11 Plus Exams Forum

wondering if there is a quick way to find the lcf of say 4,5 ,6 7,

DD multiplies them all getting 840, though this isn't the lcd

Is there a quick way to get to 420 rather than writing out all the factors of each one and seeing which ones match up?

Thanks
Sleepyhead

I think you mean lowest common multiple.
The mistake she made was with the 4 and the 6.
4 x 6 = 24 but the common multiples of 4 and 6 are 12s
She can look at the non prime numbers and use their common multiples.
12 x 5 x 7
This is a reasonably easy calculation too: 12 x 5 = 60, 6 x 7 x 10 = 420
Each time a non-prime is involved it is necessary to see where they overlap, e.g.
The common multiple of 3, 5 and 15 is 15 rather than 225.

To find this at KS3/4 students look at the prime factors:
4 (2,2), 5 (5), 6 (2,3), 7 (7)
Each prime factor that appears more than once is only counted in the biggest group. Here 2 occurs in both 4 and 6 so we keep 2 x 2 but don't use the 2 in 2 x 3
So to get the final answer
2 x 2 x 5 x 3 x 7

There is no quick way to find a Lowest Common Multiple that works reliably for all numbers I'm afraid. Sorry.
So, as you suggest in your post sleepyhead, for 4,5,6,7 you would list the prime factors of each number with the results:
4 = 2 x 2
5 = 5
6 = 2 x 3
7 = 7
You then take each of these factors and include them in a long multiplication sum. If a factor occurs in more than one of the above lists only include it once. If it occurs more than once in any list list then it needs to be included more than once in the multiplication sum.
For 4,5,6,7 we have:
Lowest Common Multiple = 2 x 2 x 5 x 3 x 7 = 420.

thank you - yes I did mean LCM

So please can I clarify

if we had instead, 4, 5, 6, and 12

then

4= 2 x 2

5 = 5

6 = 2 x 3

12 = 2 x 2 x 3

So what would be the multiplication sum here, how would we treat the two 2s in 12 and also the repeated 3.

Thanks
Sleepyhead

Hi Sleepyhead,

12 has already been 'used up' by the 6 and 4.
As such each set of factors only occurs once. If your DD is only 9 or 10 then I would always concentrate on looking at pairs of factors before going on to the abstract that we are covering here.
A factor only counts once (unless like with 4 it, in this case 2, occurs more than once in a number).

The common multiple of 2 and 6 is 6 because
2 (2)
6 (2 x 3)
We have used the first 2 in the 2 and don't need it in the 6
so we are left with 2 x 3

4, 6, 12
4 (2 x 2)
6 (2 x 3)
12 (2 x 2 x 3)
The first 2 can be crossed off the other numbers
4 (2 x 2)
6 (3)
12 (2 x 3)
The second 2 can then be crossed off the other numbers
4 (2 x 2)
6 (3)
12 (3)
All the prime factors from 4 are used up
Next we can cross out the 3 from 6
4 (2 x 2)
6 (3)
So we are left with
2 x 2 x 3 = 12

I hope the colour helps with the elimination process.

Taking this to your list of 4, 5, 6, 12
4 (2 x 2)
5 (5)
6 (2 x 3)
12 (2 x 2 x 3)
Cross out the first two from all numbers exept 4
Cross out the second two from all numbers except 4
Only one five - no crossing out necessary
Cross out the three from all numbers except 6
We are left with
2 x 2 x 5 x 3 = 60

great got it.

thanks

so in the question that DD did different light houses flashed every 4,5,6,7 seconds starting at midnight, At what time did all the light houses flash at the same time

In an exam uf DD listed all the common factors she;d be there for ages.. should I teach her this method. She's sitting her 11+ in January

thanks
SH

It depends on her. If she likes abstract numbers then teach her the prime factor method. If she prefers something that is intuitive then teach her to look for non primes and see if they overlap with the other numbers such as the 4 and 6 which both go into 12.

we were taught to do it in one step (ancient method!)

Using a long 'L' or upside down division 'bus-stop put all the numbers in it so 4, 5, 6 and 7 would be written (due to lack of a long 'L', I'm using a '/' sign for demonstrating division. All numbers are divided and any left-overs or undivided numbers are carried over)

4, 5, 6, 7 / 2

2, 5, 3, 7 / 2

1, 5, 3, 7 / 3

1, 5, 1, 7 / 5

1, 1, 1, 7 / 7

1, 1, 1, 1

(then take all the divisors and multiply them) 2 X 2 X 3 X 5 X 7 = 420.

L.C.M of 42, 50, 45, 12 could be worked out as follows -

42, 50, 45, 12 / 2
21, 25, 45, 6 / 2
21, 25, 45, 3 / 3
7, 25, 15, 1 / 3
7, 25, 5, 1 / 5
7, 5, 1, 1 / 5
7, 1, 1, 1 / 7
1, 1, 1, 1

L.C.M = 2 x 2 x 3 x 3 x 5 x 5 x 7 = 6300

Forgotten that method!
I use 'trees' a lot with children to find prime factors but cannot draw them on here.
The division method achieves the same thing and is much tidier.

Thanks for all your replies - explained both methods to Dd and she's understood them.

Finally if dd uses a method that is not on the ks2 syllabus in an exam , would she be penalised ?