asks for the small­est one that two or more num­bers share.

What a com­mon mul­ti­ple is

The acad­e­my's def­i­n­i­tion is short. It says this: the L.C.M of two (or) more nat­ural num­bers is the least com­mon mul­ti­ple of the given num­bers. Take it one word at a time.

Nat­ural num­bers are the count­ing num­bers, $1, 2, 3$ and on. A mul­ti­ple of $8$ is a num­ber that $8$ divides into exactly. A com­mon mul­ti­ple of $8$ and $12$ is a num­ber both of them divide into. The least of those is the L.C.M.

Num­bers like that are not rare. $24$, $48$ and $72$ are all com­mon mul­ti­ples of $8$ and $12$. The list goes on with­out end.

Each row of that grid says the same thing twice. The num­ber is so many lots of $8$. It is also so many lots of $12$. That is what makes it com­mon to them both.

The list of com­mon mul­ti­ples has no end. But it does have a start. That start is the L.C.M.

Remem­ber. A com­mon mul­ti­ple of some num­bers is a num­ber they all divide into exactly. The L.C.M is the small­est com­mon mul­ti­ple there is.

1st method (By writ­ing the mul­ti­ples)

Give each num­ber a line of its own. Write out its mul­ti­ples along that line. Keep going until one num­ber has turned up on every line. The first num­ber that does is the L.C.M.

Find the L.C.M of $8, 12$

$$\begin{aligned}\text{Multiples of } 8 &= 8, 16, 24, 32, 40, 48, 56, 64, 72, \ldots \\ \text{Multiples of } 12 &= 12, 24, 36, 48, 60, 72, \ldots \\ \text{Common multiples} &= 24, 48, 72, \ldots \\ \text{L.C.M} &= 24\end{aligned}$$

Only a num­ber stand­ing on both lines can be com­mon to $8$ and $12$. The small­est of those is $24$. Look at the three the sheet lists: $24$, then $48$, then $72$. Each one is $24$ more than the one before.

That is no acci­dent. Add another $24$ and you add three more lots of $8$. You add two more lots of $12$ as well. So you land on a num­ber both of them still divide into. That is why $24 = 8 \times 3$, $48 = 8 \times 6$ and $72 = 8 \times 9$.

Find the L.C.M of $8, 9, 12$

$$\begin{aligned}\text{Multiples of } 8 &= 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, \ldots \\ \text{Multiples of } 9 &= 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, \ldots \\ \text{Multiples of } 12 &= 12, 24, 36, 48, 60, 72, 84, \ldots \\ \text{Common multiples} &= 72, \ldots \\ \text{L.C.M} &= 72\end{aligned}$$

This time only $72$ stands on all three lines. Take care here. Two lines out of three is not enough. Look at $24$. It is on the first line and the third. But $24$ is not divis­i­ble by $9$. Divis­i­ble means it divides in exactly. Noth­ing is left over. Now look at $36$. It is on the sec­ond line and the third. But $36$ is not divis­i­ble by $8$.

The method is slow and hon­est. It shows you just what an L.C.M is. That is why it comes first.

It also shows you its own limit. Try to reach the L.C.M of $32, 56, 80$ and $154$ this way. You would be writ­ing lists for a very long time. That is what the next two meth­ods are for.

Remem­ber. The L.C.M is the first num­ber that turns up in every one of the lists. Turn­ing up in most of them counts for noth­ing.

The pic­ture below shows the two lists for $8$ and $12$ as dots on num­ber lines. The green dots are the num­bers that stand on both lines.

A small one to try first

Find the L.C.M of $4, 6$

$$\begin{aligned}\text{Multiples of } 4 &= 4, 8, 12, 16, 20, 24, \ldots \\ \text{Multiples of } 6 &= 6, 12, 18, 24, \ldots \\ \text{Common multiples} &= 12, 24, \ldots \\ \text{L.C.M} &= 12\end{aligned}$$

Notice that the L.C.M is not sim­ply $4 \times 6 = 24$. That prod­uct is a com­mon mul­ti­ple, but it is not the least one. The two num­bers share a fac­tor of $2$, and that is why a smaller com­mon mul­ti­ple exists.

2nd method (Prime fac­tor­iza­tion method)

The acad­emy sets the method out like this: we prime fac­tor­ize the given num­bers. Then the prod­uct of the high­est pow­ers of all the fac­tors that occur in any of the given num­bers is the L.C.M. That is a lot of words at once. Take them a few at a time.

Start with prod­uct. A prod­uct is what you get when you mul­ti­ply num­bers together. So the L.C.M here is one long mul­ti­pli­ca­tion. Your job is to find out what goes into it.

Next comes occur in any of. A fac­tor occurs in a num­ber when it turns up inside it. So a fac­tor that occurs in any of the given num­bers turns up inside at least one of them. It does not have to be in all of them. One is enough.

