Least common multiple

The least common multiple of two integers a and b is denoted as lcm(a, b). Some older textbooks use [a, b],[2][3] while the programming language J uses a*.b .

LCM(4, 6)

Multiples of 4 are:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, ...

Multiples of 6 are:

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, ...

Common multiples of 4 and 6 are the numbers that are in both lists:

12, 24, 36, 48, 60, 72, ....

So, from this list of the first few common multiples of the numbers 4 and 6, their least common multiple is 12.

Suppose there are two meshing gears in a machine, having m and n teeth, respectively, and the gears are marked by a line segment drawn from the center of the first gear to the center of the second gear. When the gears begin rotating, the number of rotations the first gear must complete to realign the line segment can be calculated by using ${\displaystyle LCM(m,n)}$. The first gear must complete ${\displaystyle LCM(m,n) \over m}$ rotations for the realignment. By that time, the second gear will have made ${\displaystyle LCM(m,n) \over n}$ rotations.

Suppose there are three planets revolving around a star which take l, m and n units of time respectively to complete their orbits. Assume that l, m and n are integers. Assuming the planets started moving around the star after an initial linear alignment, all the planets attain a linear alignment again after ${\displaystyle LCM(l,m,n)}$ units of time. At this time, the first, second and third planet will have completed ${\displaystyle LCM(l,m,n) \over l}$, ${\displaystyle LCM(l,m,n) \over m}$ and ${\displaystyle LCM(l,m,n) \over n}$ orbits respectively around the star.[4]

The following formula reduces the problem of computing the least common multiple to the problem of computing the greatest common divisor (GCD), also known as the greatest common factor:

${\displaystyle \operatorname {lcm} (a,b)={\frac {|a\cdot b|}{\gcd(a,b)}}.}$

This formula is also valid when exactly one of a and b is 0, since gcd(a, 0) = |a|. However, if both a and b are 0, this formula would cause division by zero; lcm(0, 0) = 0 is a special case.

There are fast algorithms for computing the GCD that do not require the numbers to be factored, such as the Euclidean algorithm. To return to the example above,

${\displaystyle \operatorname {lcm} (21,6)={21\cdot 6 \over \gcd(21,6)}={21\cdot 6 \over \gcd(3,6)}={21\cdot 6 \over 3}={\frac {126}{3}}=42.}$

Because gcd(a, b) is a divisor of both a and b, it is more efficient to compute the LCM by dividing before multiplying:

${\displaystyle \operatorname {lcm} (a,b)=\left({|a| \over \gcd(a,b)}\right)\cdot |b|=\left({|b| \over \gcd(a,b)}\right)\cdot |a|.}$

This reduces the size of one input for both the division and the multiplication, and reduces the required storage needed for intermediate results (overflow in the a×b computation). Because gcd(a, b) is a divisor of both a and b, the division is guaranteed to yield an integer, so the intermediate result can be stored in an integer. Done this way, the previous example becomes:

${\displaystyle \operatorname {lcm} (21,6)={21 \over \gcd(21,6)}\cdot 6={21 \over \gcd(3,6)}\cdot 6={21 \over 3}\cdot 6=7\cdot 6=42.}$

The unique factorization theorem indicates that every positive integer greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number.

For example:

${\displaystyle 90=2^{1}\cdot 3^{2}\cdot 5^{1}=2\cdot 3\cdot 3\cdot 5.}$

Here the composite number 90 is made up of one atom of the prime number 2, two atoms of the prime number 3 and one atom of the prime number 5.

This fact can be used to find the LCM of a set of numbers.

Example: lcm(8,9,21)

Factor each number and express it as a product of prime number powers.

${\displaystyle {\begin{aligned}8&=2^{3}\\9&=3^{2}\\21&=3^{1}\cdot 7^{1}\end{aligned}}}$

The lcm will be the product of multiplying the highest power of each prime number together. The highest power of the three prime numbers 2, 3, and 7 is 2³, 3², and 7¹, respectively. Thus,

${\displaystyle \operatorname {lcm} (8,9,21)=2^{3}\cdot 3^{2}\cdot 7^{1}=8\cdot 9\cdot 7=504.}$

This method is not as efficient as reducing to the greatest common divisor, since there is no known general efficient algorithm for integer factorization.

