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Is a*b*c divisible by 24 ? 1. a, b, and c are consecutive [#permalink]
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28 Nov 2007, 07:36
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Is a*b*c divisible by 24 ?
1. a, b, and c are consecutive even integers
2. a*b is divisible by 12
Is A positive?
1. x^2  2x + A is positive for all x
2. Ax^2 + 1 is positive for all x
Last edited by JDMBA on 28 Nov 2007, 08:04, edited 1 time in total.



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Is a*b*c divisible by 24 ?
Question really means: Are there 2, 2, 2 and 3 in the factor box of product ABC? Answer is A
1. a, b, and c are consecutive even integers
SUFF. 3 consecutive even integers mean there are three 2s in the prime box of abc. Now we just need a 3. Every three consecutive even numbers will be a multiple of 3.
Now we have all primes required to make it to 24
2. a*b is divisible by 12
INSUFF. We can only establish that the product ABC has 2, 2 and 3 in its factor box. Can't tell if C is an even number.
Is A positive? B.
1. x2  2x + A is positive for all x (do you mean x^2?)
SUFF. When x = 1 or 1 we get:
x^2  2x + A
12=1+A = positive
so A must be positive to yield a positive number
2. Ax2 + 1 is positive for all x[/quote]
INSUFF.
A could be 1, x could be 5:
1 + 25 + 1 = positive
but A could also be 0.5:
0.5 + 25 + 1 = positive



Manager
Joined: 02 Oct 2007
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The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer?
For number 2. If x=100 A can be a #.
10000200+(1000)= 8800
What am I missing here?



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Joined: 26 Jul 2007
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JDMBA wrote: The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer?
I'm pretty sure it is. Would the answer be E then?



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rajpdsouza wrote: JDMBA wrote: The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer? I'm pretty sure it is. Would the answer be E then?
The CR given is A. I picked C because B eliminates the possibility of zero when statement 1 and 2 are combined.



Senior Manager
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I knew it was too easy... yeah, 0 is even. duh! Answer must be C.
From 1: BC are even, so we have 2, 2
From 2: AB gives we have 2, 2, 3
From both, we have 2, 2 from stat1, and 2, 2, 3 from stat2. We could have some overlapping, but we can break it down like this:
C gives us 2
AB gives us 2, 2, 3



Senior Manager
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After looking at the work, I think C is correct as well.



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asdert wrote: CR?
The correct answer given is A, A.
I disagree on both.
Why is A correct for #2. A can definetely be negative for any number X greater than 2



SVP
Joined: 05 Jul 2006
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First question is A
SECOND
Is A positive? is A>0
1. x^2  2x + A is positive for all x
2. Ax^2 + 1 is positive for all x
from one............ X^2  2X+A >0 IE:A>2XX^2 > X(2X), X COULD BE Anything..insuff
from2
Ax^2+1 >0
Ax^2>1
A could be anything
both together
E



CEO
Joined: 29 Mar 2007
Posts: 2559

JDMBA wrote: yezz wrote: First question is A
SECOND
Is A positive? is A>0
1. x^2  2x + A is positive for all x 2. Ax^2 + 1 is positive for all x
from one............ X^2  2X+A >0 IE:A>2XX^2 > X(2X), X COULD BE Anything..insuff
from2
Ax^2+1 >0
Ax^2>1
A could be anything
both together
E Is zero an even integer?
Yes 0 is an even integer. Did u forget to type that a,b,c are consecutive positive even integers? or that a,b,c are non zero integers?
If not then the answer is C for question 1. no doubt.
Question 2: I say E.



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JDMBA wrote: The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer?
For number 2. If x=100 A can be a #.
10000200+(1000)= 8800
What am I missing here?
1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A.
2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2  2x + A for all x.
It's true that A can be negative and still make x^2  2x + A positive for some x. But let's suppose, as in the example here, that A = 1000. If x = 100, x^2  2x + A is positive. But if x = 1, it's not. If A = 1000, x^2  2x + A is not positive for all x.
In particular, x can always be 0. If x^2  2x + A > 0 when x = 0, then A > 0. Statement I is sufficient.
But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A.



