How H.C.F and L.C.M work together

H.C.F and L.C.M are usu­ally taught one after the other, as if they had noth­ing to do with each other. In fact they are two halves of one pic­ture. The H.C.F answers ques­tions like "what is the largest equal share?" and the L.C.M answers ques­tions like "when will these repeat­ing events hap­pen together again?". Bells, traf­fic lights, bus timeta­bles and cut­ting cloth all lead to one or the other.

Two ideas, one pic­ture

You know how to find the H.C.F of some num­bers. You know how to find their L.C.M too. This les­son puts the two side by side. They are not strangers. For two num­bers there is an exact link between them. Know one, and it tells you the other.

The same two ideas then stretch to frac­tions. The les­son ends with three prob­lems. Each one tells you a story in words. You have to say which of the two it wants.

How the H.C.F and the L.C.M of two num­bers are linked

Here is the rule, in the acad­e­my's own words. The prod­uct of two num­bers is always equal to the prod­uct of their H.C.F and their L.C.M. A prod­uct is what you get when you mul­ti­ply. So the rule says this. Mul­ti­ply the two num­bers together. You get the same answer as their H.C.F times their L.C.M. Now look at the word two. The rule is about a pair. A pair is where you use it.

The acad­emy uses the pair $15$ and $24$. Break each one into primes.

$15$ and $24$ bro­ken into primes

$$\begin{aligned}15 &= 3 \times 5 \\ 24 &= 2 \times 2 \times 2 \times 3\end{aligned}$$

They share only one prime, and that is $3$. Each of them holds just one $3$. So the H.C.F of $15$ and $24$ is $3$.

Now for the L.C.M. Look at every prime that appears at all. From each one, take the larger count. That gives three $2$s from $24$, one $3$, and one $5$ from $15$.

The L.C.M of $15$ and $24$

$$2 \times 2 \times 2 \times 3 \times 5 = 120$$

Now look at what those two answers took. The two num­bers share one prime. The H.C.F took the smaller count of it: a sin­gle $3$. The L.C.M took the larger count of every prime: three $2$s, one $3$ and one $5$.

Set the H.C.F and the L.C.M side by side. Every prime inside $15$ is there. Every prime inside $24$ is there too. Each one is counted once, and no more. The $3$ and the $5$ make $15$. The three $2$s and the $3$ make $24$.

A Venn dia­gram of the prime fac­tors makes this easy to see. The shared prime sits in the over­lap. The H.C.F is just the over­lap. The L.C.M is every­thing in both cir­cles, each prime counted once.

That is the rea­son, and it works for any pair. Think about one prime at a time. The H.C.F takes the smaller count of copies, because that is all both num­bers can spare. The L.C.M takes the larger count of copies, because both num­bers have to go into it a whole num­ber of times.

Smaller and larger cover every copy in both num­bers. No copy is counted twice. So mul­ti­ply the H.C.F by the L.C.M. You put back just what the two num­bers were made of.

Now test the rule on the num­bers. The two num­bers mul­ti­ply to $15 \times 24 = 360$. The H.C.F and the L.C.M mul­ti­ply to $3 \times 120 = 360$. The two agree. The rule says they must.

Remem­ber. Mul­ti­ply two num­bers together. You get the same answer as their H.C.F times their L.C.M. On $15$ and $24$, both come to $360$.

What the rule is for

Four num­bers sit in this rule. They are the two num­bers, their H.C.F and their L.C.M. Know any three of them. Then you can work out the fourth. So once you have the H.C.F of a pair, one divi­sion gives you the L.C.M. You do not have to fac­torise a sec­ond time.

The L.C.M of $15$ and $24$, once you know their H.C.F is $3$

$$\begin{aligned}(15 \times 24) \div 3 &= 360 \div 3 \\ &= 120\end{aligned}$$

It works the other way round as well. Say two num­bers mul­ti­ply to $360$. Say their L.C.M is $120$. Then their H.C.F is $360 \div 120 = 3$.

A sec­ond worked exam­ple: 18 and 42

Here is the whole rou­tine on a new pair.

$$\begin{aligned}18 &= 2 \times 3 \times 3 \\ 42 &= 2 \times 3 \times 7 \\ \text{H.C.F} &= 2 \times 3 = 6 \\ \text{L.C.M} &= 2 \times 3 \times 3 \times 7 = 126\end{aligned}$$

Check the rule: $18 \times 42 = 756$ and $6 \times 126 = 756$. They agree.

