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easy explanation pls

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easy explanation pls

by deltaforce » Tue Jun 30, 2009 7:19 am
p^a * q^b * r^c * s^d = x, where x is a perfect square. If p, q, r, and s are prime integers, are they distinct?

(1) 18 is a factor of ab and cd
(2) 4 is not a factor of ab and cd

[spoiler]Answer: B[/spoiler]
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by DanaJ » Tue Jun 30, 2009 7:41 am
Very interesting question... I wonder what the source is?
You should start off knowing this: in order for x to be a square, then the factors of its prime factorization have to be raised at even powers. This makes sense if you think of any example. Say x = 2^6 - it will indeed be the square of 2^3, since (2^3)^2 = 2^6. But if you consider x = 3^3, then you're not looking at a perfect square. 3^3 = (3^2)*3, so the square root of 3^2 will be 3*sqrt(3).

So you have two possibilities here:
a. p, q, r and s are distinct, but a, b, c and d are ALL even
b. p, q, r and s are not distinct, but their powers are "built" in such a way that when you add them up you get even powers for all prime factors. Example to clear things up:
x = (3^2)*(2^5)*(3^6)*(2^7) - group according to bases to get that x = (3^8)*(2^12) = [(3^4)*(2^6)]^2 - so x will be the square of (3^4)*(2^6).

Now, let's move on to the statements.

1. could be tackled algebraically, but I have a feeling that if I'd do so you won't understand anything. So I'll just go the classical way of examples. Note that if all powers (i.e. a, b, c and d) would be even (as mentioned in case a above), then the product of any two will be (even)*(even), which will result in a number that's divisible by 4.
Also note that 18 by itself is NOT divisible by 4.
So now we break it into examples:

Example I
ab = cd = 18 - so 18 is indeed a factor of both ab and cd
Let's say you have a = 2, b = 9, c = 6 and d = 3. In this case:

x = (p^2)*(q^9)*(r^6)*(s^3)

In order for x to be a perfect square, then you'd need to take care of that 9 and 3 in that product, so q = s. In this case, the numbers ARE NOT DISTINCT.

Example II
ab = cd = 36 - again, 18 is a factor of ab and cd
Get some values for the powers: a = 18, b = 2, c = 6 and d = 6. X will be:

x = (p^18)*(q^2)*(r^6)*(s^6)

In this case, it's no longer required for p, q, r and s to be equal in any way. They could very well NOT BE.

Since you get two outcomes from two perfectly valid examples, 1 is insufficient.


2. tells you exactly what you need to hear: since neither ab nor cd are divisible by 4, then you can safely say that NOT ALL the powers are even, so you find yourself in case b outlined above. If all powers were even, then ab and cd will both be something like (even)*(even), so they'd be divisible by 4.
Since you're in case b, p, q, r and s ARE NOT DISTINCT. 2 is sufficient.

Hope this makes sense to you... Will be hanging around waiting for other methods to solve this one...
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by deltaforce » Tue Jun 30, 2009 8:51 am
Hi Dana,

Excellent explanation. I was too focused on p, q value and was complicating it.

At the least both ab or cd should come up to 4 to justify a perfect square. I spent abt 10 minutes before i posted it on here.

Source is https://gmat-maths.blocked/



Thanks :lol:
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by DanaJ » Tue Jun 30, 2009 9:30 am
Well, don't be too hard on yourself for not figuring it out... IMHO, this is an upper-bin-type question, that would probably pop up only when you're already scoring somewhere around 49. At least that's my impression... It's because it got me thinking, you know! It is solvable in a the 2-minute time constraint, but only if you're pretty advanced and can spot the algebraic solution right off the bat, or at least I think so.
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by cameronwu » Tue Jun 30, 2009 11:16 am
Yes, very good reply to the poster above that answered the question. Those seem like the most proper and concise deductions to make.
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Re: easy explanation pls

by ssmiles08 » Tue Jun 30, 2009 12:09 pm
This seemed like a problem where you have to know your number properties inside out.

My approach was similar to DanaJ's, only involving a lot more odds and evens.

Rule 1: all perfect squares have an odd number of total factors.
Rule 2: The prime factorization of a perfect square contains only even powers of prime.

so working backwards;

2) 4 is not a factor of ab and cd, which means they only have one 2 or no 2's in common.

If we take rule#2 into consideration, we find that at least two of (p, q, r, s) are not distinct, b/c the prime factorization of a number has to be even.

for example: a = 2, b = 1, c = 3, d = 4

the common factor of ab and cd is 2, but for the product x to be a square, q and r have to be the same. (if q and r are the same, the power would be 4 (3+1), which would be a even power.)

so this statement would be sufficient since they are not all distinct.

1) for ab and cd to have a factor of 18, it must have at least two 3's and one 2.

for example if we take a = 18, b = 18, c = 18 and d = 18, and consider them all distinct(p,q,r,s), product x would have 19^4 total factors.

all perfect squares have an odd number of total factors. We know the product of 19^4 = O*O*O*O = O, this would be sufficient in this scenario.

however, if they were all the same (p,q,r,s), this would be sufficient for this scenario as well b/c the prime factorization of this product would be even, which follows rule 2. it would have (18*4) = Even powers of prime.

so this statement is insufficient b/c both scenarios are possible.

(B)
1. could be tackled algebraically,
Hey DanaJ, could you tells us how you did it algebraically, I want to know if it was a faster approach that what you showed us. Thanks.
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by DanaJ » Tue Jun 30, 2009 12:38 pm
Nope, don't think it would be faster. Just involves taking the general cases, with:

ab = cd = 18*k, and here k can be odd or even and you treat it separately as such.

The examples I've provided are basically a simpler version of the general demonstration.
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by tohellandback » Tue Jun 30, 2009 9:57 pm
my take:
if X is not distinct a+b has to be even. (can be C+d or watever, u can suppose anything)
so if ab=18k
a+b can be even or odd depending upon k
so Not SUFF

2) already explained

so B
The powers of two are bloody impolite!!
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