greMS15 wrote:

x² is divisible by both 40 and 75. if x has exactly 3 distinct prime factors, which of the following could be the value of x?

Indicate all values that apply.

A) 30

B) 60

C) 200

D) 240

E) 420

-----ASIDE---------------------

A lot of integer property questions can be solved using prime factorization.

For questions involving divisibility, divisors, factors and multiples, we can say:

If N is divisible by k, then k is "hiding" within the prime factorization of NConsider these examples:

24 is divisible by

3 because 24 = (2)(2)(2)

(3)Likewise, 70 is divisible by

5 because 70 = (2)

(5)(7)

And 112 is divisible by

8 because 112 = (2)

(2)(2)(2)(7)

And 630 is divisible by

15 because 630 = (2)(3)

(3)(5)(7)

-----ONTO THE QUESTION!---------------------

x² is divisible by 4040 = (2)(2)(2)(5)

So, x² = (2)(2)(2)(5)(?)(?)(?)...

Note: the (?)'s represent additional prime factors that COULD be in the prime factorization of x²Question: If x² = (2)(2)(2)(5)(?)(?)(?)..., what can we say about x?

Well, if we were to take the product (2)(2)(2)(5)(?)(?)(?) and break it into two EQUIVALENT products (x and x), we can conclude that each individual product must have at least TWO 2's and ONE 5.

In other words,

the prime factorization of x must have least TWO 2's and ONE 5 We can write: x = (2)(2)(5)(?)(?)...

x² is divisible by 7575 = (3)(5)(5)

So, x² = (3)(5)(5)(?)(?)(?)...

So,

the prime factorization of x must have least ONE 3 and ONE 5 We can write: x = (3)(5)(?)(?)...

When we COMBINE both pieces of information, we know that:

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5 In other words: x = (2)(2)(3)(5)(?)(?)...

If x has exactly 3 distinct prime factors, which of the following could be the value of x?This tells us that

2, 3 and 5 are the ONLY prime factors of x.

Now let's examine the answer choices....

A) 30 = (2)(3)(5)

This BREAKS the condition that

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5 So, x cannot equal 30

B) 60 = (2)(2)(3)(5)

60 works perfectly!

It satisfies the condition that

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, and it satisfies the condition that

2, 3 and 5 are the ONLY prime factors of x C) 200 = (2)(2)(2)(5)(5)

This BREAKS the condition that

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5 So, x cannot equal 200

D) 240 = (2)(2)(2)(2)(3)(5)

240 works perfectly!

It satisfies the condition that

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, and it satisfies the condition that

2, 3 and 5 are the ONLY prime factors of x E) 420 = (2)(2)(3)(5)(7)

This satisfies the condition that

the prime factorization of x must have least TWO 2's, ONE 3 and ONE 5, BUT it BREAKS the condition that

2, 3 and 5 are the ONLY prime factors of x Answer: B and D

Cheers,

Brent

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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