Now use the rule as a short­cut. If you had found only the H.C.F, one divi­sion would give the L.C.M: $756 \div 6 = 126$.

Two quick facts that fol­low

• The H.C.F always divides the L.C.M. In every exam­ple so far, $3$ divides $120$ and $6$ divides $126$. This is a handy check on your work­ing.
• For coprime num­bers the H.C.F is $1$, so the L.C.M is sim­ply the prod­uct. The L.C.M of $8$ and $15$ is $120$.

Remem­ber. The prod­uct rule is for two num­bers. For three num­bers such as $2$, $4$ and $6$, the prod­uct is $48$ but the H.C.F times the L.C.M is $2 \times 12 = 24$.

H.C.F and L.C.M of frac­tions

Frac­tions have an H.C.F and an L.C.M too. The words still mean what they always meant. The H.C.F of some frac­tions is the largest frac­tion that goes into all of them a whole num­ber of times. The L.C.M is the small­est frac­tion that all of them go into a whole num­ber of times.

You do not need a new method for either of them. Both are built from the H.C.F and the L.C.M of whole num­bers. Those you can find already. You work on the numer­a­tors, the top num­bers, on one side. You work on the denom­i­na­tors, the bot­tom num­bers, on the other.

First, write each frac­tion in its low­est terms. Low­est terms means the top and the bot­tom share no fac­tor big­ger than $1$. Take $\displaystyle \frac{6}{8}$. It is not in low­est terms, because the H.C.F of $6$ and $8$ is $2$. Halve both and you have $\displaystyle \frac{3}{4}$. That is the same frac­tion, writ­ten prop­erly.

Do this first, because the two rules below read the top and the bot­tom just as they stand. Hand them $\displaystyle \frac{6}{8}$ when you meant $\displaystyle \frac{3}{4}$. They will give you a dif­fer­ent answer for the same frac­tion.

The two rules are mir­ror images. There is a rea­son for that. A frac­tion gets smaller when its top num­ber gets smaller. It also gets smaller when its bot­tom num­ber gets big­ger.

The H.C.F has to be the small one. So on top it takes the smaller choice, the H.C.F of the numer­a­tors. Under­neath it takes the big­ger choice, the L.C.M of the denom­i­na­tors. The L.C.M has to be the large one. So it takes them the other way round.

Find­ing the L.C.M of two frac­tions

Take $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{9}{10}$. The L.C.M of the numer­a­tors $3$ and $9$ is $9$. The H.C.F of the denom­i­na­tors $4$ and $10$ is $2$.

Find the L.C.M of $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{9}{10}$

$$\displaystyle \begin{aligned}\text{L.C.M} &= \frac{\text{L.C.M of } 3 \text{ and } 9}{\text{H.C.F of } 4 \text{ and } 10} \\ &= \frac{9}{2}\end{aligned}$$

Now check it from the other end. $\displaystyle \frac{3}{4} \times 6 = \frac{18}{4} = \frac{9}{2}$. And $\displaystyle \frac{9}{10} \times 5 = \frac{45}{10} = \frac{9}{2}$. Each frac­tion goes into $\displaystyle \frac{9}{2}$ a whole num­ber of times. So $\displaystyle \frac{9}{2}$ is a com­mon mul­ti­ple of the two. The rule gives the least of them.

Find­ing the H.C.F of two frac­tions

Now take the same two frac­tions. Swap the two rules over. The numer­a­tors are $3$ and $9$. Their H.C.F is $3$. The denom­i­na­tors are $4$ and $10$. Their L.C.M is $20$.

Find the H.C.F of $\displaystyle \frac{3}{4}$ and $\displaystyle \frac{9}{10}$

$$\displaystyle \begin{aligned}\text{H.C.F} &= \frac{\text{H.C.F of } 3 \text{ and } 9}{\text{L.C.M of } 4 \text{ and } 10} \\ &= \frac{3}{20}\end{aligned}$$

Check that it goes into both a whole num­ber of times. $\displaystyle \frac{3}{20} \times 5 = \frac{15}{20} = \frac{3}{4}$. So it goes into $\displaystyle \frac{3}{4}$ five times. And $\displaystyle \frac{3}{20} \times 6 = \frac{18}{20} = \frac{9}{10}$. So it goes into $\displaystyle \frac{9}{10}$ six times. A whole num­ber of times each way. That is what a com­mon fac­tor has to do.

