The GCD of three numbers are 3 and 1008 respectively. If two numbers are 48 and 72, find the least possible value of the third number?

Answer 1

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Let the third number be x.

Given, GCD of 48, 72 and x is 3. So we can write 48 as 3 * 2^4 and 72 as 3^2 * 2^3.

As GCD of 48, 72 and x is 3, the prime factorization of x must have at least one factor of 3 and no factors of 2^4 or 3^2.

Hence, the prime factorization of x can be written as 3^n * p, where p is a product of primes other than 2 and 3, and n is a non-negative integer.

Now, GCD of x and 1008 is 3. So, 1008 can be written as 2^4 * 3^2 * 7.

Since the GCD of x and 1008 is 3, the prime factorization of x must have at least one factor of 3 and no factors of 2^5, 3^3 or 7.

Hence, the prime factorization of x can be written as 3^1 * p, where p is a product of primes other than 2, 3 and 7.

So, x must be a multiple of 3, but cannot have any factors of 2, 3^2 or 7.

Therefore, the least possible value of x is 3 * 5 = 15 (where p = 5, since it is a product of primes other than 2, 3 and 7).

Hence, the least possible value of the third number is 15.