The LCM of 48,72 and another number, x, is 576 . Which of the values given below can be the maximum value of x ?

Correct option is B

Given:

LCM of 48, 72, and x = 576.

We need to find the maximum value of x such that the LCM of 48, 72, and x remains 576.

Concept Used:

The LCM of a set of numbers is determined by the highest powers of all prime factors involved in those numbers. To ensure the LCM of 48, 72, and x equals 576, x must be a number that complements the prime factors of 48 and 72 without exceeding the powers in 576.

Solution: ​

Prime factorization of the given numbers:
​$$48 = 2^4 × 3 \\\ \\ 72 = 2^3 × 3^2 \\\ \\ 576 = 2^6 × 3^2$$​​
The LCM is determined by the highest powers of the prime factors:$$2^6 × 3^2.$$​

The value of x must complement 48 and 72 to ensure the LCM remains 576.

If x introduces factors of $$2^k × 3^m:$$​​
- The power of 2 in x must not exceed 6.
- The power of 3 in x must not exceed 2.
x must also be a divisor of 576 to ensure the LCM condition is satisfied.

Testing the options:

1. Option A (x = 288): Prime factorization = $$2^5 × 3^2$$​. LCM of 48, 72, and 288 = 576. Valid.

But , LCM of 48 , 72 , 288= 288 so 288 is not correct because LCM of the numbers = 576
2. Option B (x = 192): Prime factorization = $$2^6 × 3.$$​ LCM of 48, 72, and 192 = 576. Valid and higher than 144.
3. Option C (x = 144): Prime factorization = $$2^4 × 3^2$$​. LCM of 48, 72, and 144 = 576. Valid but less than 192.
4. Option D (x = 96): Prime factorization = $$2^5 × 3$$​. LCM of 48, 72, and 96 = 576. Valid but less than 192.

The maximum value of x that satisfies the condition is Option B: 192.