On this pair the two prod­ucts agree. $\displaystyle \frac{3}{4} \times \frac{9}{10} = \frac{27}{40}$. And $\displaystyle \frac{3}{20} \times \frac{9}{2} = \frac{27}{40}$ as well. But take care here. The rule ear­lier in the les­son was about two whole num­bers. So notice it on this pair. Do not use it as a rule for all frac­tions.

Remem­ber. For frac­tions, the H.C.F is the H.C.F of the numer­a­tors over the L.C.M of the denom­i­na­tors. The L.C.M is the L.C.M of the numer­a­tors over the H.C.F of the denom­i­na­tors. Small over large gives you the small one. Large over small gives you the large one.

The acad­e­my's three prob­lems

Here is a quick way to decide. Ask your­self whether the answer should be smaller than the given num­bers or big­ger. Words like "largest num­ber that divides", "great­est length", or "max­i­mum num­ber of equal groups" point to an answer no big­ger than the small­est given num­ber: that is the H.C.F. Words like "small­est num­ber divis­i­ble by", "next time together", or "least num­ber of" point to an answer at least as big as the largest given num­ber: that is the L.C.M.

The sheet ends with three prob­lems. None of them says H.C.F. None of them says L.C.M. Each one tells you a story. You have to work out which of the two it is ask­ing for. So go back to the two mean­ings every time. The H.C.F is the great­est num­ber that goes into all the given num­bers. The L.C.M is the least num­ber that all the given num­bers go into. Watch the direc­tion. One num­ber goes into them all. They all go into the other num­ber.

1. The largest num­ber which can exactly divide $10$, $20$ and $30$

A num­ber that divides all three is a com­mon fac­tor of all three. The prob­lem asks for the largest one. That is word for word what the H.C.F means. So the answer is the H.C.F of $10$, $20$ and $30$. The acad­emy also calls it the G.C.D.

The 1st method (By writ­ing all fac­tors).

Four num­bers appear in all three rows: $1, 2, 5$ and $10$. The great­est of them is $10$.

The 2nd method (Prime fac­tor­iza­tion method).

$10$, $20$ and $30$ bro­ken into primes

$$\begin{aligned}10 &= 2 \times 5 \\ 20 &= 2 \times 2 \times 5 \\ 30 &= 2 \times 3 \times 5\end{aligned}$$

Every one of them holds one $2$ and one $5$. Noth­ing else is in all three. So the H.C.F is $2 \times 5 = 10$.

The answer is $10$. It goes into $10$ once. It goes into $20$ twice, and into $30$ three times. Noth­ing larger can work, because noth­ing larger than $10$ goes into $10$ at all.

2. Three plan­ets

The time peri­ods of three plan­ets are $100$ days, $250$ days and $300$ days. A plan­et's time period is how long it takes to go round once. After how many days do the plan­ets come back to where they stand now?

The first planet is back where it stands today after $100$ days. Then again after every fur­ther $100$ days. The sec­ond is back after $250$ days, and every $250$ days after that. The third is back after $300$ days, and every $300$ days after that.

So all three stand where they stand today on one kind of day. That day has to be a mul­ti­ple of all three peri­ods at once. The first such day is their least com­mon mul­ti­ple.

Prime fac­torise the three peri­ods.

$100$, $250$ and $300$ bro­ken into primes

$$\begin{aligned}100 &= 2 \times 2 \times 5 \times 5 \\ 250 &= 2 \times 5 \times 5 \times 5 \\ 300 &= 2 \times 2 \times 3 \times 5 \times 5\end{aligned}$$

Take the largest count of each prime. There are two $2$s, from $100$ and from $300$. There is one $3$, from $300$. There are three $5$s, from $250$.

The L.C.M of $100$, $250$ and $300$

$$2 \times 2 \times 3 \times 5 \times 5 \times 5 = 1500$$

So the plan­ets are back where they stand now after $1500$ days. Count the jour­neys and you can see it is right. $1500 \div 100 = 15$, $1500 \div 250 = 6$ and $1500 \div 300 = 5$. In those $1500$ days the first planet goes round fif­teen times. The sec­ond goes round six times. The third goes round five times. All three arrive home together.

