Carcass wrote:

\(x^2\) is divisible by both 40 and 75. If x has exactly three distinct prime factors, which of the following could be the value of x?

Indicate

all values that apply.

❑ 30

❑ 60

❑ 200

❑ 240

❑ 420

\(40 = 2*2*2*5\) and \(75 = 3*5*5\)

For \(x^2\) to be divisible by 40 and 75, its prime-factorization must include at least three 2's (since there are three 2's within 40), at least one 3 (since

there is one 3 within 75), and at least two 5's (since there are two 5's within 75):

\(2^3 * 3^1 * 5^2\)

However, since \(x^2\) is a perfect square, its prime-factorization must have an EVEN NUMBER of every prime factor.

Since the prime-factorization of x must include \(2^3\), \(3^1\) and \(5^2\) -- but \(x\) must have an even number of each of these prime factors -- the least possible option for \(x^2\) is as follows:

\(2^4 * 3^2 * 5^2\)

Since the least possible option for \(x^2 = 2^4 * 3^2 * 5^2\), the least possible option for \(x = 2^2 * 3 * 5 = 60\).

Implication:

\(x\) must be a MULTIPLE OF 60.

In addtion, since \(x\) must have exactly three distinct prime factors, it cannot be divisible by any prime number other than 2, 3 and 5.

Since 30 and 200 are not divisible by 60, eliminate A and C.

Since 420 is divisible by 7 -- a prime number other than 2, 3 and 5 -- eliminate E.

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