Two more words come from ear­lier lessons. The first is prime fac­tori­sa­tion. That means writ­ing a num­ber as a prod­uct of primes. So you write it as some primes mul­ti­plied together. Primes are the num­bers picked out in Prime and com­pos­ite num­bers.

A com­pos­ite num­ber is one you can make by mul­ti­ply­ing smaller num­bers together. A com­pos­ite num­ber breaks into primes in one way only. Prime fac­tori­sa­tion and the stan­dard form of a num­ber shows that. So the primes you find below are the num­ber's own. They are not an acci­dent of where you started.

The other word is power. A power is the short way of writ­ing $2 \times 2 \times 2$ as $2^{3}$. It is taught in Let­ters that stand for num­bers. So the high­est power of a prime means this. It is the most times that prime turns up inside any one of the given num­bers.

Here is why the high­est power is the right thing to take. Break­ing a num­ber into primes shows what it is built from. Take $8 = 2 \times 2 \times 2$. So any num­ber that $8$ divides into must carry three $2$s inside it.

In the same way $9 = 3 \times 3$. So any num­ber that $9$ divides into must carry two $3$s. A num­ber that both of them divide into needs three $2$s and two $3$s. Take fewer and one of them will not fit. Take more and the num­ber is big­ger than it had to be.

The L.C.M of $8, 12$

First break both num­bers up: $8 = 2 \times 2 \times 2$ and $12 = 2 \times 2 \times 3$.

Only two primes turn up any­where: $2$ and $3$. The most $2$s in any one of the num­bers is three. Those three are inside $8$. The most $3$s is one, inside $12$. Mul­ti­ply those together and noth­ing else.

L.C.M of $8, 12$

$$\begin{aligned}&2 \times 2 \times 2 \times 3 \\ &= 24\end{aligned}$$

A Venn dia­gram shows the same work­ing at a glance. The primes that $8$ and $12$ share sit in the mid­dle. The ones that belong to only one num­ber sit on its own side. The L.C.M takes every prime in the pic­ture, each one once.

The L.C.M of $8, 9, 12$

Break all three up first. $8 = 2 \times 2 \times 2$. Then $9 = 3 \times 3$ and $12 = 2 \times 2 \times 3$.

The most $2$s in any one of them is three, in $8$. The most $3$s is two, in $9$. Now look at $12$. It is built from two $2$s and a $3$. Both of those are inside the counts already. So $12$ adds noth­ing new to the answer.

L.C.M of $8, 9, 12$

$$\begin{aligned}&2 \times 2 \times 2 \times 3 \times 3 \\ &= 72\end{aligned}$$

It is worth test­ing an answer you have not seen listed. $72 = 8 \times 9$ and $72 = 12 \times 6$. So all three num­bers divide into it exactly. It is the same $72$ the first method reached by the longer road.

Remem­ber. Every prime that turns up in any of the num­bers must turn up in the L.C.M. Count how many times it appears in each num­ber. Then take the biggest of those counts.

One more: the L.C.M of $6, 10, 15$

Each of these num­bers is a prod­uct of two dif­fer­ent primes: $6 = 2 \times 3$, $10 = 2 \times 5$ and $15 = 3 \times 5$. Only three primes turn up any­where, and each turns up at most once in any one num­ber.

$$\begin{aligned}\text{L.C.M} &= 2 \times 3 \times 5 \\ &= 30\end{aligned}$$

Check it: $30 = 6 \times 5 = 10 \times 3 = 15 \times 2$. Mul­ti­ply­ing the three num­bers together gives $900$, which is a com­mon mul­ti­ple too, but thirty times too big.

3rd method (Divi­sion method)

Write all the given num­bers in a line, with com­mas between them. Find a num­ber that divides at least two of them. Write that num­ber out to the left. It is the divi­sor for the row.

Now go along the row. Under each num­ber the divi­sor divides, write the answer to that divi­sion. $8$ divided by $2$ is $4$, so $4$ goes under the $8$. That answer is called the quo­tient. Any num­ber the divi­sor does not divide comes straight down, unchanged.

Then start a new row and do the same again. Stop when no num­ber big­ger than $1$ divides two of the num­bers left along the bot­tom. Num­bers that share noth­ing but $1$ are called coprime.

The L.C.M is then the prod­uct of every­thing round the out­side. So you mul­ti­ply the divi­sors down the side by the coprimes along the bot­tom. Those rows of divi­sions make a shape like a lad­der. In the lad­ders below, the upright line only splits the divi­sor from the row it divides. It is not a sum to work out.

Find the L.C.M of $8, 12$

$$\begin{aligned}2 &\mid 8, \; 12 \\ 2 &\mid 4, \; 6 \\ &2, \; 3\end{aligned}$$

Look at what hap­pened. Both $8$ and $12$ can be halved. So $2$ went out to the side, and $4, 6$ went under­neath. Both of those halve again. That leaves $2$ and $3$.