3. The largest five-digit num­ber divis­i­ble by $2$, $3$, $5$ and $11$

A num­ber divis­i­ble by all four is a com­mon mul­ti­ple of all four. Look back at $8$ and $12$ in Least Com­mon Mul­ti­ple (L.C.M). Their com­mon mul­ti­ples were $24$, $48$, $72$ and on. Every one of them is a mul­ti­ple of $24$. And $24$ was their L.C.M. Com­mon mul­ti­ples always sit like that.

So every num­ber that answers this prob­lem is a mul­ti­ple of one num­ber. That num­ber is the L.C.M of $2$, $3$, $5$ and $11$. No other num­ber can answer it.

$2$, $3$, $5$ and $11$ are four dif­fer­ent primes. So they share noth­ing at all. None of them goes into any of the oth­ers. Their L.C.M is just all four of them mul­ti­plied together.

The L.C.M of $2$, $3$, $5$ and $11$

$$2 \times 3 \times 5 \times 11 = 330$$

Now find the largest mul­ti­ple of $330$ that still has five dig­its. The largest five-digit num­ber of all is $99999$. Divide that by $330$. Then look at what is left over.

$99999 \div 330$

$$\begin{aligned}99999 &= 330 \times 303 + 9 \\ &= 99990 + 9\end{aligned}$$

The remain­der is $9$. So $99999$ sits $9$ above the last mul­ti­ple of $330$. Take that $9$ off.

The largest five-digit num­ber divis­i­ble by $2$, $3$, $5$ and $11$

$$99999 - 9 = 99990$$

Now test the answer against the divis­i­bil­ity rules. Take one rule at a time. The rules for $2$, $3$ and $5$ are in Divis­i­bil­ity rules for 2, 3, 4 and 5. The rule for $11$ is in Divis­i­bil­ity rules for 6, 7, 8, 9, 10 and 11.

• Divis­i­ble by $2$, because its unit digit is $0$.
• Divis­i­ble by $5$, for the same rea­son.
• Divis­i­ble by $3$, because its dig­its add to $9 + 9 + 9 + 9 + 0 = 36$, and $36$ is divis­i­ble by $3$.
• Divis­i­ble by $11$ as well. Num­ber the places from the right. The first place from the right is place one. The next is place two, and so on. Places one, three and five are the odd places. Places two and four are the even places. The dig­its in the odd places are $0$, $9$ and $9$. They add to $18$. The dig­its in the even places are $9$ and $9$. They add to $18$. The dif­fer­ence between the two sums is $0$.

And $99990 \div 330 = 303$. So it is a mul­ti­ple of the L.C.M, as it had to be.

It is the largest one, not just one that works, because $99999$ is not divis­i­ble by $330$. Nor is any of the eight num­bers from $99991$ to $99998$. Come down from $99999$. The first num­ber you reach that is divis­i­ble by all four is $99990$.

Your turn

Check the rule

For each pair, find the H.C.F and find the L.C.M. Then mul­ti­ply the two num­bers together. Then mul­ti­ply the H.C.F by the L.C.M. Check that the two answers agree.

Use the rule

1. The H.C.F of two num­bers is $6$ and their L.C.M is $36$. One of the num­bers is $12$. What is the other?
2. The H.C.F of two num­bers is $9$ and their L.C.M is $90$. One of the num­bers is $18$. What is the other?
3. Two num­bers mul­ti­ply to $360$ and their H.C.F is $6$. What is their L.C.M?
4. Two num­bers mul­ti­ply to $588$ and their L.C.M is $84$. What is their H.C.F?

H.C.F and L.C.M of frac­tions

Find both the H.C.F and the L.C.M of each set. Every frac­tion here is already in its low­est terms.

1. $$\displaystyle \frac{2}{3}, \; \frac{8}{9}$$
2. $$\displaystyle \frac{5}{6}, \; \frac{10}{9}$$
3. $$\displaystyle \frac{6}{7}, \; \frac{4}{21}$$
4. $$\displaystyle \frac{9}{10}, \; \frac{12}{25}$$
5. $$\displaystyle \frac{1}{2}, \; \frac{3}{4}, \; \frac{5}{8}$$
6. $$\displaystyle \frac{2}{3}, \; \frac{4}{9}, \; \frac{8}{27}$$

Before the word prob­lems, one more exam­ple of the rea­son­ing. Two ropes are $36$ m and $48$ m long. They must be cut into pieces all of the same length, as long as pos­si­ble, with none left over. The length must divide both, and be as large as pos­si­ble, so it is the H.C.F of $36$ and $48$, which is $12$ m. There will be $3 + 4 = 7$ pieces.

