Question

Four bells are ringing at intervals $$12,16,24$$ and $$36$$ minutes. They start ringing simultaneously at $$12$$o’clock. Find when they will ring again together? What is the L.C.M. of the ringing of the bells?

Answer 1

Hint: Observe that the first, second, third and fourth bell will ring when multiples of $$12,16,24$$ and $$36$$ have passed from the time of ringing of the first bell. This time interval can be found from the L.C.M. of these numbers. So for the second time also the bells will ring after a time interval equal to the L.C.M. after the first time they ring together.

Complete step by step solution:
The first time the bells will ring together is when the L.C.M. of $$12,16,24$$ and $$36$$ passed after the ringing of the first bell. It is given that they ring together for the first time at $$12$$ o’clock. So for the second time they will ring at $$12+L.C.M.$$ hrs.
First find the L.C.M. of $$12,16,24$$ and $$36$$:
We find the L.C.M. by tabular method,
$$\begin{array}{c}2|\underset{―}{12,16,24,36}\\ 2|\underset{―}{06,08,12,18}\\ 3|\underset{―}{03,04,06,09}\\ 2|\underset{―}{01,04,02,03}\\ 01,02,01,03\end{array}$$
$$\therefore$$ L.C.M. $$=$$ $$2\times 2\times 2\times 2\times 3\times 3$$
                 $$=144$$
Hence the time when they will ring together for the second time is $$12:00+144$$mins
Now $$144$$ minutes $$=2$$ hrs $$24$$ mins. [$$1$$ hr $$=$$ $$60$$ mins.]
$$\therefore$$ Required time $$=$$ $$12:00+2:24$$ $$=$$ $$14:24$$ $$=2:24$$ p.m.
Hence, the bells will ring together again at $$2:24$$ p.m.

L.C.M. of the intervals of the ringing of the bells is $$144$$minutes.

Note:
L.C.M. means the lowest common multiple. Alternatively the L.C.M. of the numbers can also be found by the prime factorization method. In the prime factorization method first we need to find the prime factors of each number. Then the L.C.M. will be the product of all prime factors with the highest degree (power).