Question

Find the L.C.M of m, 2m, 3m, 4m and 5m where m is any positive integer.

Answer 1

HEY!!

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➡️The given terms can be prime factorize as, 

⏺️m = 1×m

⏺️2m=2×m

⏺️3m=3×m

⏺️4m=2×2×m

⏺️5m=5×m

⏩So LCM can be given as =  1×2× 3×5×m=30m

Answer 2

Answer:  The required L.C.M. of the given numbers is 60m,

Step-by-step explanation:  We are given to find the L.C.M. of m, 2m, 3m, 4m and 5m, where m is any positive integer.

We know that

L.C.M. stands for the least common multiple.

Since m is common in all the five numbers, so we get

$L.C.M.(m,2m,3m,4m,5m)\\\\=m\times L.C.M.(1,2,3,4,5)\\\\=m\times 12\times5\\\\=60m.$

Thus, the required L.C.M. of the given numbers is 60m,