Least common multiple for 3 or more numbers?

You can compute the LCM of more than two numbers by iteratively computing the LCM of two numbers, i.e.

You can compute the LCM of more than two numbers by iteratively computing the LCM of two numbers, i.e. Lcm(a,b,c) = lcm(a,lcm(b,c)).

Thanks. That done it for me – paan Sep 29 '08 at 4:53 1 Ooooh textbook recursion :) – Peter Wone Sep 30 '08 at 13:24.

In Python (modified primes. Py): def gcd(a, b): """Return greatest common divisor using Euclid's Algorithm. """ while b: a, be = b, a % be return a def lcm(a, b): """Return lowest common multiple.""" return a * be // gcd(a, b) def lcmm(*args): """Return lcm of args.

""" return reduce(lcm, args) Usage: >>> lcmm(100, 23, 98) 112700 >>> lcmm(*range(1, 20)) 232792560 reduce() works something like that: >>> f = lambda a,b: "f(%s,%s)" % (a,b) >>> print reduce(f, "abcd") f(f(f(a,b),c),d).

– paan Sep 29 '08 at 4:49 3 Given a function f and a list l = a,b,c,d, reduce(f,l) returns f(f(f(a,b),c),d). It's the functional implementation of "lcm can be computed by iteratively computing the lcm of the current value and the next element of the list." – A. Rex Sep 29 '08 at 4:53 +1 for showing a solution that can adapt to more than three parameters – OnesimusUnbound Aug 4 '11 at 14:26.

Here's an ECMA-style implementation: function gcd(a, b){ // Euclidean algorithm var t; while (b! = 0){ t = b; be = a % b; a = t; } return a; } function lcm(a, b){ return (a * be / gcd(a, b)); } function lcmm(args){ // Recursively iterate through pairs of arguments // i.e. Lcm(args0, lcm(args1, lcm(args2, args3))) if(args.

Length == 2){ return lcm(args0, args1); } else { var arg0 = args0; args.shift(); return lcm(arg0, lcmm(args)); } }.

It feels bad that I don't understand what you mean by "ECMA-style" =/ – freitass yesterday en.wikipedia.org/wiki/ECMAScript – Virgil Disgr4ce 2 hours ago.

I just figured this out in Haskell: lcm' :: Integral a => a -> a -> a lcm' a be = a`div`(gcd a b) * be lcm :: Integral a => a -> a lcm (n:ns) = foldr lcm' n ns I even took the time to write my own gcd function, only to find it in Prelude! Lots of learning for me today :D.

You can do it another way - Let there be n numbers. Take a pair of consecutive numbers and save its lcm in another array. Doing this at first iteration program does n/2 iterations.

Then next pick up pair starting from 0 like (0,1) , (2,3) and so on. Compute their LCM and store in another array. Do this until you are left with one array.

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