Question

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm, and 45 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

Answer 1

Hint:
Here, we need to find the minimum distance each should walk so that each can cover the same distance in complete steps. We will write the given numbers as a product of their prime factors. Then, you will calculate the L.C.M. of the three numbers. The lowest common multiple is the product of the prime factors with the greatest powers.

Complete step by step solution:
The minimum distance each should walk is the lowest common multiple of their distances covered in each step.
Thus, the minimum distance each should walk so that each can cover the same distance in complete steps is the L.C.M. of the distances 40 cm, 42 cm, and 45 cm.
We will find the L.C.M. of 40, 42, 45 using the fundamental theorem of arithmetic.
First, we will write the given numbers as a product of their prime factors.
We know that 40 is the product of 8 and 5.
Therefore, we can write 40 as
$$40=8\times 5$$
8 is the cube of the prime number 2. Thus, we get
$$\Rightarrow 40=2^3\times 5$$
Now, we know that 42 is the product of 6 and 7.
Therefore, we can write 42 as
$$42=6\times 7$$
6 is the product of the prime numbers 2 and 3. Thus, we get
$$\Rightarrow 42=2\times 3\times 7$$
Next, we know that 45 is the product of 9 and 5.
Therefore, we can write 45 as
$$45=9\times 5$$
9 is the square of the prime number 3. Thus, we get
$$\Rightarrow 45=3^2\times 5$$
Therefore, we have
$$40=2^3\times 5$$
$$42=2\times 3\times 7$$
$$45=3^2\times 5$$
Now, in the product of primes, we can observe that the greatest power of 2 is 3, greatest power of 3 is 2, greatest power of 5 is 1, and the greatest power of 7 is 1.
Thus, the prime factors with the greatest powers are $$2^3$$, $$3^2$$, 5, and 7.
The lowest common multiple of the numbers 40, 42, 45 is the product of the prime factors with the greatest powers.
Therefore, we get
$$L.C.M.=2^3\times 3^2\times 5\times 7$$
Simplifying the expression, we get
$$\begin{array}{l}\Rightarrow L.C.M.=8\times 9\times 5\times 7\\ \Rightarrow L.C.M.=72\times 35\end{array}$$
Multiplying the terms, we get
$$\Rightarrow L.C.M.=2520$$
$$\therefore$$ The L.C.M. of 40, 42, 45 is 2520.

Thus, the minimum distance each should walk so that each can cover the same distance in complete steps is 2520 cm.

Note:
Here, we need to remember that all the prime factors with the greatest powers are selected, irrespective of whether that power appears in the prime factorization of all the three numbers 24, 60, and 150. For example, $$2^3$$ does not appear in the prime factorization of 60 and 150. But it should be included while calculating L.C.M. because it has the highest power. Another common mistake is to use the common factors with the lowest powers to calculate the L.C.M. That is incorrect because it will give you the H.C.F. and not the L.C.M. of the numbers.
We used the fundamental theorem of arithmetic in the solution. The fundamental theorem of arithmetic states that every composite number can be written as a product of its prime factors in a unique way. A prime factor is a factor of a number which is divisible by 1 and itself only.