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Three friends plan to go for a morning walk. They step off together and their steps measures 48 cm, 52 cm and 56 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times?
Three friends plan to go for a morning walk. They step off together and their steps measures 48 cm, 52 cm and 56 cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ten times?
This Question has 1 answers.
The three friends take steps of lengths $48$ cm, $52$ cm, and $56$ cm respectively.
To find the minimum distance each should walk so that they cover the same distance in complete steps, we need to find the Least Common Multiple (LCM) of $48$, $52$, and $56$.
First, we find the prime factorization:
$48 = 2^4 \times 3$
$52 = 2^2 \times 13$
$56 = 2^3 \times 7$
The LCM is given by taking the highest powers of all prime factors:
$LCM(48, 52, 56) = 2^4 \times 3 \times 7 \times 13$
Calculating step-by-step:
$2^4 = 16$
$16 \times 3 = 48$
$48 \times 7 = 336$
$336 \times 13 = 4368$
Thus, the minimum distance each should walk is $4368$ cm.
Since they need to cover this distance ten times, the total distance each friend should walk is:
$10 \times 4368 = 43680$ cm
To find the minimum distance each should walk so that they cover the same distance in complete steps, we need to find the Least Common Multiple (LCM) of $48$, $52$, and $56$.
First, we find the prime factorization:
$48 = 2^4 \times 3$
$52 = 2^2 \times 13$
$56 = 2^3 \times 7$
The LCM is given by taking the highest powers of all prime factors:
$LCM(48, 52, 56) = 2^4 \times 3 \times 7 \times 13$
Calculating step-by-step:
$2^4 = 16$
$16 \times 3 = 48$
$48 \times 7 = 336$
$336 \times 13 = 4368$
Thus, the minimum distance each should walk is $4368$ cm.
Since they need to cover this distance ten times, the total distance each friend should walk is:
$10 \times 4368 = 43680$ cm
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