Question

A=2^3×3 b=2×3×5 c=3^n×5 and lcm (abc) = 2^3×^2×5 then find n

Answer 1

Given,

a = 2ˆ3 x 3
=> a = 2ˆ3 x 3ˆ1 x 5ˆ0 ………………….(1)

b = 2 x 3 x 5
=> b = 2ˆ1 x 3ˆ1 x 5ˆ1 ……………………(2)

c = 3ˆn x 5
=> c = 2ˆ0 x 3ˆn x 5ˆ1 …………………….(3)

LCM(a,b,c) = 2ˆ3 x 3ˆ2 x 5ˆ1 ………….(4)

Now taking highest exponents for each of the primes in equation (1), (2) and (3), the LCM can be written as -

LCM(a,b,c) = 2ˆ3 x 3ˆn x 5^1

comparing this with equation (4) we get,

n = 2.