1. LCM involving algebra

Three lighthouses flash at intervals of 12,45 and x minutes respectively. If all of them flash at the same time at 12 hours interval, find the smallest / largest possible of x.

I have prime factorized the numbers - please advise how to proceed??
12 = 2^2 x 3
45 = 3^2 x 5
12 hours = 720 mins = 2^4 x 3^2 x 5

2. Re: LCM involving algebra

Hi Drdj,
Firstly lets observe the numbers
$\displaystyle 12 = 2^2.3$
$\displaystyle 45 = 3^2.5$
$\displaystyle 720 = 2^4.3^2.5$
so we observe that the LCM lacks a term that supplies $\displaystyle 2^4$ i.e if had taken the LCM of just 12,45 we would have ended up in $\displaystyle 180 = 2^2.3^2.5$
so the least value of x = 16 in that case we have all the factors we would need.

Now other feasible values of x are those that have $\displaystyle 2^4$ along with other factors like $\displaystyle 48 = 2^4.3$, $\displaystyle 90 = 2^4.5$,....,and last(and definitely not the least ) 720 itself so the maximum values of x = 720.

Kalyan.

3. Re: LCM involving algebra

Originally Posted by kalyanram
Hi Drdj,
Firstly lets observe the numbers
$\displaystyle 12 = 2^2.3$
$\displaystyle 45 = 3^2.5$
$\displaystyle 720 = 2^4.3^2.5$
so we observe that the LCM lacks a term that supplies $\displaystyle 2^4$ i.e if had taken the LCM of just 12,45 we would have ended up in $\displaystyle 180 = 2^2.3^2.5$
so the least value of x = 16 in that case we have all the factors we would need.

Now other feasible values of x are those that have $\displaystyle 2^4$ along with other factors like $\displaystyle 48 = 2^4.3$, $\displaystyle 90 = 2^4.5$,....,and last(and definitely not the least ) 720 itself so the maximum values of x = 720.

Kalyan.
Just a question.. why the minimum value of x can't be be 1?

4. Re: LCM involving algebra

Well take $\displaystyle x = 1$ and compute the LCM. You will see $\displaystyle lcm(12,45,1) = 180$ and not $\displaystyle 720$ which is actually required.

Kalyan.