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I am not very much convinced with strategy mentioned for option 1. Example has used x = 100 and Y as 40. Since both are not divisible by 3 that means term X^16 - Y^8 is not divisible by 15. (X^2-Y) and (X^2 + Y) are factors of X^16 - Y^8 . And if plug in the value of X and Y in X^2-Y you get the number = 100^2 - 40^1 = 9960 and this is divisible by both 3 and 5. I tried couple of combinations and could find term X^16 - Y^8 divisible by both 3 & 5. I reached to conclusion that substituting numbers is not the right method for this.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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20 Sep 2010, 22:30

14

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Expert's post

If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?

First of all as \(345y^2\) is divisible by 15 (this term won't affect the remainder), we can drop it.

The question becomes: is \(x^{16}-y^8\) divisible by 15?

(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20 --> they both could be multiples of 15 as well (eg x=25*15 and y=20*15) and in this case \(x^{16}-y^8=15*(...)\) will be divisible by 15 OR one could be multiple of 15 and another not (eg x=25*15 and y=20) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we can not factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\) --> \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 0 is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.

Answer: B.

Notes for statement (1):

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

So according to above info that \(x\) is a multiple of 25, and \(y\) is a multiple of 20 tells us nothing whether \(x^{16}-y^8\) is divisible by 15.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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25 Mar 2009, 20:01

3

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Actually I came up with a slightly different explanation. I may be biased because I saw the answer accidentally before solving the problem.

Is x^16 -y^8+345 y^2 divisible by 15 - This can be broken down to :

x^16 -y^8+345 y^2 is divisible by 5 and 3 .

Statement 1: x is a multiple of 25 and y is a multiple of 20. So lets say x=25k y =20l now we have :

is (25k)^16 - (20l)^8 + 345*(20l)^2 divisible by 5 and 3.

This term is definitely divisible by 5 because all the individual terms end up in a 0 in the units place. However it cannot be determined if they are divisible by 3. So this is not sufficient.

Statement 2; y=x^2. So the question now becomes

Is x^16 - x^16 + 345 * x^4 divisible by 5 and 3

Is 345 * x^4 divisible by 5 and 3

Yes, because 345 is divisible by both 3 and 5 and any multiple of 345 has to be divisible by 3 and 5. Hence sufficient.

Answer B.

Though the explanation looks long, it took me under 2 minutes to solve.

-pradeep _________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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25 Mar 2009, 15:07

I see your point.

Yes, the 100 and 40 numbers are not a good example here as you pointed out. Let's try 50 and 60 for that. It gets really simpler. \(50^2 - 60 = 2500 - 60 = 2440\), \(50^2 + 60 = 2500 + 60 = 2560\), neither of which is divisible by 3. So this example proves that S1 is not sufficient. It was trickier than I thought.

I'll have to change the OE a bit. I still think that picking numbers is the fastest method here. You're right that we need to use different numbers. +1 Kudos to you!

asthanap wrote:

I am not very much convinced with strategy mentioned for option 1. Example has used x = 100 and Y as 40. Since both are not divisible by 3 that means term X^16 - Y^8 is not divisible by 15. (X^2-Y) and (X^2 + Y) are factors of X^16 - Y^8 . And if plug in the value of X and Y in X^2-Y you get the number = 100^2 - 40^1 = 9960 and this is divisible by both 3 and 5. I tried couple of combinations and could find term X^16 - Y^8 divisible by both 3 & 5. I reached to conclusion that substituting numbers is not the right method for this.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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01 Jul 2009, 02:12

dzyubam wrote:

I see your point.

Yes, the 100 and 40 numbers are not a good example here as you pointed out. Let's try 50 and 60 for that. It gets really simpler. \(50^2 - 60 = 2500 - 60 = 2440\), \(50^2 + 60 = 2500 + 60 = 2560\), neither of which is divisible by 3. So this example proves that S1 is not sufficient. It was trickier than I thought.

I'll have to change the OE a bit. I still think that picking numbers is the fastest method here. You're right that we need to use different numbers. +1 Kudos to you!

asthanap wrote:

I am not very much convinced with strategy mentioned for option 1. Example has used x = 100 and Y as 40. Since both are not divisible by 3 that means term X^16 - Y^8 is not divisible by 15. (X^2-Y) and (X^2 + Y) are factors of X^16 - Y^8 . And if plug in the value of X and Y in X^2-Y you get the number = 100^2 - 40^1 = 9960 and this is divisible by both 3 and 5. I tried couple of combinations and could find term X^16 - Y^8 divisible by both 3 & 5. I reached to conclusion that substituting numbers is not the right method for this.

Is there any other explanation for this question?

Also in the explanation to the answer for Point (1) - the suggestion is that; as the two examples provided are not divisible by 3 - one cannot determine whether the expression is divisible by 15. this is slightly misleading. if the expression is dfntly not divisible by 3 then information is sufficient.

by proving that the expression may be divisible by 3 and may not not divisible by 3 - the information is insufficient. _________________

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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04 Sep 2009, 00:04

Did not get the highlighted part..

tejal777 wrote:

OE for stmt 1 is: We only need to check if expressions from first two parentheses are divisible by 3.

Can somebody explain this?

Note: x and y are +ve integers:

x^16 - y^8 + 345y^2 (x^16 - y^8) + 345y^2

In the above expression, 345y^2 is already divisible by 15 no matter the value of y. Now what we need to make sure whether (x^16 - y^8) is divisble by 15. To make sure (x^16 - y^8) is divible by 15, the difference of x^16 and y^8 should be divisible by 15.

