# Prove gcf times the lcm of any two numbers equals the products of those two numbers

Given two numbers, l and m, prove that:

gcf(l, m)lcm(l, m) = lm

Note by Varun Iyer
4 years, 9 months ago

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Assuming that the prime factorisations of $$l$$ and $$m$$ are as follow,

$l=2^{a_1}\times3^{a_2}\times...\times P^{a_n}$

$m=2^{b_1}\times3^{b_2}\times...\times P^{b_n}$

where $$\left \{ a_n \right \}\in \mathbb{N}\cup \left \{ 0 \right \}$$, $$\left \{ b_n \right \}\in \mathbb{N}\cup \left \{ 0 \right \}$$ and $$P$$ is a prime.

WLOG, assuming that $$a_1\leq b_1, a_2\leq b_2,...,a_n\leq b_n$$,

The $$\operatorname{lcm}(l,m)$$ can be obtained by choosing the highest power of each prime factor from either of the numbers. $\operatorname{lcm} (l,m)=2^{b_1}\times 3^{b_2}\times...P^{b_n}$

On the other hand, the $$\gcd (l,m)$$ is obtained by choosing the smallest power of each prime factor from either of the numbers.

$\gcd (l,m)=2^{a_1}\times3^{a_2}\times...\times P^{a_n}$

Notice that

$\gcd (l,m)\times \operatorname {lcm}(l,m)= 2^{a_1+b_1}\times 3^{a_2+b_2}\times ... \times P^{a_n+b_n}$

which is equivalent to

$l \times m=2^{a_1+b_1}\times 3^{a_2+b_2}\times...\times P^{a_n+b_n}$

$\therefore\gcd(l,m)\times \operatorname {lcm}(l,m)=l\times m$

I may have made some mistakes in my proof. Please do correct me if you spotted any.

- 4 years, 9 months ago

You can refer to the GCD/LCM writeup in the Olympiad section.

This is a basic interesting fact which relates these two.

Is there a similar equation for the three variable case? Why, or why not?

Staff - 4 years, 9 months ago

See http://www.proofwiki.org/wiki/ProductofGCDandLCM for a much neater, albeit trickier, proof of the above.

- 4 years, 9 months ago