LCM

As ever in maths, just when you think you’ve caught on to an idea like HCF, the subject leaps up and slaps you in the face. *Least Common Multiple i*s a number bigger than the two numbers that they are *both* factors of. The LCM is the smallest number exactly divisible by both of the given numbers.

Examples:

LCM (144, 72) = 144 144=1x144; 144=2x72 this is the one extreme

LCM (144,128) = 1152 1152=144x8; 1152=128x9

LCM (144, 56) = 1008 1008=144x7; 1008=56x18

LCM (144, 49) = 7056 7057 = 144x49 this is the other extreme.

Many students find LCM with a calculator and find it difficult to do so without.

This is usually taught by persuading students to write down (list) multiples of each number and then spotting which numbers appear in both lists, Choosing the smallest one such number.

Theory: LCM(A,B) x HCF(A,B) = AxB Test this for yourself, and try to prove it. Many teachers think this is so obvious they cannot explain it well. I may be one of them. However, you could test your theories on the numbers in the exercises for HCF, on the previous page.

When you have found the LCM of two numbers, so that LCM (A,B) = Axa = Bxb then the pair (a,b) are relatively prime and so HCF(a,b)=1. If this is not so, then the LCM is not correct. I say that this is enough help.

It is at about this point that many students simply lose the will to engage in maths for a while. Too many letters on the board, fr too many words, numbers that everyone else can ‘see’ a connection between (and they can’t), far too much terminology when they weren’t following the earlier explanation - they all add up to losing perhaps a third of the class. If the teacher knows this goes wrong, they should be able to counter that trend. I reckon I lost one or two whatever new topic I came up with, people who simply couldn’t concentrate that day. So, once a load of people *have*’got it’ and a similar load have not ‘got it’, there is a problem. It may be solved by having students help each other (that usually works), but in my opinion, the biggest problem is the pointlessness of the two ideas, LCM and HCF. the history of wanting to do this came from that time when people thought numbers were magic, even holy. At such a time big numbers were a problem and so any methods for simplifying problems to do with multiplication or division were considered wonderful. Now we have calculators so readily available, those old reasons are invalid. What is still true, though is that this is integer arithmetic and no decimals or fractions shoud appear anywhere (or you have a wrong answer). While examiners want to have non-calculator exam papers, this will remain in the syllabus mostly because it uses a set of skills that readily separates the mathematically able from the less able. So, in a sense, hard luck. In another sense, though this is a valuable opportunity to cause other bits of arithmetic to make sense.

With calculators, one can test answers. To test LCM(144,56)=1008 for example, 1008=144x7 and 1008=56x18. 7 and 18 are relatively prime, so this is indeed the *lowest* common multiple.

I usually find the LCM by dividing the product by the HCF.

LCMs could be of more than two numbers. For example, to find LCM (36,45,60) we should notice that all three have 3 as a factor, suggesting to me that 36/3x45/3 = 180 is a likely answer.

One technique shown on the right here does successive division by primes wherever possible until the last row is all relatively prime or absolutely prime The resulting 2x2x3x5 x 3 =180 is the LCM.

I demonstrate this with LCM (144,56) as earlier. You could stop when the row is relatively prime, at 18 and 7 or continue (assuming that this state is too hard to spot reliably) to having only primes in the last row, labelled abs(olutely) prime. You might test this on some of the examples above.

1. Find the LCM (48,90)

2. Find LCM (90, 156)

3. Find the LCM of 48,90,156 in a similar way to finding the HCF on the previous page.

4. The LCM and HCF of two positive numbers are 300 and 30 respectively. If one number is divided by 4, the quotient is 15, then what is the other number?

5. F..... oh, I lost the plot, again.

DJS 20181003

1. LCM (48,90)=6 LCM (48,90) = 48x90/5 = 48x15 = 720

Check HCF(720/48, 720,90) = HCF(15,8) = 1 i.e. 15 and 8 are relatively prime. √

2. HCF (90, 156) = 6 90/6=15, 156/6=26, so LCM (90, 156) = 6x15x26 = 2340

Check; 15 and 26 are relatively prime. √

3. LCM(48,90,156)=2x2x2x3x3x(2X5x13) = 16x9x5x13 = 9360. Check by testing to see if 9360 divided by each of the three original numbers gives a set that are relatively prime, by confirming that HCF(195,104,60) = 1 √

4. Obviously one number is 4x15=60, so the other must be 30 x 300 / 60 = 150