Word prob­lems

Before you work each one out, say what it asks for. Is it the H.C.F or the L.C.M? Say how you can tell. One ques­tion uses the word inter­val. An inter­val is the gap of time between one ring and the next.

1. Find the largest num­ber which can exactly divide $12$, $18$ and $30$.
2. Find the small­est num­ber which is exactly divis­i­ble by $8$, $9$ and $12$.
3. Three bells ring at inter­vals of $6$ min­utes, $8$ min­utes and $12$ min­utes. They have just rung together. After how many min­utes will they next ring together?
4. Find the largest four-digit num­ber which is divis­i­ble by $2$, $3$, $5$ and $11$.
5. Find the largest five-digit num­ber which is divis­i­ble by $2$, $3$, $5$ and $7$.

Com­mon mis­takes

• Using the prod­uct rule on three or more num­bers. It holds for a pair only.
• For frac­tions, putting the H.C.F on the bot­tom of the H.C.F. It is H.C.F of tops over L.C.M of bot­toms.
• For­get­ting to write frac­tions in low­est terms before using the frac­tion rules.
• Choos­ing H.C.F in a "next time together" prob­lem. Events that repeat meet at a com­mon mul­ti­ple.
• In the largest-num­ber prob­lems, sub­tract­ing the quo­tient instead of the remain­der.
• Giv­ing an H.C.F that does not divide the L.C.M. If it does not, some­thing went wrong.

Key terms

Prod­uct

The answer when num­bers are mul­ti­plied.

H.C.F (G.C.D)

The great­est num­ber that goes into all the given num­bers.

L.C.M

The least num­ber that all the given num­bers go into.

Prime fac­tori­sa­tion

Writ­ing a num­ber as a prod­uct of primes.

Low­est terms

A frac­tion whose top and bot­tom share no fac­tor big­ger than $1$.

Inter­val

The gap of time between one event and the next.

Remain­der

What is left over after divid­ing.

Answers

Check the rule

1. $12, 18$: H.C.F $6$, L.C.M $36$; $12 \times 18 = 216 = 6 \times 36$.
2. $9, 15$: H.C.F $3$, L.C.M $45$; $9 \times 15 = 135 = 3 \times 45$.
3. $14, 21$: H.C.F $7$, L.C.M $42$; $14 \times 21 = 294 = 7 \times 42$.
4. $8, 20$: H.C.F $4$, L.C.M $40$; $8 \times 20 = 160 = 4 \times 40$.
5. $16, 24$: H.C.F $8$, L.C.M $48$; $16 \times 24 = 384 = 8 \times 48$.
6. $25, 35$: H.C.F $5$, L.C.M $175$; $25 \times 35 = 875 = 5 \times 175$.

Use the rule

1. $(6 \times 36) \div 12 = 18$.
2. $(9 \times 90) \div 18 = 45$.
3. $360 \div 6 = 60$.
4. $588 \div 84 = 7$.

H.C.F and L.C.M of frac­tions

1. H.C.F $\displaystyle \frac{2}{9}$, L.C.M $\displaystyle \frac{8}{3}$.
2. H.C.F $\displaystyle \frac{5}{18}$, L.C.M $\displaystyle \frac{10}{3}$.
3. H.C.F $\displaystyle \frac{2}{21}$, L.C.M $\displaystyle \frac{12}{7}$.
4. H.C.F $\displaystyle \frac{3}{50}$, L.C.M $\displaystyle \frac{36}{5}$.
5. H.C.F $\displaystyle \frac{1}{8}$, L.C.M $\displaystyle \frac{15}{2}$.
6. H.C.F $\displaystyle \frac{2}{27}$, L.C.M $\displaystyle \frac{8}{3}$.

Word prob­lems

1. H.C.F, because one num­ber must go into all three: $6$.
2. L.C.M, because all three must go into it: $72$.
3. L.C.M of the inter­vals, because the rings repeat: $24$ min­utes.
4. L.C.M first, $330$. Then $9999 = 330 \times 30 + 99$, so the answer is $9999 - 99 = 9900$.
5. L.C.M of $2, 3, 5, 7$ is $210$. Then $99999 = 210 \times 476 + 39$, so the answer is $99999 - 39 = 99960$.