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Chapter 1 · Real Numbers
Why the HCF-times-LCM shortcut works for two numbers but breaks for three
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For two numbers, HCF times LCM is their product - and that is not a coincidence, it is what happens when a column holding two indices hands both of them back. For three numbers the same argument stops working, and it stops working in a way you can predict to the exact factor before you divide anything: the middle index is the one that gets dropped.
The idea
HCF times LCM equalling the product is not a fact about numbers; it is a fact about counting two indices. For each prime, the smaller index and the larger index between them use up exactly the two indices that the product contains, so nothing is lost and nothing is double counted. Add a third number and the smaller and the larger still take two indices — the middle one is simply dropped, and the identity falls short by exactly what that middle index contributes. That also says precisely when three numbers do satisfy it: when no middle index has anything to contribute. So the lesson is not "beware of three", it is that you can see in advance which way any such identity will go.
What you should be able to do
- Verify the two-number identity on a worked pair and state what it lets you compute
- Explain the identity by showing that the lower and higher indices of a prime reassemble its two indices in the product
- Use the identity in both directions: HCF and product to LCM, and HCF and LCM to the product
- Show, on the chapter's own three numbers, that the identity fails, and measure the size of the failure
- Attribute that failure to the middle index, prime by prime, and predict the discrepancy before dividing
- Give three numbers for which the identity nevertheless holds, and say what is special about them
- Apply the corrected three-number formulas from the chapter's closing note and check them against a known answer
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| HCF | the largest number dividing each of the given numbers | printed in §1.2, p. 4 |
| LCM | the smallest positive number that every one of the given numbers divides | printed in §1.2, p. 4 |
| positive integers | the whole numbers greater than zero, the setting for the whole discussion | printed in §1.1, p. 1, and in the closing note on p. 9 |
| greatest power | the highest index a prime reaches across the numbers | printed in §1.2, p. 4 |
| smallest power | the lowest index a prime reaches across the numbers being compared | printed in §1.2, p. 4 |
| middle index | with three numbers, the index that is neither the lowest nor the highest in a prime's column | an added term; the chapter never separates it out |
| shortfall factor | how many times larger three numbers multiplied together are than their HCF times their LCM | an added coinage; not printed in this chapter |
| pairwise coprime | no two of the numbers share any prime at all | an added compound; the chapter prints "coprime" for two numbers in §1.3, p. 7, but does not use this three-number form |
Where people slip up
- "HCF × LCM = product is a law of numbers." It is a consequence of there being exactly two indices in each column. State it that way once and the three-number case stops being a surprise rule to memorise.
- "So for three numbers the identity is always wrong." No — 17, 23, 29 and 8, 9, 25 both satisfy it. The chapter's remark is written about the example in front of it; the defensible general statement is that the identity is not guaranteed for three numbers.
- "You can find the HCF of three numbers by pairing them and using the identity twice." The closing note exists precisely because that does not work; it needs all three pairwise HCFs and the three-way one together.
- "The shortfall is random." It is 24 for 6, 72, 120 and 3 for 12, 15, 21, and both are read off the columns before dividing. Predict it, then divide, and let the prediction be confirmed.
- "If the HCF is 1 the LCM is the product, for any count of numbers." True only when no two of them share a prime. Three numbers can have HCF 1 while two of them still share a factor — 6, 10, 15 is the standard trap, and their LCM is 30, not 900.
- "Q4 wants the factorisations of 306 and 657." It hands over the HCF so that the identity does the work. Reaching for factor trees there is a sign the identity has not landed.
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Worked answers: Exercise 1.1 · Exercise 1.2 · this video explains Exercise 1.1 Q4
Transcript1,686 words
Here is something you might notice and then pass straight over. The HCF of 6 and 20 is 2. Their LCM is 60. Now multiply those two answers together. 2 times 60 is 120. And 6 times 20 is also 120. The HCF times the LCM came out as the product of the two numbers you started with. That is either a coincidence about 6 and 20, or it is true of every pair, and those are very different situations.
It is worth knowing which, because if it always holds it is a shortcut, and if it does not, it is a trap. So take one prime column at a time. In the 2 column, 6 has index 1 and 20 has index 2. The HCF takes the lower of those, which is 1. The LCM takes the higher, which is 2. Now look carefully at what just happened. The column held two indices, 1 and 2.
Lower and higher handed back 1 and 2. The same two numbers, in the same quantities. Nothing was lost and nothing was counted twice. And multiplying the HCF by the LCM adds those indices back together: 1 plus 2 is 3, which is what the column held all along. That is not something special about the 2 column. Take any two whole numbers, and any prime at all. It has some index in the first number and some index in the second.
And the lower of two numbers and the higher of two numbers are just those two numbers again, sorted. So they always add up to what you started with. Every column, every time. Multiply the HCF by the LCM and each column reassembles itself, so the answer has to be the product. It is not a coincidence, and it is not a fact about 6 and 20. It is a fact about a column holding exactly two indices.
Which makes it genuinely useful. 96 is 2 to the fifth times 3. 404 is 2 squared times 101. Their HCF is 4, and that part is quick: the 2 column holds 5 and 2, and the lower is 2. Now suppose you want the LCM. You could go back through the columns and take the higher index in each one. Or you could multiply 96 by 404, which is 38784, and divide that by 4.
Which gives 9696. One multiplication and one division, instead of a second pass over the columns. It runs the other way just as well. Suppose someone hands you 306 and 657, tells you their HCF is 9, and asks for the LCM. You do not need to factorise either of them. 306 times 657 is 201042. Divide by 9, and you get 22338. That is the LCM, and you never found out which primes are in either number.
