两个以上数字的最小公倍数

Question

I want to find out Least Common Multiple(LCM) of more than two numbers. I know the formula of lcm(a,b) = (a * b) / gcd(a,b) . Let's say, I have an array of numbers: [2, 6, 8, 13] and the the lcm should be modulo M = 1000000007.

I have seen below code to calculate the LCM of multiple numbers but I am not understanding how the calculation is going on with both the loops.

int arr[] = {2, 6, 8, 13}, n = 4
long long int ans=1;
long long int M=1000000007;
for(int i=0;i<n;i++)   // Calculating LCM
{
    for(int j=i+1;j<n;j++)
    {
        arr[j]=arr[j]/__gcd(arr[i],arr[j]);
    }
    ans=((ans%M)*(arr[i]%M))%M;
}
return (ans)%M;

Can anyone please help me to understand the LCM calculation in the above code?

Answer 1

Knowing that gcd(a, b) represents the product of all the prime factors shared by a and b , what is the significance of a / gcd(a,b) ?

a / gcd(a, b) is equal to the prime factors of a that are not in b .

Therefore, when you multiple that quantity by b , you get a product of all the prime factors of b and all the prime factors of a that are not in b . This is precisely lcm(a, b) .

Let's extend that to an arbitrary number of integers.

The lcm(a, b) is a times all the prime factors of b not in a or:

a * (b / gcd(a, b)) = (a * b) / gcd(a, b)

Easy enough, you knew that already.

But if we have a third number, lcm(a, b, c) is a times all the prime factors of b not in a times all the prime factors of c in neither a nor b . Well, the first part is straight forward, it's the same as above:

lcm(a, b, c) = lcm(a, b) * (all the prime factors of c in neither a nor b)

How to calculate all the prime factors of c in neither a nor b might not be obvious at first, but it's not overly complicated:

all the prime factors of c in neither a nor b = c / (gcd(a, c) * gcd(b, c))

Which means that

lcm(a, b, c) = lcm(a, b) * c / (gcd(a, c) * gcd(b, c))
lcm(a, b, c) = (a * b * c) / (gcd(a, b) * gcd(a, c) * gcd(b, c))

And now, you can generalize easily:

lcm(a[0], ..., a[N]) = prod(a[0], ..., a[N]) / pairwise_gcd(a[0], ..., a[N])

But a more relevant formulation is the recursive version:

lcm(a[0], ..., a[N]) = lcm(a[0], ..., a[N-1]) * a[N] / (gcd(a[0], a[N]) * ... * gcd(a[N-1], a[N]))

Or:

lcm(a[0], ..., a[N]) = a[0] * lcm(a[1] / gcd(a[0], a[1]), ..., a[N] / gcd(a[0], a[N]))

Here's an attempt at translating your code snippet to psuedocode

Compare this to the last definition of lcm on an array, I tried to make them appear similar.

given int array = arrayOfNums
int product := 1
for number in arrayOfNums
    remove all prime factors of number from all subsequent array elements
    product = product * number
product is now the lcm of arrayOfNums

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Hopefully, that wasn't too confusing; I admit it may not be much of an explanation, but it is a starting point. Please let me know if anything is still unclear.