Predict, test and explain: HCF and LCM patterns

In this layer you are the mathematician. Instead of being told a rule, you will make a guess, test it on lots of examples, look for a pattern, and then ask the most important question in mathematics: is it always true, or only sometimes?

Every chapter starts with a question and a prediction. Commit to an answer before you read on. Being wrong is part of the method: a surprising result is the best clue that there is something to explain.

The table shows a clean rule: if a divides b, then HCF(a, b) = a and LCM(a, b) = b. When the smaller number does not divide the larger, the HCF is smaller than both, and the LCM is bigger than both.

This gives a quick test for tricky questions. "The HCF of two numbers is 15 and one of them is 15. What can you say?" The other number must be a multiple of 15, and the LCM is that other number.

In the first chapter you noticed that the product divided by the LCM was always the HCF. That is the same as saying HCF × LCM = a × b. Before believing it, a good investigator tries to break it: big numbers, primes, equal numbers, numbers where one divides the other, numbers with lots of shared factors.

It works for (2, 3, 5), (3, 4, 5) and (5, 7, 9). In each of those, every pair is co-prime: no two of the numbers share a factor. It fails as soon as any two share a factor, even when the HCF of all three is 1. Look at (6, 10, 15): HCF = 1 and LCM = 30, but the product is 900. Each pair shares a prime (6 and 10 share 2, 6 and 15 share 3, 10 and 15 share 5), and each of those shared primes is counted twice in the product, but it is not in the HCF of all three, so nothing makes up for it.

So the honest statement is: for three numbers, HCF × LCM = a × b × c only when the numbers are pairwise co-prime (and then both sides equal a × b × c with HCF = 1). The Deepen layer shows a correct three-number formula.

Some of the most famous textbook problems use the same remainder trick. Investigate this one:

Find the smallest number that leaves remainder 2 when divided by 3, 4 and 5.

Try listing numbers that leave remainder 2 when divided by 5: 7, 12, 17, 22, 27, 32, 37, 42, 47, 52, 57, 62, … Now test each with 3 and 4. The first one that works is 62.

Why 62? Take 2 away and you get 60, which is divisible by 3, 4 and 5. In fact 60 = LCM(3, 4, 5). So the rule is: smallest number = LCM + remainder. And the next ones are 60 × 2 + 2 = 122, 60 × 3 + 2 = 182, and so on.

How many numbers from 1 to 100 are divisible by both 4 and 6?

Your first guess might be to count multiples of 4 (there are 25) and multiples of 6 (there are 16) and do something with them. But a number divisible by both 4 and 6 is a common multiple, and every common multiple is a multiple of the LCM, which is 12. So the question is really "how many multiples of 12 are there up to 100?" The answer is 100 ÷ 12 = 8 remainder 4, so 8: 12, 24, 36, 48, 60, 72, 84, 96.

A common wrong answer is to count multiples of 4 × 6 = 24: that gives only 4, and misses 12, 36, 60 and 84.

Investigating is not just for pure numbers. Take the courtyard from Discover, 240 cm × 180 cm, which needed 60 cm tiles. What if the courtyard changes?

• Make it 250 cm × 180 cm (10 cm longer). Now HCF(250, 180) = 10 cm. A tiny change made the best tile six times smaller, and you would need 450 tiles instead of 12.
• Make it 300 cm × 180 cm. HCF(300, 180) = 60 cm, back to large tiles: 15 tiles.

This is why tilers and carpenters love "round" measurements like 240, 300 and 360: they have lots of factors, so many tile sizes fit.

Investigating also means asking "what if?" about a word problem. A sweet shop packs 24 laddoos with some barfis into identical boxes, using the most boxes possible. How does the number of boxes change with the number of barfis?

The number of boxes is HCF(24, barfis). With 36 barfis it is 12; with 30 barfis 6; with 25 barfis only 1, a single big box, because 24 and 25 are neighbours. With 48 barfis it is 24, because 24 divides 48. Tiny changes in a count can change the answer enormously.

Here is a classic investigation that seems to have nothing to do with HCF, until it does.

Draw a rectangle on squared paper, m squares wide and n squares tall. Draw a straight line from the bottom-left corner to the top-right corner. How many squares does the line pass through (through the inside, not just touching a corner)?

Try a 2 × 3 rectangle: the diagonal passes through 4 squares. A 4 × 6 rectangle: 8 squares, not twice as many. A 5 × 5 square: just 5, straight along the diagonal squares.