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johnrb wrote: JDMBA wrote: The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer?
For number 2. If x=100 A can be a #.
10000200+(1000)= 8800
What am I missing here? 1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A. 2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2  2x + A for all x. It's true that A can be negative and still make x^2  2x + A positive for some x. But let's suppose, as in the example here, that A = 1000. If x = 100, x^2  2x + A is positive. But if x = 1, it's not. If A = 1000, x^2  2x + A is not positive for all x. In particular, x can always be 0. If x^2  2x + A > 0 when x = 0, then A > 0. Statement I is sufficient. But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A.
I see the trick in question 1.
but i still dont see why S1 in question 2 is sufficient.



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GMATBLACKBELT wrote: johnrb wrote: JDMBA wrote: The reason I didnt pick A for number 1 is because of 0.
Is 0 an even integer?
For number 2. If x=100 A can be a #.
10000200+(1000)= 8800
What am I missing here? 1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A. 2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2  2x + A for all x. It's true that A can be negative and still make x^2  2x + A positive for some x. But let's suppose, as in the example here, that A = 1000. If x = 100, x^2  2x + A is positive. But if x = 1, it's not. If A = 1000, x^2  2x + A is not positive for all x. In particular, x can always be 0. If x^2  2x + A > 0 when x = 0, then A > 0. Statement I is sufficient. But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A. I see the trick in question 1. but i still dont see why S1 in question 2 is sufficient.
How's this:
Yezz is right to say that A > x(2x). But the point is that this has to hold for all values of x, not just for one particular value of x. So think about how this inequality behaves for different values of x.
If x is less than 0, then x (2  x) is negative.
If x = 0 or 2, then x (2  x) is zero.
If x> 0 and less than 2, then x (2  x) is positive.
If x > 2, then x (2  x) is negative.
A is greater than x (2  x) whatever x is. So it must be greater than zero.
You could also think about it geometrically. The problem is equivalent to saying that the graph of y = A is always above the graph of y = x (2  x). That means A must be more than the maximum value of x (2  x), not just more than some particular value of x (2  x). The maximum value of x (2  x) is 1. So A must be greater than 1.



Senior Manager
Joined: 26 Jul 2007
Posts: 371

I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: 4,2,0 2,0,2 0,2,4. None of those are divisible by 24.



CEO
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Posts: 2559

rajpdsouza wrote: I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: 4,2,0 2,0,2 0,2,4. None of those are divisible by 24.
Yes they are as Johnrb pointed out.
0/24=0 This will be the case for any integer. So 4(2)(0)=0/24=0 it is divisible.
Quick question what is 0/0? Is this still undefined or just 0?



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GMATBLACKBELT wrote: Yes they are as Johnrb pointed out.
0/24=0 This will be the case for any integer. So 4(2)(0)=0/24=0 it is divisible.
Long day at work, not paying attention. Don't mind my stupidity.



Senior Manager
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Posts: 466

rajpdsouza wrote: I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: 4,2,0 2,0,2 0,2,4. None of those are divisible by 24.
Do we have to consider negative number when problem talks about consecutive even integers?
I thought we only consider positive integers...



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Joined: 09 Mar 2003
Posts: 463

asdert wrote: rajpdsouza wrote: I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: 4,2,0 2,0,2 0,2,4. None of those are divisible by 24. Do we have to consider negative number when problem talks about consecutive even integers? I thought we only consider positive integers...
No, you definitely need to consider them. Integers, by definition, are all whole numbers, positive and negative.
0/0 is undefined. 0/(anything else) is 0, so it's all good for that one.
As for the second one, I'm on board with A. It's a hellish problem, but you see by plugging in some numbers for x, the equation is negative around the number +1. So A has to be a positive number to counteract that. Under all other conditions, A could be negative and it wouldn't matter, but since we're told the equation is ALWAYS positive, A must be positive.
The second statement is A times some x^2. x^2 is always positive, but that doesn't mean A has to be. As long as Ax^2 > 1, this will work out, so A could be some negative number to make that happen.



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Posts: 466

I need to stop posting when I'm busy doing something else... I thought we were talking about negative prime numbers, duh!
I've read the thread carefully and yes, I agree. Neg., positive integers must be considered unless otherwise indicated,



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JDMBA wrote: Is A positive?
1. x^2  2x + A is positive for all x 2. Ax^2 + 1 is positive for all x
1. (x1)^2 + (A1) > 0 for all x
A > 1 (x1)^2 for all x
Thus A > 1 SUFF
2. A could be 0. NOT SUFF







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