If you do factorise them, the answer is the same: 306 is 2 times 3 squared times 17, and 657 is 3 squared times 73. But the whole point is that none of that was necessary. And there is a third way to lean on it, which is the one people usually miss. Because the four quantities are tied together, any three of them give you the fourth. Suppose you know the HCF of two numbers is 4, their LCM is 9696, and one of the numbers is 96.
Then the product of the two numbers has to be 4 times 9696, which is 38784. And 38784 divided by 96 is 404. So the other number was never really unknown. It was forced. That works for every pair, not just this one, and it is the same one line of reasoning each time. So here is the obvious next question. Does this work for three numbers? The tempting answer is yes.
It worked so cleanly for two, and the argument sounded like it was about columns rather than about pairs. That is exactly the kind of guess that is worth testing instead of assuming. And testing it is easy, because you can simply compute both sides and look. Although before computing anything, go back to the argument, because it already tells you what to expect. With two numbers, a column holds two indices, and lower and higher use both of them.
With three numbers, a column holds three indices. The HCF still takes the lowest one. The LCM still takes the highest one. And the middle one is taken by neither. It is simply dropped. So the two sides cannot match in general. The product carries all three indices. The HCF times the LCM carries only two of them. Take 6, 72 and 120. 6 is 2 times 3. 72 is 2 cubed times 3 squared. 120 is 2 cubed times 3 times 5.
Their HCF is 6, and their LCM is 360. Multiply all three numbers together and you get 51840. Multiply the HCF by the LCM and you get 2160. Those are not the same number, and they are not close. 51840 divided by 2160 is 24, exactly. So it does not merely fail. It falls short by a factor of 24. And now the part that makes this worth knowing. That 24 was predictable from the columns, before any of the dividing.
The 2 column holds 1, 3 and 3. Those total 7. The lowest and the highest are 1 and 3, which uses 4 of the 7. So 3 are dropped. The 3 column holds 1, 2 and 1. Those total 4, and 1 and 2 use 3 of them, so 1 is dropped. The 5 column holds 0, 0 and 1, and dropping the middle one there costs nothing at all.
Dropped altogether: three 2s and one 3. 2 cubed times 3 is 24. The same 24, named before it was measured. Try it once more on 12, 15 and 21, so that it is not luck. Their HCF is 3 and their LCM is 420, so the HCF times the LCM is 1260. Their product is 3780. And 3780 divided by 1260 is 3. Predicted from the columns: the 3 column holds 1, 1 and 1, and dropping the middle one loses a single factor of 3.
In every other column only one of the three numbers contributes anything, so there is no middle index to lose. A shortfall of exactly 3, again named before it was measured. Now be careful, because the identity does not always fail for three numbers. Take 17, 23 and 29. Their HCF is 1 and their LCM is 11339, and 1 times 11339 is 11339, which is exactly their product. It holds.
Take 8, 9 and 25, and notice that not one of those is prime. Their HCF is 1 and their LCM is 1800, and their product is 1800 as well. It holds again. In both cases every prime appears in only one of the three numbers, so every middle index is 0, and there is nothing to drop. Which is not the same as saying the HCF is 1. Take 6, 10 and 15.
Nothing divides all three except 1, so their HCF really is 1. But their LCM is 30, not 900. Their product is 900, so this one falls short by a factor of 30. The columns say why immediately. 6 and 10 both carry a 2. 6 and 15 both carry a 3. 10 and 15 both carry a 5. Every column has a middle index of 1. So what matters is not whether all three of them share something. It is whether any two of them do.
So the honest statement is not that the identity fails for three numbers. It is that it is not guaranteed. It holds exactly when no two of the three share a prime, and otherwise it falls short by exactly the dropped middle indices, multiplied together. That was checked on every triple of whole numbers up to 40. There are 11480 of them. The identity held on 3008 and failed on 8472.
So neither of the two easy things to say about it is true. And on every single one of the 11480, the shortfall read off the columns was the shortfall you get by dividing. Which raises a practical question. If the identity is what you use to get an LCM cheaply, what do you do with three numbers? The natural move is to use it anyway. Multiply all three, divide by the HCF, and hope.
On 6, 72 and 120, that gives 51840 divided by 6, which is 8640. The real LCM is 360. So it does not just miss. It overshoots by a factor of 24, which is the same 24 as before, arriving from the other side. And this is not a bad case picked to make a point. Tried on every triple of whole numbers up to 30, all 4960 of them, that shortcut is wrong on 3645.
It is wrong far more often than it is right. So here is the repair, and it is worth seeing because it is not arbitrary. What went missing was a middle index, and what fixes it is information about pairs. Multiply the three numbers together, multiply by their three-way HCF, then divide by the HCF of each of the three pairs. On 6, 72 and 120, those pairwise HCFs are 6, 24 and 6.
So 51840 times 6 is 311040, and 6 times 24 times 6 is 864. And 311040 divided by 864 is 360. The LCM, exactly as before. And the same shape swaps the two roles. The product, times the three-way LCM, divided by the three pairwise LCMs, gives you the HCF instead. For 6, 72 and 120 those pairwise LCMs are 72, 360 and 120. 51840 times 360 is 18662400, and 72 times 360 times 120 is 3110400.
Divide one by the other and you get 6, which is the HCF you already had. None of which is a rule to memorise, once you know that a middle index is the thing that goes missing. It is just bookkeeping, arranged to get the count right.
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
- Reading HCF and LCM straight off two factorisationsClass 10 · Ch 1, Real Numbers
- Why a composite number has one prime factorisation and no otherClass 10 · Ch 1, Real Numbers
Either side of this one
- A prime dividing a square must divide the number itselfClass 10 · Ch 1, Real Numbers