This method can be illustrated with a Venn diagram as follows with the prime factorization of each of the two numbers demonstrated in each circle and all factors they share in common in the intersection. The LCM can be found by multiplying all of the prime numbers in the diagram.

Here is an example:

48 = 2 × 2 × 2 × 2 × 3,

180 = 2 × 2 × 3 × 3 × 5,

sharing two "2"s and a "3" in common:

Least common multiple = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

Greatest common divisor = 2 × 2 × 3 = 12

This also works for the greatest common divisor (GCD), except that instead of multiplying all of the numbers in the Venn diagram, one multiplies only the prime factors that are in the intersection. Thus the GCD of 48 and 180 is 2 × 2 × 3 = 12.

This method works easily for finding the LCM of several integers.

Let there be a finite sequence of positive integers X = (x₁, x₂, ..., x_n), n > 1. The algorithm proceeds in steps as follows: on each step m it examines and updates the sequence X^(m) = (x₁^(m), x₂^(m), ..., x_n^(m)), X⁽¹⁾ = X, where X^(m) is the mth iteration of X, that is, X at step m of the algorithm, etc. The purpose of the examination is to pick the least (perhaps, one of many) element of the sequence X^(m). Assuming x_k0^(m) is the selected element, the sequence X^(m+1) is defined as

x_k^(m+1) = x_k^(m), k ≠ k₀

x_k0^(m+1) = x_k0^(m) + x_k0⁽¹⁾.

In other words, the least element is increased by the corresponding x whereas the rest of the elements pass from X^(m) to X^(m+1) unchanged.

The algorithm stops when all elements in sequence X^(m) are equal. Their common value L is exactly LCM(X).

For example, if X = X⁽¹⁾ = (3, 4, 6), the steps in the algorithm produce:

X⁽²⁾ = (6, 4, 6)

X⁽³⁾ = (6, 8, 6)

X⁽⁴⁾ = (6, 8, 12) - by choosing the second 6

X⁽⁵⁾ = (9, 8, 12)

X⁽⁶⁾ = (9, 12, 12)

X⁽⁷⁾ = (12, 12, 12) so LCM = 12.

This method works for any number of numbers. One begins by listing all of the numbers vertically in a table (in this example 4, 7, 12, 21, and 42):

The process begins by dividing all of the numbers by 2. If 2 divides any of them evenly, write 2 in a new column at the top of the table and the result of division by 2 of each number in the space to the right in this new column. If a number is not evenly divisible, just rewrite the number again. If 2 does not divide evenly into any of the numbers, repeat this procedure with the next largest prime number, 3 (see below).

Now, assuming that 2 did divide at least one number (as in this example), check if 2 divides again:

Once 2 no longer divides any number in the current column, repeat the procedure by dividing by the next larger prime, 3. Once 3 no longer divides, try the next larger primes, 5 then 7, etc. The process ends when all of the numbers have been reduced to 1 (the column under the last prime divisor consists only of 1's).

Now, multiply the numbers in the top row to obtain the LCM. In this case, it is 2 × 2 × 3 × 7 = 84.

As a general computational algorithm, the above is quite inefficient. One would never want to implement it in software: it takes too many steps, and requires too much storage space. A far more efficient numerical algorithm can be obtained by using Euclid's algorithm to compute the gcd first, and then obtaining the lcm by division.

According to the fundamental theorem of arithmetic a positive integer is the product of prime numbers, and, except for their order, this representation is unique:

${\displaystyle n=2^{n_{2}}3^{n_{3}}5^{n_{5}}7^{n_{7}}\cdots =\prod _{p}p^{n_{p}},}$

where the exponents n₂, n₃, ... are non-negative integers; for example, 84 = 2² 3¹ 5⁰ 7¹ 11⁰ 13⁰ ...

