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I'm having trouble with the rule behind this concept. This [#permalink]
07 Sep 2008, 15:24

1

This post was BOOKMARKED

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain... _________________

Factorials were someone's attempt to make math look exciting!!!

Re: Last minute help....(test in 2 days) [#permalink]
07 Sep 2008, 16:07

brokerbevo wrote:

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain...

Yes, the rule does only apply to the product of three consecutive integers. This is because one of those consecutive integers, by definition, is divisible by 3 by itself.

Re: Last minute help....(test in 2 days) [#permalink]
07 Sep 2008, 16:12

zoinnk wrote:

brokerbevo wrote:

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain...

Yes, the rule does only apply to the product of three consecutive integers. This is because one of those consecutive integers, by definition, is divisible by 3 by itself.

Ok good, that's what I thought but didn't know for sure. So, in that case, how do you solve the above problem? _________________

Factorials were someone's attempt to make math look exciting!!!

Re: Last minute help....(test in 2 days) [#permalink]
07 Sep 2008, 16:41

brokerbevo wrote:

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain...

you are almost there.

(1) y = 2x^3 + 9x^2 - 5x = x (2x^2 + 9x - 5) = x (2x - 1) (x + 5)

plug-in: any integer value for x in the above expression is divisible by 3. if x = 2: x (2x - 1) (x + 5) = 2 (2x2 - 1) (2+5) = 2 x 3 x 7. suff. if x = 4: x (2x - 1) (x + 5) = 4 (2x4 - 1) (4+5) = 4 x 7 x 9. suff. if x = 5: x (2x - 1) (x + 5) = 5 (2x5 - 1) (5+5) = 4 x 9 x 10. suff. if x = 7: x (2x - 1) (x + 5) = 7 (2x7 - 1) (7+5) = 7 x 13 x 12. suff.

i believe every integer value for x result y divisible by 3.

Re: Last minute help....(test in 2 days) [#permalink]
07 Sep 2008, 18:45

brokerbevo wrote:

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain...

Well, it doesn't *only* apply when the integers are consecutive. Say x is an integer. You know already that one of x, x+1 or x+2 is a multiple of 3. Well, that guarantees that one of, say, x, x+4, and x+8 is divisible by 3 (because if x+1 is, then x+1+3 = x+4 is, and if x+2 is, so is x+2 + 6 = x+8), and many other combinations besides.

You can apply that logic here if you want. We know that one of these numbers: x, x+1 or x+2 is a multiple of 3:

-If x is a multiple of 3, then x is a multiple of 3 -If x+1 is a multiple of 3, then 2x+2 is also a multiple of 3, and (subtract 3) 2x-1 is also a multiple of 3 -If x+2 is a multiple of 3, then x+2+3 = x+5 is also a multiple of 3.

So one of x, 2x-1 or x+5 will always be a multiple of 3, and since y = x (2x - 1) (x + 5), y must be a multiple of 3.

There are other ways to go about the question that might be preferable here (modular arithmetic, for example); I just wanted to show how you can extend the logic about 'consecutive numbers' to this situation. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: Last minute help....(test in 2 days) [#permalink]
18 Sep 2008, 23:30

GMAT TIGER wrote:

brokerbevo wrote:

I'm having trouble with the rule behind this concept. This problem is a good example of my confusion:

If y is an integer, is y divisible by 3? (1) y = 2x^3 + 9x^2 - 5x (2) x is indivisible by 3

Please explain your rationale behind your answer.

Here is what I did:

(1) factor out an x --> x (2x^2 + 9x - 5) factor the quadratic --> x (2x - 1) (x + 5) so, y is the product of 3 integers and I wrote down that since 3 integers are multiplied together, it must be that it is divisible by 3. HOWEVER, I thought that this rule ONLY applies if the integers that are multiplied together are consecutive!!!

For example, x (x - 1) (x -4) is not divisible by 3 (even though it is the product of 3 numbers) because the parts are not consecutive.

Please explain...

you are almost there.

