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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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18 Aug 2007, 10:20

kevincan wrote:

bkk145 wrote:

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

If they are distinct, a b c and d must be positive even integers. (1) Not sufficient x could be (pqrs)^18 and p,q r s may or may not be distinct

(2) As ab is not divisible by 4, a and b are not both even. Thus p,q,r,s are not distinct SUFF

Further Explanation...

Will go for B...
In 2nd
As ab and cd is not divisible by 4, a and b both are not even no.Similarly c and d both are not even no. But in a and b, one would be even number.Also in c and d, one would be even number.So 2 variables (In p,q,r,s)can be different,whose squares are even but 2 should be same to make it a perfect square..

Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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19 Aug 2007, 12:09

2

This post received KUDOS

bkk145 wrote:

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

OK let me begin with statement 2..its the easiest to work out

if X is a perfect square then P^a...s^d all have to be raised to an even power..

4 is not a factor of ab clearly tells me that one of these powers either a or b or c or d is not even..if thats the case then one of the primes is repeated..

therefore 2 is sufficient

I have trouble with 1...

so again same approach

18 is a factor..well which means that if 18*2 is 36...which is an even power then we can have possible repeat of prime factors...and possibly not..say p^6 Q^6 r^6s^6 is a perfect square...now p and q =2 or could be 3 or 2...we dont know..INSUFF...

Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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20 Aug 2007, 18:13

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1

This post was BOOKMARKED

OA=B

Answer
When a perfect square is broken down into its prime factors, those prime factors always come in "pairs." For example, the perfect square 225 (which is 15 squared) can be broken down into the prime factors 5 × 5 × 3 × 3. Notice that 225 is composed of a pair of 5's and a pair of 3's.

The problem states that x is a perfect square. The prime factors that build x are p, q, r, and s. In order for x to be a perfect square, these prime factors must come in pairs. This is possible if either of the following two cases hold:

Case One: The exponents a, b, c, and d are even. In the example 3^2 5^4 7^2 11^6, all the exponents are even so all the prime factors come in pairs.

Case Two: Any odd exponents are complemented by other odd exponents of the same prime. In the example 3^1 5^4 3^3 11^6, notice that 3^1 and 3^3 have odd exponents but they complement each other to create an even exponent (3^4), or "pairs" of 3's. Notice that this second case can only occur when p, q, r, and s are NOT distinct. (In this example, both p and r equal 3.)

Statement (1) tells us that 18 is a factor of both ab and cd. This does not give us any information about whether the exponents a, b, c, and d are even or not.

Statement (2) tells us that 4 is not a factor of ab and cd. This means that neither ab nor cd has two 2's as prime factors. From this, we can conclude that at least two of the exponents (a, b, c, and d) must be odd. As we know from Case 2 above, if paqbrcsd is a perfect square but the exponents are not all even, then the primes p, q, r and s must NOT be distinct.

The correct answer is B: Statement (2) alone is sufficient, but statement (1) alone is not sufficient.

Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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24 Jan 2008, 10:47

fresinha12 wrote:

bkk145 wrote:

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

OK let me begin with statement 2..its the easiest to work out

if X is a perfect square then P^a...s^d all have to be raised to an even power..

4 is not a factor of ab clearly tells me that one of these powers either a or b or c or d is not even..if thats the case then one of the primes is repeated..

therefore 2 is sufficient

I have trouble with 1...

so again same approach

18 is a factor..well which means that if 18*2 is 36...which is an even power then we can have possible repeat of prime factors...and possibly not..say p^6 Q^6 r^6s^6 is a perfect square...now p and q =2 or could be 3 or 2...we dont know..INSUFF...

B it is..

thanks. very helpful _________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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11 Dec 2014, 08:16

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Re: p^a*q^b*r^c*s^d=x, where x is a perfect square. If p, q, r, and s are [#permalink]

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11 Dec 2014, 10:27

Expert's post

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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?

First of all: a perfect square always has even powers of its prime factors. So, if \(p\), \(q\), \(r\), and \(s\) ARE distinct primes, then in order \(x\) to be a perfect square \(a\), \(b\), \(c\), and \(d\) MUST be even.

(1) 18 is a factor of ab and cd --> we cannot get whether \(a\), \(b\), \(c\), and \(d\) are even or odd. For example we can have following two cases: \(p^a*q^b*r^c*s^d=2^3*3^6*2^3*3^6\): in this case \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. \(p^a*q^b*r^c*s^d=2^2*3^{18}*5^2*7^{18}\): in this case \(p\), \(q\), \(r\), and \(s\) are distinct primes. Not sufficient.

(2) 4 is not a factor of ab and cd --> which means that at least one from \(a\) and \(b\), and at least one from \(c\) and \(d\) is NOT even (if for example \(a\) and \(b\) were BOTH even then \(ab\) would be a multiple of 4) --> \(p\), \(q\), \(r\), and \(s\) are NOT distinct primes. Sufficient.

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