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Divisibility/Primes: If r and s are positive integers, is

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by II » Fri Aug 08, 2008 3:44 pm
Divisibility/Primes: If r and s are positive integers, is r/s an integer ?

(1) Every factor of s is also a factor of r

(2) Every prime factor of s is also a prime factor of r.

Please can you illustrate your logic/thinking in getting to the answer.

Thanks in advance.
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by Senator 153 » Fri Aug 08, 2008 4:04 pm
The question stem: "Divisibility/Primes: If r and s are positive integers, is r/s an integer ? "

Well, the answer (A) is that r/s is an integer IF and only if s is a factor of r. That is, you need to be able to divide s entirely into r.

(1) "Every factor of s is also a factor of r" <- perfect. So r can be a really big number or small, but s divides neatly into it.

(2) "Every prime factor of s is also a prime factor of r. " The trouble with this info is that "s" may contain repeat factors that won't divide into "r" even though the prime factors do. Integer "s" could even be a bigger number than r. Say... s=60, r=30.
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Re: Divisibility/Primes: If r and s are positive integers, i

by Ian Stewart » Fri Aug 08, 2008 5:22 pm
II wrote:Divisibility/Primes: If r and s are positive integers, is r/s an integer ?

(1) Every factor of s is also a factor of r

(2) Every prime factor of s is also a prime factor of r.

Please can you illustrate your logic/thinking in getting to the answer.

Thanks in advance.
Senator's explanation is good. It's even easier to see that 1) is sufficient: since s is a factor of s, 1) tells us that s is a factor of r. Thus r/s is an integer- that's exactly what 's is a factor of r' means.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by II » Sat Aug 09, 2008 1:59 am
Thanks for feedback.

Statement 2 was the one which I was trying to get to grips with.
I think it helps to plug in numbers to see this:

let r = 18 (has prime factors 2, 3, 3)
let s = 6 (has prime factors 2,3)

Since the prime factors of 6 (2, 3) are also the prime factors of 18, we can say in this case that r/s is an integer

let r = 18 (has prime factors 2, 3, 3)
let s = 8 (has prime factors 2, 2, 2)

All of the prime factors of s (2, 2, 2) are not the prime factors of 18, so we can say that r/s is not an integer in this case.

So statement (2) is insufficient.

Any comments on this.

Thanks.
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by georgy » Tue Dec 02, 2008 10:27 pm
an easier explanation .
any number is a factor of itself. eg factor of 2 is 1,2. factor of 6 is 1,2,3,6. so , using condition (1), r/s = s*a/s = a, where a is any factor of r .
using (2), r/s maybe an integer ( as in 100/10, since 100 & 10 can be expressed in prime factors of 2 & 5 ), or maybe not ( as in 10/1000)
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by cramya » Tue Dec 02, 2008 11:09 pm
I am sure u got Stmt I already

Stmt II

Take the case of 2 and 4(prime factor of both is 2)

If r=4 and s=2 r/s is an integer
If r=2 s=4 then r/s is not an integer

INSUFF
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by volderam » Thu Jul 09, 2009 7:11 am
hm...I'm sort of confused...
For example, let's consider r=6 and s=18
Then r/s = 6/18 = 2*3/2*3*3
Every factor of s is also a factor of r, but r/s is not integer. Therefore 1) is insufficient.
Can you tell me, please, where is the flaw in my reasoning?
Thanks in advance.
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by volderam » Thu Jul 09, 2009 8:19 am
oh...I see
because in this case 9 (that is 3*3) is not a factor of 6
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by clawhammer » Tue Nov 16, 2010 5:48 am
Kind of caught-up with the meaning.

In statement (2), when it's said 'EVERY' Prime Factor of s, why should we consider this as every 'unique' prime factor, bot not 'all' prime factors?

To consider statement 2 insufficient, we could try: 18/8
s= 8: how many prime factors does 8 have? 3 or 1?
I thought s can be divided by 3 (same) prime factors (three 2's). So every 2 is not in r (18), and hence this shouldn't be an example for Statement 2?
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by deepakteja1988 » Thu Jul 07, 2011 6:30 am
Guess at this point of time, everyone is clear with why statement 1 is sufficient. Coming to statement 2,

Every prime factor of s is also a prime factor of r

Any number can be broken into powers of prime numbers. The phrase 'Every prime factor' means each and every prime factor. Say the all the prime factors of 4 is 1,2,2 but not 1,2.

So there is problem with statement 2 which is insufficient.

Correct me if am wrong.
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by mirantdon » Thu Jul 07, 2011 9:29 am
+1 for A .
Since in option B .
For eg :we could have this case (2^3/2^4) --> resulting in a fraction although they contain the same prime factors .
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by whats_in_the_store » Mon Sep 10, 2012 6:11 am
I have same doubt, what if s has exactly same prime factors occurring more than once? Experts please enlighten me..
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by hsharif » Wed Sep 26, 2012 7:31 pm
@whats_in_the_store, I think you've already answered your question...

If s has MORE prime factors than r

e.g. r = 2*2 AND s = 2*2*2

then r/s will not be an integer.

BUT

if they have the SAME prime factors in the SAME amounts

e.g. r = 2*2 AND s = 2*2

then r/s will be an integer.

Makes sense?

I think I was initially tripped by the wording "EVERY", which in this case does not mean ALL.

I think what really tripped me (and maybe others) is that I thought that the same logic that nullifies Statement 2 could also apply to Statement 1

e.g. r = 6 AND s = 6 * 6

but what I didn't consider is that s, in this case, would actually be 36 so r would HAVE to be 36 in order to satisfy Statement 1.
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