Can LCM of n numbers be obtained by taking LCM of 2 numbers and then taking LCM of this resulting LCM with the next number in the queue?

Question

Can LCM of n numbers be obtained by taking LCM of 2 numbers and then taking LCM of this resulting LCM and the next number in the queue until all the numbers are exhausted by repeating this process?

In other words, Is the following formula valid?

LCM of (a,b,c,d,....) = LCM of ((LCM of ((LCM of (a,b)),c)),d)...

Yes, it is the case. Have you tried to attempt a proof? – Sarvesh Ravichandran Iyer Feb 06 '20 at 13:07 — Sarvesh Ravichandran Iyer, Feb 06 '20 at 13:07

score 1 · Accepted Answer · answered Feb 06 '20 at 13:54

Taking the lowest common multiple corresponds to taking the maximum of the exponents for each prime in the prime factorizations of the numbers. The maximum operation satisfies

$$ \max(a,b,c,d,\ldots)=\ldots\max(\max(\max(a,b),c),d),\ldots\;, $$

and the lower common multiple operation inherits this property.

Note that taking the greatest common divisor corresponds to taking the minimum of the exponents, and the minimum operation also has this property, so the greatest common divisor operation also inherits it.

Can LCM of n numbers be obtained by taking LCM of 2 numbers and then taking LCM of this resulting LCM with the next number in the queue?

1 Answers1