Noth­ing but $1$ divides both $2$ and $3$. So they are coprime, and the lad­der stops. Now mul­ti­ply the side by the bot­tom: $2 \times 2 \times 2 \times 3 = 24$. That is the answer both of the other meth­ods gave.

That $2$ down the side was a fac­tor $8$ and $12$ were both car­ry­ing. Writ­ing it once on the out­side is how it gets counted once instead of twice. What is left along the bot­tom is the part of each num­ber the other did not share.

The divi­sor need not be a prime. Look at $4$. It divides both $8$ and $12$. So you could have taken $4$ at the first step. You would end with $4 \times 2 \times 3 = 24$, the same answer. But tak­ing one small fac­tor at a time keeps the rows easy to read.

Find the L.C.M of $8, 9, 12$

$$\begin{aligned}2 &\mid 8, \; 9, \; 12 \\ 2 &\mid 4, \; 9, \; 6 \\ 3 &\mid 2, \; 9, \; 3 \\ &2, \; 3, \; 1\end{aligned}$$

The first $2$ halves $8$ and $12$. It leaves $9$ where it is. That is because $9$ is not divis­i­ble by $2$. Halv­ing it would not come out exactly. So $9$ comes straight down. The sec­ond $2$ halves $4$ and $6$.

Then $3$ divides $9$ and $3$. This time it is the $2$ that comes down untouched. The bot­tom row is $2, 3, 1$. No num­ber big­ger than $1$ divides any two of those. So they are coprime. The lad­der is fin­ished: $2 \times 2 \times 3 \times 2 \times 3 \times 1 = 72$.

Remem­ber. The lad­der is the rows of divi­sions you have built. It stops when the num­bers along the bot­tom are coprime. Those num­bers are the quo­tients. They are what is left of each num­ber after the divid­ing. Coprime means noth­ing but $1$ divides any two of them. Then mul­ti­ply every­thing down the side and along the bot­tom.

The three meth­ods on one ques­tion

All three land on the same num­ber. A set of num­bers has only one least com­mon mul­ti­ple. What changes is how much writ­ing each one costs. Each row below is only a short reminder. The full work­ing for every method is fur­ther up the page.

Use the first when you want to see what is going on. Use the sec­ond when the num­bers break into primes eas­ily. Use the third when there are three or four num­bers at once.

Using the L.C.M

Back to the bells

The bells ring together at nine o'clock. The first rings at $8, 16, 24, \ldots$ min­utes past. The sec­ond rings at $12, 24, \ldots$ min­utes past. The first time on both lists is the L.C.M, $24$. So the bells next ring together at $24$ min­utes past nine, and after that every $24$ min­utes.

A check for two num­bers

For two num­bers only, there is a neat link with the H.C.F from the les­son before. The H.C.F of $8$ and $12$ is $4$. Mul­ti­ply it by the L.C.M.

$$\begin{aligned}\text{H.C.F} \times \text{L.C.M} &= 4 \times 24 \\ &= 96 \\ &= 8 \times 12\end{aligned}$$

The H.C.F times the L.C.M of two num­bers always equals the two num­bers mul­ti­plied together. Use it as a check. Take care: it does not work for three or more num­bers.

Your turn

Find the L.C.M of each set. Check each answer before you move on. Divide it by each of the given num­bers in turn. Every one of those divi­sions should come out exactly. That much proves your num­ber is a com­mon mul­ti­ple.

You still have to know it is the least one. So make sure no smaller com­mon mul­ti­ple passes the same test. With the divi­sion method, check one thing. Was the bot­tom row really coprime before you mul­ti­plied?

The twelve sets under the next two head­ings are not on the acad­emy sheet. They are here to give you smaller num­bers to work with. Walk the first two meth­ods through them. The acad­e­my's own exer­cises come after.

Extra prac­tice — by writ­ing the mul­ti­ples

Extra prac­tice — by prime fac­tori­sa­tion

The acad­e­my's own exer­cises

These three are the sheet's own. It asks for the first by the divi­sion method by name. Take the other two whichever way you like.

1. $32, \; 56, \; 80, \; 154$
2. $24, \; 50, \; 75$
3. $212, \; 246, \; 214$

Take the last set slowly. The method does not mind large num­bers. But the mul­ti­pli­ca­tion at the end needs care.