(1) y = 2x^3 + 9x^2 - 5x = x (2x^2 + 9x - 5) = x (2x - 1) (x + 5)

plug-in: any integer value for x in the above expression is divisible by 3. if x = 2: x (2x - 1) (x + 5) = 2 (2x2 - 1) (2+5) = 2 x 3 x 7. suff. if x = 4: x (2x - 1) (x + 5) = 4 (2x4 - 1) (4+5) = 4 x 7 x 9. suff. if x = 5: x (2x - 1) (x + 5) = 5 (2x5 - 1) (5+5) = 4 x 9 x 10. suff. if x = 7: x (2x - 1) (x + 5) = 7 (2x7 - 1) (7+5) = 7 x 13 x 12. suff.

i believe every integer value for x result y divisible by 3.

2: x = 3k is not sufficient.

So //A//

GMAT TIGER, why are we trying only integer values of x here ? The stem does not mention anything about x, it could be a fraction also. _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: Last minute help....(test in 2 days) [#permalink]
18 Sep 2008, 23:35

IanStewart wrote:

-If x is a multiple of 3, then x is a multiple of 3 -If x+1 is a multiple of 3, then 2x+2 is also a multiple of 3, and (subtract 3) 2x-1 is also a multiple of 3 -If x+2 is a multiple of 3, then x+2+3 = x+5 is also a multiple of 3.

So one of x, 2x-1 or x+5 will always be a multiple of 3, and since y = x (2x - 1) (x + 5), y must be a multiple of 3.

There are other ways to go about the question that might be preferable here (modular arithmetic, for example); I just wanted to show how you can extend the logic about 'consecutive numbers' to this situation.

If x is a multiple of 3, then x is a multiple of 3 . I guess there is a typo or am I missing the concept ?

Ian how do we apply modular arithmetic here ? Please guide.

Thanks _________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Re: Last minute help....(test in 2 days) [#permalink]
19 Sep 2008, 06:13

amitdgr wrote:

If x is a multiple of 3, then x is a multiple of 3 . I guess there is a typo or am I missing the concept ?

No typo- I was just pointing out that one of x, x+5 or 2x-1 must be a multiple of 3, and if x is a multiple of 3, then it's just obvious that one of the three expressions is a multiple of 3.

I can explain how to approach this with modular arithmetic, but be aware that you won't need modular arithmetic on the GMAT. It would take me too long to explain the theory, so if you don't know modular arithmetic already, don't worry about it!

We want to know if

2x^3 + 9x^2 - 5x ~ 0 (mod 3)

We don't need to bother simplifying, but it makes things slightly easier: since 9 ~ 0 mod 3, and -5 ~ 1 mod 3, we just want to know if:

2x^3 + x ~ 0 (mod 3)

Now it's easy to check that if x ~ 0, 1 or 2 (mod 3), that 2x^3 + x is always congruent to 0 (mod 3), so it must be divisible by 3. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: Last minute help....(test in 2 days) [#permalink]
19 Sep 2008, 06:21

GMAT TIGER wrote:

scthakur wrote:

But, the question stem does not specifically mention that x is an integer....hence I will go with C.

Do you have any non-integer value for x that yeilds an integer value for y?

any example?

y = 2x^3 + 9x^2 - 5x is a continuous function with an unrestricted range, so y can take on any value at all (graph it on the co-ordinate plane and you'll see why). I've ignored the possibility, in my posts above, that x is not an integer, because on the GMAT you would never see a question like this unless you were told in the question that x is an integer. I assumed there was a typo in the question, and that it was meant to begin "If x is an integer" rather than "If y is an integer". If the question doesn't mention this, it's really testing whether you know properties of continuous functions, something you learn in calculus, not in GMAT prep.

Still, if we accept the possibility that x is not an integer, the answer is not C, since Statement 2 is not at all helpful in this case; the answer would be E. But I'm sure the intended answer is A, and that the question intends for x to be an integer. _________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

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Re: Last minute help....(test in 2 days)
[#permalink]
19 Sep 2008, 06:21

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