Question

If a $a= 2^{3} * 3,b=2*3*5,c= 3^{n} *5$ and LCM (a,b,c)= $a= 2^{3} * 3^{2}*5$, then n=
(a) 1
(b) 2
(c) 3
(d) 4

Answer 1

SOLUTION :  

Option (b) is Correct : 2

Given : a = 2³ × 3 , b = 2 × 3 × 5 , c = 3ⁿ × 5 &

LCM (a,b,c) = 2³ × 3² × 5¹ ………….. (1)

Factors of a,b,c are as follows :  

a = 2³ × 3¹  

b = 2¹ × 3¹ × 5¹  

c = 3ⁿ × 5¹

LCM(a,b,c) = 2³ × 3ⁿ × 5¹ ……………(2)

In Comparing eq 1 & eq 2,  

2³ × 3² × 5¹ = 2³ × 3ⁿ × 5¹

3² = 3ⁿ

n = 2  

Hence, the value of n = 2.

★★LCM = LCM of two or more numbers =  product of the greatest power of each common prime factor involved in the numbers with highest power.

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Answer 2

Answer:

2

Step-by-step explanation:

Given:

⇒ a = 2³ * 3.

⇒ b = 2 * 3 * 5

⇒ c = 3ⁿ * 5.

LCM(a,b,c) = 2³ * 3² * 5.

On comparing both, we get

⇒ 3ⁿ = 3²

⇒ n = 2.

Therefore, the value of n = 2.

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