Com­mon mis­takes

• Mul­ti­ply­ing the num­bers together and call­ing that the L.C.M. $4 \times 6 = 24$, but the L.C.M of $4$ and $6$ is $12$.
• Stop­ping at a num­ber that is on most of the lists but not all. $24$ is a mul­ti­ple of $8$ and $12$ but not of $9$.
• Tak­ing the low­est power of a prime instead of the high­est. That gives the H.C.F, not the L.C.M.
• Leav­ing out a prime that turns up in only one of the num­bers.
• Stop­ping the divi­sion lad­der before the bot­tom row is coprime, or for­get­ting to mul­ti­ply in the num­bers along the bot­tom.

Key terms

Mul­ti­ple

A num­ber that a given num­ber divides into exactly, such as $24$ for $8$.

Com­mon mul­ti­ple

A num­ber that every one of the given num­bers divides into exactly.

L.C.M

The least com­mon mul­ti­ple: the small­est of the com­mon mul­ti­ples.

Prime fac­tori­sa­tion

Writ­ing a num­ber as a prod­uct of primes, such as $12 = 2 \times 2 \times 3$.

Power

A short way to write repeated mul­ti­ply­ing, such as $2^{3} = 2 \times 2 \times 2$.

Divi­sor

The num­ber writ­ten at the side of each row in the divi­sion method.

Quo­tient

The answer to a divi­sion, writ­ten under the num­ber that was divided.

Coprime

Num­bers that share no fac­tor except $1$.

Answers

Extra prac­tice by writ­ing the mul­ti­ples

1. $4, 6$: the mul­ti­ples of $4$ are $4, 8, 12$ and of $6$ are $6, 12$. L.C.M $= 12$.
2. $6, 9$: $6, 12, 18$ and $9, 18$. L.C.M $= 18$.
3. $5, 10$: $5, 10$ and $10$. L.C.M $= 10$.
4. $3, 8$: $3, 6, 9, 12, 15, 18, 21, 24$ and $8, 16, 24$. L.C.M $= 24$.
5. $6, 8$: $6, 12, 18, 24$ and $8, 16, 24$. L.C.M $= 24$.
6. $10, 15$: $10, 20, 30$ and $15, 30$. L.C.M $= 30$.

Extra prac­tice by prime fac­tori­sa­tion

1. $36 = 2^{2} \times 3^{2}$, $48 = 2^{4} \times 3$. L.C.M $= 2^{4} \times 3^{2} = 144$.
2. $45 = 3^{2} \times 5$, $75 = 3 \times 5^{2}$. L.C.M $= 3^{2} \times 5^{2} = 225$.
3. $16 = 2^{4}$, $40 = 2^{3} \times 5$. L.C.M $= 2^{4} \times 5 = 80$.
4. $18 = 2 \times 3^{2}$, $24 = 2^{3} \times 3$. L.C.M $= 2^{3} \times 3^{2} = 72$.
5. $20 = 2^{2} \times 5$, $24 = 2^{3} \times 3$, $30 = 2 \times 3 \times 5$. L.C.M $= 2^{3} \times 3 \times 5 = 120$.
6. $9 = 3^{2}$, $12 = 2^{2} \times 3$, $15 = 3 \times 5$. L.C.M $= 2^{2} \times 3^{2} \times 5 = 180$.

The acad­e­my's own exer­cises

1. L.C.M of $32, 56, 80, 154$ by the divi­sion method

$$\begin{aligned}2 &\mid 32, \; 56, \; 80, \; 154 \\ 2 &\mid 16, \; 28, \; 40, \; 77 \\ 2 &\mid 8, \; 14, \; 20, \; 77 \\ 2 &\mid 4, \; 7, \; 10, \; 77 \\ 7 &\mid 2, \; 7, \; 5, \; 77 \\ &2, \; 1, \; 5, \; 11\end{aligned}$$

The bot­tom row $2, 1, 5, 11$ is coprime. L.C.M $= 2 \times 2 \times 2 \times 2 \times 7 \times 2 \times 1 \times 5 \times 11 = 112 \times 110 = 12320$.

2. L.C.M of $24, 50, 75$

$$\begin{aligned}2 &\mid 24, \; 50, \; 75 \\ 3 &\mid 12, \; 25, \; 75 \\ 5 &\mid 4, \; 25, \; 25 \\ 5 &\mid 4, \; 5, \; 5 \\ &4, \; 1, \; 1\end{aligned}$$

L.C.M $= 2 \times 3 \times 5 \times 5 \times 4 = 600$. By primes: $2^{3} \times 3 \times 5^{2} = 600$.

3. L.C.M of $212, 246, 214$

$$\begin{aligned}2 &\mid 212, \; 246, \; 214 \\ &106, \; 123, \; 107\end{aligned}$$

Here $106 = 2 \times 53$, $123 = 3 \times 41$ and $107$ is prime, so no two of the bot­tom num­bers share a fac­tor. L.C.M $= 2 \times 106 \times 123 \times 107 = 2 \times 1395066 = 2790132$. By primes: $2^{2} \times 3 \times 41 \times 53 \times 107 = 2790132$.