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It is not mentioned in the question that "s" is a positive integer. Then why are we discardin the possibility of s=0?

Because general rule is "all positive integers are divisors of zero".

Then S could be 0 or 24 and hence answer should be E not A.

Can someone please validate this understanding.
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It is not mentioned in the question that "s" is a positive integer. Then why are we discardin the possibility of s=0?

Because general rule is "all positive integers are divisors of zero".

Then S could be 0 or 24 and hence answer should be E not A.

Can someone please validate this understanding.

Great points!!
Interestingly enough, when it comes to Integer Properties questions, the GMAT test-makers almost always restrict the values to positive integers, but they haven't done that here.
You're right to say that all positive integers are divisors of 0.
However that doesn't change the correct answer.

For statement 1, even if we include the possibility that s = 0, the answer to the target question is still the same: YES, 24 is a divisor of s
So statement 1 is still sufficient.

Cheers,
Brent
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oh, that's right.

Yes, I forgot to apply the same reasoning on 24 as well. Thank you Brent for clarifying (Sorry, don't know how to tag a person here).
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oh, that's right.

Yes, I forgot to apply the same reasoning on 24 as well. Thank you Brent for clarifying (Sorry, don't know how to tag a person here).

Glad to help!

Cheers,
Brent
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hello chetan2u
in option 2, how can i understand that the we should take LCM of 4 and 6. I'm completely lost on this
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If s is an integer, is 24 a divisor of s ?

Prime factorization of \(24 = 2^3*3\)

(1) Each of the numbers 3 and 8 is a divisor of s.

\(8 = 2^3\)
We have \(2^3\) and \(3\), answering our question. SUFFICIENT.

(2) Each of the numbers 4 and 6 is a divisor of s.
\(4 = 2^2\)
\(6 = 2*3\)

We can't say for certainty -- 2 could overlap. We could have 12 instead of 24 here. INSUFFICIENT.

Answer is A.
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If a number has 2 numbers as its divisors, it also has their LCM as its divisor.

1) LCM (3,8) = 24
So, S = 24n -> So 24 is definitely a divisor of S

2) LCM (4,6) = 12
So, S = 12n -> 24 is NOT a divisor of 12 (and of 36, 60, 84, etc..)

Thus only 1 is sufficient -> A
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If s is an integer, is 24 a divisor of s ?

(1) Each of the numbers 3 and 8 is a divisor of s.
(2) Each of the numbers 4 and 6 is a divisor of s.
------ASIDE----------------------------
ASIDE: A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:

If k is a divisor of N, then k is "hiding" within the prime factorization of N

Consider these examples:
3 is a divisor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a divisor of 70 because 70 = (2)(5)(7)
And 8 is a divisor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a divisor of 630 because 630 = (2)(3)(3)(5)(7)
------ONTO THE QUESTION----------------------------
24 = (2)(2)(2)(3)
So, we can rephrase the target question as....
REPHRASED target question: Are there three 2's and one 3 "hiding" in the prime factorization of s?

Statement 1: Each of the numbers 3 and 8 is a divisor of s.
This tells us that 3 is hiding in the prime factorization of s
And, since 8 = (2)(2)(2), we also now know that three 2's are hiding in the prime factorization of s
So, the answer to the REPHRASED target question is YES, there three 2's and one 3 "hiding" in the prime factorization of s
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Each of the numbers 4 and 6 is a divisor of s
Since 4 = (2)(2), we now know that two 2's are hiding in the prime factorization of s
Since 6 = (2)(3), we now know that one 2 and one 3 are hiding in the prime factorization of s
So, all we can be certain of is that there are two 2's and one 3 hiding in the prime factorization of s
Consider these two possible cases:
Case a: s = 12, in which case 24 is NOT a divisor of s
Case b: s = 24, in which case 24 IS a divisor of s
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

BrentGMATPrepNow When we say "divisor" we basically mean "factor" right? So if we say "is K a divisor of N" we are basically seeing if K goes completely into N OR simply saying "Is K a factor of N" Right?
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BrentGMATPrepNow When we say "divisor" we basically mean "factor" right? So if we say "is K a divisor of N" we are basically seeing if K goes completely into N OR simply saying "Is K a factor of N" Right?
That's correct.
In fact, that are many several different ways to say "y is a factor of x"

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If s is an integer, is 24 a divisor of s ?

Statement 1: Each of the numbers 3 and 8 is a divisor of s.
This tells us that 3 is hiding in the prime factorization of s
And, since 8 = (2)(2)(2), we also now know that three 2's are hiding in the prime factorization of s
So, the answer to the REPHRASED target question is YES, there three 2's and one 3 "hiding" in the prime factorization of s
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Each of the numbers 4 and 6 is a divisor of s
Since 4 = (2)(2), we now know that two 2's are hiding in the prime factorization of s
Since 6 = (2)(3), we now know that one 2 and one 3 are hiding in the prime factorization of s
So, all we can be certain of is that there are two 2's and one 3 hiding in the prime factorization of s
Consider these two possible cases:
Case a: s = 12, in which case 24 is NOT a divisor of s
Case b: s = 24, in which case 24 IS a divisor of s
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Doesn the 4 and 6 also have three 2s and one 3? Like 3 and 8.
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BrentGMATPrepNow
gmatt1476
If s is an integer, is 24 a divisor of s ?

Statement 1: Each of the numbers 3 and 8 is a divisor of s.
This tells us that 3 is hiding in the prime factorization of s
And, since 8 = (2)(2)(2), we also now know that three 2's are hiding in the prime factorization of s
So, the answer to the REPHRASED target question is YES, there three 2's and one 3 "hiding" in the prime factorization of s
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: Each of the numbers 4 and 6 is a divisor of s
Since 4 = (2)(2), we now know that two 2's are hiding in the prime factorization of s
Since 6 = (2)(3), we now know that one 2 and one 3 are hiding in the prime factorization of s
So, all we can be certain of is that there are two 2's and one 3 hiding in the prime factorization of s
Consider these two possible cases:
Case a: s = 12, in which case 24 is NOT a divisor of s
Case b: s = 24, in which case 24 IS a divisor of s
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent

Doesn the 4 and 6 also have three 2s and one 3? Like 3 and 8.
Not quite, since 4 and 6 could be "sharing" a 2.
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Not quite, since 4 and 6 could be "sharing" a 2.

Pardon my ignorance but can you elaborate on that? Or point to a resource where I could learn more about it?
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BrentGMATPrepNow


Not quite, since 4 and 6 could be "sharing" a 2.

Pardon my ignorance but can you elaborate on that? Or point to a resource where I could learn more about it?

Sorry, I didn't explain that very well.

Statement 2 says: Each of the numbers 4 and 6 is a divisor of s
Since 4 = (2)(2), we can be certain that there are at least two 2's in the prime factorization of s.
Since 6 = (2)(3), we can be certain that there's at least one 2 and one 3 in the prime factorization of s.
So, it could be the case that s = (2)(2)(3) = 12. As we can see we have least two 2's in the prime factorization of s, AND we have least one 2 and one 3 in the prime factorization of s. In this case, 24 is NOT a divisor of s.

That said, it could ALSO be the case that s = (2)(2)(2)(3) = 24. As we can see we have least two 2's in the prime factorization of s, AND we have least one 2 and one 3 in the prime factorization of s. In this case, 24 IS a divisor of s.

Does that help?
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