In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?
Given: Steps of three persons measure 80 cm, 85 cm and 90 cm respectively.
To find: Here we have to find the minimum distance each should walk so that he can cover the distance in complete steps.
Solution:
Required distance will be the LCM of the measure of steps of each person.
Calculating the LCM of 80, 85 and 90 using prime factorization method:
Writing the numbers as a product of their prime factors:
Prime factorisation of 80:
- $2\ \times\ 2\ \times\ 2\ \times\ 2\ \times\ 5\ =\ 2^4\ \times\ 5^1$
Prime factorisation of 85:
- $5\ \times\ 17\ =\ 5^1\ \times\ 17^1$
Prime factorisation of 90:
- $2\ \times\ 3\ \times\ 3\ \times\ 5\ =\ 2^1\ \times\ 3^2\ \times\ 5^1$
Multiplying the highest power of each prime number:
- $2^4\ \times\ 5^1\ \times\ 17^1\ \times\ 3^2\ =\ 12240$
LCM(80, 85, 90) $=$ 12240
Which means required distance $=$ 12240 cm $=$ 122 m 40 cm (as, 100 cm $=$ 1 m)
So, the minimum distance each should walk so that he can cover the distance in complete steps is 12240 cm or 122 m 40 cm.
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