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# Three boys step off together from the same spot. Their steps measure $ 63 \mathrm{~cm}, 70 \mathrm{~cm} $ and $ 77 \mathrm{~cm} $ respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Given:

Three boys step off together from the same spot. Their steps measure \( 63 \mathrm{~cm}, 70 \mathrm{~cm} \) and \( 77 \mathrm{~cm} \) respectively.

To do:

We have to find the minimum distance each should cover so that all can cover the distance in complete steps.

Solution:

Required distance will be the LCM of the measure of steps of each boy.

__Calculating the LCM of 63, 70 and 77 using prime factorization method__:

Writing the numbers as a product of their prime factors:

Prime factorisation of 63:

- $63=3\ \times\ 3\ \times\ 7\ =\ 3^2\ \times\ 7^1$

Prime factorisation of 70:

- $70=2\ \times\ 5\ \times 7=\ 2^1\ \times\ 5^1\ \times\ 7^1$

Prime factorisation of 77:

- $77=7\ \times\ 11\=\ 7^1\ \times\ 11^1$

Multiplying the highest power of each prime number:

- $2^1\ \times\ 3^2\ \times\ 5^1\ \times 7^1\ \times 11^1=\ 6930$

LCM(63, 70, 77) $=$ 6930

Which means required distance $=$ 6930 cm

$=$ 69 m 30 cm (as, 100 cm $=$ 1 m)

So, the minimum distance each should walk so that each can cover the distance in complete steps is 6930 cm or 69 m 30 cm.

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