# LCM(a,b,c) and abc

#### Hartlw

Theorem: LCM(a,b,c)·F=abc
F=GCD(a,b)·GCD(ab/GCD(a,b),c)

Proof:
1) LCM(a,b)·GCD(a,b)=ab
2) LCM(a,b,c)=LCM(LCM(a,b),c) or see below
3) Let H=gcd(a,b), then
4) LCM(a,b)=ab/H
4) LCM(a,b,c)=LCM(ab/H,c)
5) LCM(a,b,c)=[abc/H]/GCD(ab/H,c)
6) LCM(a,b,c)·H·GCD(ab/H,c)=abc,

EXAMPLE
LCM(4,6,15)x2xGCD(2x6,15)=4x6x15
LCM(4,6,15)x2x3=4x6x15
LCM(4,6,15)=60
You get same answer for any order of a,b,c
---------------------------------------------------------
LCM(a,b,c)=LCM(LCM(a,b),c), by prime factorization which is easy to see for specific cases but awkard to type in the general case, or:
7) Let Q=LCM(LCM(a,b),c)
8) a|LCM(a,b), LCM(a,b)|Q, → a|Q
9) b|LCM(a,b), LCM(a,b)|Q, → b|Q
10) c|Q

So Q is a common multiple of (a,b,c). Now you have to show Q divides every common multiple of (a,b,c). Then Q =LCM(a,b,c)

CM(a,b,c) =ra=sb=tc
ra=sb =CM(a,b)
CM(a,b,c)=CM(a,b)=tc
Q|CM(a,b), Q|c → Q|CM(a,b,c)

The beauty of the proof is its easy to type.

#### HallsofIvy

MHF Helper
Theorem: LCM(a,b,c)·F=abc
F=GCD(a,b)·GCD(ab/GCD(a,b),c)
You can't prove this, it is not true.
Counter-example: a= 6, b= 8, c= 12. LCM(6, 8, 12)= 2 and 6(8)(12)= 576 so F= 576/2= 488.
But GCD(6, 8)= 2 and (6)(8)/GCD(6, 8)= 48/2= 24. GCD(24, 12)= 12. There product is 24, not 488.

Proof:
1) LCM(a,b)·GCD(a,b)=ab
2) LCM(a,b,c)=LCM(LCM(a,b),c) or see below
3) Let H=gcd(a,b), then
4) LCM(a,b)=ab/H
4) LCM(a,b,c)=LCM(ab/H,c)
5) LCM(a,b,c)=[abc/H]/GCD(ab/H,c)
6) LCM(a,b,c)·H·GCD(ab/H,c)=abc,

EXAMPLE
LCM(4,6,15)x2xGCD(2x6,15)=4x6x15
LCM(4,6,15)x2x3=4x6x15
LCM(4,6,15)=60
You get same answer for any order of a,b,c
---------------------------------------------------------
LCM(a,b,c)=LCM(LCM(a,b),c), by prime factorization which is easy to see for specific cases but awkard to type in the general case, or:
7) Let Q=LCM(LCM(a,b),c)
8) a|LCM(a,b), LCM(a,b)|Q, → a|Q
9) b|LCM(a,b), LCM(a,b)|Q, → b|Q
10) c|Q

So Q is a common multiple of (a,b,c). Now you have to show Q divides every common multiple of (a,b,c). Then Q =LCM(a,b,c)

CM(a,b,c) =ra=sb=tc
ra=sb =CM(a,b)
CM(a,b,c)=CM(a,b)=tc
Q|CM(a,b), Q|c → Q|CM(a,b,c)

The beauty of the proof is its easy to type.

• 1 person

#### Hartlw

The proof is correct. I tried various a,b,c. There may be a typo, or you made a mistake. Will clear it up tomorrow morning. I do tend to screw up details of typing. Thanks for checking

• 1 person

#### Hartlw

You can't prove this, it is not true.
Counter-example: a= 6, b= 8, c= 12.
Au contraire mon ami.

If (a,b,c) = 6,8,12
GCD(6.8) = 2
F = 2xGCD(6x8/2,12) = 2x12
LCM(6,8,12)x2x12 = 6x8x12
LCM(6,8,12) = 24

It’s really a simple derivation. All you need is LCM(a,b,c)=LCM(LCM(a,b),c), LCM(a,b)xGCD(a,b)=ab, and some simple algebra. It’s probably easier to do yourself than follow mine.

The formula can be recursive:
LCM(a,b,c,d) = LCM(LCM(a,b,c),d), to give
LCM(a,b,c,d)xF = abcd,
and F depends only on GCD’s of two numbers, which are easy to work with, but formula may be a little messy to write out.
So in general,
LCM(a,b,c,d,….)xF=abcd….

#### Hartlw

HallsofIvy, Also note the order of a,b,c doesn’t matter:

Theorem: LCM(a,b,c)•F=abc
F=GCD(a,b)•GCD(ab/GCD(a,b),c)

If (a,b,c) = (12,6,8)
GCD(12,6) = 6
F = 6xGCD(12x6/6,8) = 6x4
LCM(12,6,8)x6x4 = 12x6x8
LCM(12,6,8) = 24

• 1 person