Find the smallest value $n$ such that the $LCM$ of $n$ and $6$ is $24$

Question

How do i find this through prime factorisation? There's a tip that says to work backwards (It's in Sec one tb) Answer is 8 but im not sure about steps.

Answer 1

If your goal is to find this through a prime factorization then find the prime factorization for the numbers you know so far: $$6=2 \cdot 3, 24 = 2^3 \cdot 3.$$ Now, the least common multiple of $6$ and some number $n$ needs to yield a prime factorization of $2^3 \cdot 3$. Note that the prime factorization of the least common multiple between two numbers will be the largest exponent of each prime between the two numbers. So, in this case, we need $n$ to have 3 factors of 2 and up to 1 factor of 3.

This implies that there are 2 such integers $n$ that will work. Namely, 8 and 24. The smaller of these two is 8.

Answer 2

Since $24$ is the $LCM$ of $6$ and $n$, $24$ is a multiple of $n$. Now:

1. $\operatorname{LCM}(6,1)=6$;
2. $\operatorname{LCM}(6,2)=6$;
3. $\operatorname{LCM}(6,3)=6$;
4. $\operatorname{LCM}(6,4)=12$;
5. $\operatorname{LCM}(6,6)=6$;
6. $\operatorname{LCM}(6,8)=24$.

So, $n=8$.

Answer 3

For any two integers $n,m$, we have $\mbox{lcm}(n,m)=\frac{nm}{\mbox{gcd}(n,m)}$ where gcd is the greatest common divisor.

Since $6=2 \cdot 3$ and $24=2^3\cdot 3$, we get $$2^3\cdot3=\frac{2\cdot3 \times n}{\mbox{gcd}(6,n)}$$

It is easy to see that $n$ must have a factor of $2^2$ so that the RHS ($6n$) is at least the LHS ($24$). But this means $\mbox{gcd}(6,n)$ has a factor of $2$, so that $n$ must have a factor of $2^3$ instead so that $\frac{2\cdot3 \times n}{\mbox{gcd}(6,n)}$ is at least $24$.

Out of all positive integers which have a factor of $2^3, n=2^3=8$ is indeed the smallest, and indeed $\mbox{lcm}(6,8)=24$