Given two positive integers ${\displaystyle \;a=\prod _{p}p^{a_{p}}}$ and ${\displaystyle \;b=\prod _{p}p^{b_{p}}}$ their least common multiple and greatest common divisor are given by the formulas

${\displaystyle \gcd(a,b)=\prod _{p}p^{\min(a_{p},b_{p})}}$

and

${\displaystyle \operatorname {lcm} (a,b)=\prod _{p}p^{\max(a_{p},b_{p})}.}$

Since

${\displaystyle \min(x,y)+\max(x,y)=x+y,}$

this gives

${\displaystyle \gcd(a,b)\operatorname {lcm} (a,b)=a\,b.}$

In fact, every rational number can be written uniquely as the product of primes if negative exponents are allowed. When this is done, the above formulas remain valid. For example:

${\displaystyle {\begin{array}{llll}4=2^{2}3^{0},&6=2^{1}3^{1},&\gcd(4,6)=2^{1}3^{0}=2,&\operatorname {lcm} (4,6)=2^{2}3^{1}=12.\\[8pt]{\tfrac {1}{3}}=2^{0}3^{-1}5^{0},&{\tfrac {2}{5}}=2^{1}3^{0}5^{-1},&\gcd({\tfrac {1}{3}},{\tfrac {2}{5}})=2^{0}3^{-1}5^{-1}={\tfrac {1}{15}},&\operatorname {lcm} ({\tfrac {1}{3}},{\tfrac {2}{5}})=2^{1}3^{0}5^{0}=2,\\[8pt]{\tfrac {1}{6}}=2^{-1}3^{-1},&{\tfrac {3}{4}}=2^{-2}3^{1},&\gcd({\tfrac {1}{6}},{\tfrac {3}{4}})=2^{-2}3^{-1}={\tfrac {1}{12}},&\operatorname {lcm} ({\tfrac {1}{6}},{\tfrac {3}{4}})=2^{-1}3^{1}={\tfrac {3}{2}}.\end{array}}}$

The positive integers may be partially ordered by divisibility: if a divides b (that is, if b is an integer multiple of a) write a ≤ b (or equivalently, b ≥ a). (Forget the usual magnitude-based definition of ≤ in this section - it is not used.)

Under this ordering, the positive integers become a lattice with meet given by the gcd and join given by the lcm. The proof is straightforward, if a bit tedious; it amounts to checking that lcm and gcd satisfy the axioms for meet and join. Putting the lcm and gcd into this more general context establishes a duality between them:

If a formula involving integer variables, gcd, lcm, ≤ and ≥ is true, then the formula obtained by switching gcd with lcm and switching ≥ with ≤ is also true. (Remember ≤ is defined as divides).

The following pairs of dual formulas are special cases of general lattice-theoretic identities.

It can also be shown[5] that this lattice is distributive; that is, lcm distributes over gcd and gcd distributes over lcm:

${\displaystyle \operatorname {lcm} (a,\gcd(b,c))=\gcd(\operatorname {lcm} (a,b),\operatorname {lcm} (a,c)),}$

${\displaystyle \gcd(a,\operatorname {lcm} (b,c))=\operatorname {lcm} (\gcd(a,b),\gcd(a,c)).}$

This identity is self-dual:

${\displaystyle \gcd(\operatorname {lcm} (a,b),\operatorname {lcm} (b,c),\operatorname {lcm} (a,c))=\operatorname {lcm} (\gcd(a,b),\gcd(b,c),\gcd(a,c)).}$

• Let D be the product of ω(D) distinct prime numbers (that is, D is squarefree).

Then[6]

${\displaystyle |\{(x,y)\;:\;\operatorname {lcm} (x,y)=D\}|=3^{\omega (D)},}$

where the absolute bars || denote the cardinality of a set.

• If none of ${\displaystyle a_{1},a_{2},\ldots ,a_{r}}$ is zero, then

${\displaystyle \operatorname {lcm} (a_{1},a_{2},\ldots ,a_{r})=\operatorname {lcm} (\operatorname {lcm} (a_{1},a_{2},\ldots ,a_{r-1}),a_{r}).}$[7][8]