mersenneforum.org Definition of “common multiple”
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 2023-02-04, 17:53 #1 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 2·229 Posts Definition of “common multiple” I am thinking of doing an Open University (OU) mathematics degree. https://www.open.ac.uk/courses/maths...athematics-q31 This is part-time, distance learning. I would need to start on the course “Essential Mathematics 1”, which is pretty basic. https://www.open.ac.uk/courses/modules/mst124 The OU fees include all the books, but I decided to buy the course books from eBay, to start reading them now. I am going through book A but are confused by this statement “A common multiple of two or more integers is a number that is a multiple of all of them. For example the common multiples of 4 & 6 are …, -36, -24, -12, 0, 12, 24, 36, … Common multiple is not a term that I have come across before, although I have come across least common multiple (lcm). I decided to Google the term common multiple and found this webpage https://www.splashlearn.com/math-voc...ommon-multiple There they list the common multiples of 6 & 7, and say 42 & 84 are the common multiples. Is this an ambiguous term, or are one of the sources giving blatantly wrong information?
 2023-02-04, 19:09 #2 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 236710 Posts Where is the conflict between the 2? Can you be specific of what number is included in one but not the other.
2023-02-04, 19:15   #3
charybdis

Apr 2020

3·317 Posts

Quote:
 Originally Posted by drkirkby Is this an ambiguous term, or are one of the sources giving blatantly wrong information?
There is no conflict. The second source is only talking about common multiples on that grid of numbers from 1 and 100. It is not claiming that 42 and 84 are the only common multiples of 6 and 7.

In practice we often work only with positive integers, so it's perfectly fine to say that the common multiples of 6 and 7 are 42, 84, 126, 168, ... ignoring 0 and the negative common multiples.

 2023-02-04, 19:36 #4 drkirkby   "David Kirkby" Jan 2021 Althorne, Essex, UK 2×229 Posts Ignoring 0 and negative integers, the OU source gives 12. But 6 x 4 = 24, not 12. So how do the OU say 12 should be in the list? I appreciate that https://www.splashlearn.com/math-voc...ommon-multiple is just not listing larger numbers. That's fair enough - the full list would be infinite in length. But the lowest value they give (42) is obtained by multiplying 6 by 7. But the OU gets a smaller number (12) than multiplying 6 and 4. Last fiddled with by drkirkby on 2023-02-04 at 19:41
 2023-02-04, 19:48 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 32×263 Posts I think you are correct. If you include 12 for 6 and 4, then there is no logical way to exclude 8 which is not included. I would let it down as a typo since there is no logical/algorithmic way of generating: 0, 12, 24, 36, without including 8. Nice catch. My old eyes/brain failed me yet once more. I would define common-multiples of integers a and b as all the multiples of a*b where the multiplier can be any integer. But I am usually wrong about everything. So take that with a grain of sugar (or is it sand). Last fiddled with by a1call on 2023-02-04 at 19:53
2023-02-04, 19:56   #6
charybdis

Apr 2020

16678 Posts

Quote:
 Originally Posted by drkirkby Ignoring 0 and negative integers, the OU source gives 12. But 6 x 4 = 24, not 12. So how do the OU say 12 should be in the list? I appreciate that https://www.splashlearn.com/math-voc...ommon-multiple is just not listing larger numbers. That's fair enough - the full list would be infinite in length. But the lowest value they give (42) is obtained by multiplying 6 by 7. But the OU gets a smaller number (12) than multiplying 6 and 4.
6 = 2*3 and 4 = 2*2. Because they have the factor 2 in common, the lowest common multiple is not 6*4 but rather 6*4/2 = 2*2*3 = 12. 6 and 7 do not have any common factors so their lowest common multiple is 6*7 = 42.

lcm(m,n) = mn/gcd(m,n) where gcd = greatest common divisor (sometimes known as highest common factor).

Quote:
 Originally Posted by a1call I think you are correct. If you include 12 for 6 and 4, then there is no logical way to exclude 8 which is not included.
8 is not a multiple of 6.

Quote:
 I would define common-multiples of integers a and b as all the multiples of a*b where the multiplier can be any integer. But I am usually wrong about everything. So take that with a grain of sugar (or is it sand).
You're right about one thing, which is that you're wrong.

Last fiddled with by charybdis on 2023-02-04 at 19:59

 2023-02-04, 20:00 #7 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 44778 Posts I stand corrected thank you charybdis. ETA: Hay (or it Hey). two things, I was correct about the grain of sugar too. ETA II: This helped me understand the algorithm: https://multiply.info/CommonMultiple...f-4-and-6.html Last fiddled with by a1call on 2023-02-04 at 20:16
2023-02-04, 20:23   #8
drkirkby

"David Kirkby"
Jan 2021
Althorne, Essex, UK

2·229 Posts

Quote:
 Originally Posted by charybdis 6 = 2*3 and 4 = 2*2. Because they have the factor 2 in common, the lowest common multiple is not 6*4 but rather 6*4/2 = 2*2*3 = 12. 6 and 7 do not have any common factors so their lowest common multiple is 6*7 = 42. lcm(m,n) = mn/gcd(m,n) where gcd = greatest common divisor (sometimes known as highest common factor). 8 is not a multiple of 6. You're right about one thing, which is that you're wrong.
So when computing the common factors of m and n, you have to factorise m and n if they are not prime? The OU doesn’t say that. Given that this is a very basic book on maths, not mentioning that seems wrong. There are 4 books for this maths course, with 3 units in each. I am currently reading unit 1. Differentiation is not introduced until unit 6, integration until unit 8. So it’s a pretty Noddy maths course.

I note that maxima gives lcm(6,4) as 12 and not 24.

2023-02-04, 20:33   #9
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

32×263 Posts

Quote:
 Originally Posted by drkirkby So when computing the common factors of m and n, you have to factorise m and n if they are not prime? The OU doesn’t say that. Given that this is a very basic book on maths, not mentioning that seems wrong. There are 4 books for this maths course, with 3 units in each. I am currently reading unit 1. Differentiation is not introduced until unit 6, integration until unit 8. So it’s a pretty Noddy maths course. I note that maxima gives lcm(6,4) as 12 and not 24.

Factoring m and n could be computationally expensive. It's generally usually much cheaper (for 2 large integers) to find their GCD instead. But, you are still stuck with having to factor their GCD.
ETA: You wouldn't need to factor the GCD, unless you wanted to include the likes of 8 in relation to 4 and 6 (which with this definition you don't).

Last fiddled with by a1call on 2023-02-04 at 20:50

2023-02-04, 20:52   #10
drkirkby

"David Kirkby"
Jan 2021
Althorne, Essex, UK

2·229 Posts

Quote:
 Originally Posted by a1call ETA II: This helped me understand the algorithm: https://multiply.info/CommonMultiple...f-4-and-6.html
Cheers, that is useful. I understand it now. Given this is such a book for such a basic mathematics course- see description at

https://www.open.ac.uk/courses/modules/mst124

I think this should have be explained better.

But thank you both for your help. I have gone from being totally confused to feeling I understand it within a couple of hours!

Last fiddled with by drkirkby on 2023-02-04 at 20:53

 2023-02-04, 21:52 #11 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 32·263 Posts My pleasure drkirkby, anytime. I guess charybdis deserves some credit too.

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