Find the smallest even value of $m$ so that $\mathrm{lcm}(48,180,m)=2160.$

Question

Find the smallest even value of $m$ so that $\mathrm{lcm}(48,180,m)=2160.$

Can anyone solve this? It's similar to the one before but I can't seem to get the point.

Answer 1

Hint: Write the prime factorization of each of the numbers in the problem.

Answer 2

Note that $$48=2^4\cdot 3^1,\quad 180=2^2\cdot 3^2\cdot 5^1,\quad 2160=2^4\cdot 3^3\cdot 5^1.$$ Hence, if $m=2^a\cdot 3^b\cdot 5^c$ with $a,b,c\geq 0$ (no prime $p>5$ divides $m$, why?) $$\mbox{lcm}(48,180,m)=2^{\max(4,a)}\cdot 3^{\max(2,b)}\cdot 5^{\max(1,c)}=2^4\cdot 3^3\cdot 5^1= 2160.$$ Can you take it from here?

Answer 3

You can use this $$lcm(a,b,c) = lcm(lcm(a,b),c).$$

So $$2160 = lcm(lcm(48,180),m) = lcm(720,m)$$

Now, you can see as follows: $2160 = 2^43^35$, and $720=2^43^25$, then the smallest even number $m$ is $2\times 3^3=54$.

Answer 4

$2160=2^4\times 3^3\times 5$

$48=2^4\times 3$ and $180=2^2\times 3^2 \times 5$

To have $\text{LCM}(48,180,x)=2160$ you just need $x=3^3=27$

As the request is for an even number you must take $2\times 27=54$

Hope this helps