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by raunekk » Thu Jul 03, 2008 11:15 am
24. If b is the product of three consecutive positive integers c, c + 1, and c + 2, is b a multiple of 24 ?

(1) b is a multiple of 3,
(2) c is odd


alternate approach..other then plugging numbers


thanks!!
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Source: — Data Sufficiency |
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by gauravkhare » Thu Jul 03, 2008 11:43 am
Answer should be "E".

product of any three consecutive numbers is always divisible by 3. so statement 1 doest not gives us any info. statement 2 also does not add any value.
gaurav
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by raunekk » Thu Jul 03, 2008 10:13 pm
sorry dude..OA= A

And i got d answer by plugging numbers..

any alternative tips or method..

anyone??
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by Ian Stewart » Fri Jul 04, 2008 5:18 am
raunekk wrote:sorry dude..OA= A

And i got d answer by plugging numbers..

any alternative tips or method..

anyone??
If the original answer is A, you haven't typed out the question correctly. You've typed:

1) b is a multiple of 3

We know b is the product of three consecutive integers from the question, and therefore know that b must be divisible by 3! = 6 (that's a very useful fact to understand: the product of n consecutive integers is always divisible by n!). As written, statement 1) gives us no new information. If the OA to the question is A, then statement 1) surely is supposed to read:

1) b is a multiple of 8

Then, since we know b is divisible by 8 and by 3, b must be divisible by LCM of 8 and 3, or 24. So with this adjustment, statement 1 is sufficient.
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by raunekk » Fri Jul 04, 2008 5:36 am
Ian Stewart,Even i couldnt understand the question properly ..

But unfortunately A is the correct answer.

Got this question from a similar kinda forum ..

even i am
confused,, :?
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by Ian Stewart » Fri Jul 04, 2008 6:06 am
To the question you've given, I guarantee you that A is not the correct answer- don't trust everything you read on internet GMAT forums! It's sufficient if you change '3' to '8' in Statement 1, but as written, it certainly is not sufficient. As explained above, statement 2 is not sufficient either: c could be 1 (and b would then not be divisible by 24), or c could be 7 (and b would be divisible by 24).
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by ildude02 » Sat Jul 05, 2008 5:12 pm
Ian, I have a question with reagrds to multiples in this same regard,

Say if X is a multiple of Y; and if in an toher statment says, X is a multiple of Z; Now when we combine both the statemetns, can we assume that X MUST be a multiple of of the LCM of Y and Z; assuming X, Y and Z are integers. I think at one point I made the mistake of thinking X will be a mutlitple of the product of YZ instead of the LCM. Just wanted to confirm this. Also, in the same reagrd, how do we treat factors. Say Y is a factor of X, Z is also a factor of X; can we derive anyhting out of it as to what can be the nfactors with respect to X, Y and Z ? Appreciate your response.

Ian Stewart wrote:
raunekk wrote:sorry dude..OA= A

And i got d answer by plugging numbers..

any alternative tips or method..

anyone??
If the original answer is A, you haven't typed out the question correctly. You've typed:

1) b is a multiple of 3

We know b is the product of three consecutive integers from the question, and therefore know that b must be divisible by 3! = 6 (that's a very useful fact to understand: the product of n consecutive integers is always divisible by n!). As written, statement 1) gives us no new information. If the OA to the question is A, then statement 1) surely is supposed to read:

1) b is a multiple of 8

Then, since we know b is divisible by 8 and by 3, b must be divisible by LCM of 8 and 3, or 24. So with this adjustment, statement 1 is sufficient.
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by Ian Stewart » Sun Jul 06, 2008 7:17 am
ildude02 wrote:Ian, I have a question with reagrds to multiples in this same regard,

Say if X is a multiple of Y; and if in an toher statment says, X is a multiple of Z; Now when we combine both the statemetns, can we assume that X MUST be a multiple of of the LCM of Y and Z; assuming X, Y and Z are integers. I think at one point I made the mistake of thinking X will be a mutlitple of the product of YZ instead of the LCM.
What you said above is entirely correct- if X is divisible by Y and by Z, that means exactly the same thing as 'X is divisible by the LCM of Y and Z'. Those who think that X will be divisible by the product YZ will fall into one of the most common traps in GMAT number theory questions. If you understand this, you're going to do well on many GMAT divisibility questions.
ildude02 wrote: Just wanted to confirm this. Also, in the same reagrd, how do we treat factors. Say Y is a factor of X, Z is also a factor of X; can we derive anyhting out of it as to what can be the nfactors with respect to X, Y and Z ? Appreciate your response.
I think I'd need to see the type of question you're thinking of here to give a more concrete answer. Without any more information about the relationship between Y and Z, you can't say much. You might have, for example, Y = 3, Z = 4, X = 12, or you might have Y = 12, Z = 12 and X=12. That is, the GCD of X, Y and Z could be 1, or it could be X, depending on the situation. Normally you'd have more information, so in some circumstances, you would be able to say something more than that.

One fact that can be useful (though it's typically only useful at the most difficult level of the GMAT), perhaps in the kind of question you're thinking about, is that the lcm(x,y)*gcd(x,y) = x*y. That is, the product of two numbers equals the product of the LCM and the GCD. If you haven't seen this before, try it with, say, 12 and 9:

GCD of 12 and 9 = 3
LCM of 12 and 9 = 36
product of 12 and 9 = 108 = 3*36 = GCD*LCM
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by ildude02 » Sun Jul 06, 2008 7:24 am
Thanks, Ian. I got to learn something new today. I didn't know about the GCF(X, Y) * LCM(X, Y) = X*Y. I will remeber that, just in case.
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by lunarpower » Fri Jul 11, 2008 3:15 am
i'm a little late to this particular party, but i wanted to point out that the following weak-looking condition:

(1) c is even
would actually be SUFFICIENT in this problem.

here's why:
we need the number to be divisible by 24, which means the product of c, c+1, and c+2 must contain three 2's and a 3.
as other posters have mentioned above, the 3 is automatic in any string of three consecutive integers, because every third integer is a multiple of 3.
here's what's sweet, though: you're guaranteed to have three 2's. this is because you have two even numbers (c and c+2), and every other even number is a multiple of 4, containing at least two 2's by itself.
specifically, if c only contains one 2, then c+2 will contain at least two 2's, and vice versa.
therefore, b contains at least one 3 and three 2's.
sufficient.
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