PS Problem

[asbestos]

P is afacot of 48 and Q is a factor of 27, which of the following could not be a value of p^2-Q^2
a. 20
b. 0
c. 72
d. 5
e. None

[pathaniaus]

I think the answer should be D. (5) (But I am debating with option A as well)
If u do the prime factorization of 48, you get 2^4*3^1.
Prime factorization of 27 is 3^3.
From here if you took 3^2 (factor of 48) and subtracted it by 3^2 (factor of 27) you end up with zero.
But if you also did 2^2(factor of 48) and subtracted it by 3^2(factor of 27) then you get -5.
Hence because I knew that 5 is an impossibility when you are subtracting q^2 from p^2 I chose it.

Is this correct?

[ssmiles08]

I would like to see someone else's take on this problem.

Hey asbestos,

Are you sure the question is asking which could not be the value or which could be the value?


because my reasoning was that the only odd factor of 48 is 3. the rest are even factors.

27 is composed of all odd factors.

So the value it can yield is 0. b/c (3^2 - 3^2)

Since the rest of the factors for 48 are even: the only possibility is EVEN^2 = EVEN

All of the factors for 27 are odd: so the only possibility for this situation is ODD^2 = ODD

so EVEN^2 - ODD^2 = EVEN - ODD = ODD.

so I don't believe either 20 or 72 has a possibility of occurring.

please correct me if I am wrong.

[DanaJ]

@ssmiles08 - I got the same result. Neither 20 nor 72 can show up (but -72 can!). However:
5 = 3^2 - 2^2
0 = 3^2 - 3^2.

[ssmiles08]

@ssmiles08 - I got the same result. Neither 20 nor 72 can show up (but -72 can!). However:
5 = 3^2 - 2^2
0 = 3^2 - 3^2.

Yes you are right DanaJ.

but 5 can occur if it is q^2 - p^2 and not p^2 - q^2. we could get -5!

Anyway I thought I was going wrong somewhere...but I think it's just the question!

Thanks!

[DanaJ]

Yeah, you're right - silly me for rushing!