Flashcards, Number Properties, #22

[claireseymour12]

I have a question about the Number Properties Flashcard #22.

"x is divisible by 42. Which of the following numbers is definitely a factor of x^2? (Choose all the apply.)"

The answer states that we can guarantee that all numbers that solely use prime factors found in x^2 are factors of x^2.

I understand this point and I understand why 36 and 63 are factors of x^2 and that 33 is not a factor...but I do not understand why 8 would then not be a factor of x^2 as 8's prime factors are 2, 2, and 2. 2 is in fact a prime factor of x^2 in this case.

Would you please clarify?

(Where:
42: 2, 3, 7
42^2: 2, 2, 3, 3, 7, 7

a) 63: 7, 3, 3
b) 33: 3, 11
c) 36: 2, 2, 3, 3
d) 8: 2, 2, 2 )

[abemartin87]

Hey Claire,

Not sure where you got the question.

"x is divisible by 42. Which of the following numbers is definitely a factor of x^2? (Choose all the apply.)"

x is divisible by 42 is to say x=42k, where k is some integer
x=42*k=2*3*7*k

therefore x^2=(2^2)(3^2)(7^2)m , where m is some integer

now as you've said:


a) 63: 7, 3, 3
b) 33: 3, 11
c) 36: 2, 2, 3, 3
d) 8: 2, 2, 2 )


63 has one 7 and two 3's. X^2 has two 7's and one 3. So 63 is divisible by 63. Same with (c) 6=36 has two 2's and two 3's. Yes, X^2 has two 2's and two 3's, hence 36 is divisible by X^2.

Now for option (d) 8 has 3 two's. Do we have the same primes AND the same or GREATER power on each prime? You have to first check to see if X^2 has the same primes, which it does X^2 has a two as its prime. But you also have to check the power on each prime. X^2 has only two 2's. We need X^2 to have three 2's at minimum for it be divisible by 8.

If X^2=(2^5)(3^2)(7^2)k

then X^2 would be divisible by 8=2^3 , because it has five 2's, more than we need for it be divisible by 8.

[claireseymour12]

Thank you !

This helps greatly, I got confused about the "same greater power" portion, and now it is clear. For your knowledge, one of the resources on this website is flashcards, I got this problem from flashcard #22 of the "Number Properties" deck.

Thanks again!

[jnelson0612]

Wonderful! Thanks Abe!