From 1: If x = 25 and y = 20, (x^16 - y^8) is not divisible by 15. If x = 75 and y = 60, (x^16 - y^8) is divisible by 15.

If x = 100 and y = 40, (x^16 - y^8) is divisible by 15. If x = 100 and y = 60, (x^16 - y^8) is not divisible by 15.

NSF...

Statement 2 is self sufficient.. _________________

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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04 Sep 2009, 02:40

Thanks GT..two questions though. 1. When I said the first two parenthesis I meant the expansion of (x^16 - y^8).I should have clarified sorry.

The OE says: Statement (1) by itself is not sufficient. For a number to be a multiple of 15, it has to be divisible by both 3 and 5. As both 25 and 20 are multiples of 5, we need to check if is divisible by 3.(x^16 - y^8) can be rewritten as (x^2-y)(x^2+y)(x^4+y^2)(x^8+y^8) . We only need to check if expressions from first two parentheses are divisible by 3. We'll pick 50 and 60 for that purpose: Neither of these numbers is divisible by 3. Therefore, isn't divisible by 3 (neither by 15) either.

2.you wrote: From 1: If x = 25 and y = 20, (x^16 - y^8) is not divisible by 15. If x = 75 and y = 60, (x^16 - y^8) is divisible by 15. Could you please show the working? _________________

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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04 Sep 2009, 11:12

tejal777 wrote:

Thanks GT..two questions though. 1. When I said the first two parenthesis I meant the expansion of (x^16 - y^8).I should have clarified sorry.

The OE says: Statement (1) by itself is not sufficient. For a number to be a multiple of 15, it has to be divisible by both 3 and 5. As both 25 and 20 are multiples of 5, we need to check if is divisible by 3.(x^16 - y^8) can be rewritten as (x^2-y)(x^2+y)(x^4+y^2)(x^8+y^8) . We only need to check if expressions from first two parentheses are divisible by 3. We'll pick 50 and 60 for that purpose: Neither of these numbers is divisible by 3. Therefore, isn't divisible by 3 (neither by 15) either.

2.you wrote: From 1: If x = 25 and y = 20, (x^16 - y^8) is not divisible by 15. If x = 75 and y = 60, (x^16 - y^8) is divisible by 15.

= (x^2 - y) (x^2 + y) = (50^2 - 20) (50^2 + 20) = (2480) (2520) 2520 is divisible by 3 and 5 (i.e. 15). Suff...

3. If x = 75 and y = 40: = (x^2 - y) (x^2 + y) = (75^2 - 20) (75^2 + 20) = 5585 (5665) Neither term is divisible by 3 or 5 pr 15. NSF........... _________________

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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22 Nov 2010, 22:27

Here is my explanation: X16 –y8+345y2: should be divisible by 15. Implies that : X16 –y8+345y2: should be divisible by 5 & 3. Or, X16 –y8+5*3*23y2: It means, for this equation to be divisible by 15, X16 –y8 must have a 3 & 5 as factor. So, now, Case 1: Statement 1: x is multiple of 25 & y is multiple of 20. Or, x is multiple of 5*5 & y is multiple of 5*4. Both terms has 5 as a factor, but for these terms to be divisible by 15, it must have 3 as a common factor; Which is not present.

Hence, we can say that it is not divisible by 15 or not. Case 2: Statement 2: Y=x2, By replacing Y=x2 in the equation, we get, X16-x16+5*3*23x4 = 5*3*23 x4 which is divisible by 15.

Hence, Statement 1 can’t answer the question. Statement 2 alone can answer the question. _________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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25 Nov 2010, 15:45

Bunuel wrote:

If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?

First of all as \(345y^2\) is divisible by 15 (this term won't affect the remainder), we can drop it.

The question becomes: is \(x^{16}-y^8\) divisible by 15?

(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20 --> they both could be multiples of 15 as well (eg x=25*15 and y=20*15) and in this case \(x^{16}-y^8=15*(...)\) will be divisible by 15 OR one could be multiple of 15 and another not (eg x=25*15 and y=20) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we can not factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\) --> \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 0 is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.

Answer: B.

Notes for statement (1):

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

So according to above info that \(x\) is a multiple of 25, and \(y\) is a multiple of 20 tells us nothing whether \(x^{16}-y^8\) is divisible by 15.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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03 Jan 2011, 10:52

for 1, simplest option is to plug in 100 and 100 for x and y since 100 is a multiple of both 25 and 20. and you know right away that x^16 - y^8 is not divisible by 15. and you are set.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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27 Nov 2011, 12:42

Bunuel wrote:

If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?

First of all as \(345y^2\) is divisible by 15 (this term won't affect the remainder), we can drop it.

The question becomes: is \(x^{16}-y^8\) divisible by 15?

(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20 --> they both could be multiples of 15 as well (eg x=25*15 and y=20*15) and in this case \(x^{16}-y^8=15*(...)\) will be divisible by 15 OR one could be multiple of 15 and another not (eg x=25*15 and y=20) and in this case \(x^{16}-y^8\) won't be divisible by 15 (as we can not factor out 15 from \(x^{16}-y^8\)). Not sufficient.

(2) \(y = x^2\) --> \(x^{16}-y^8=x^{16}-(x^2)^8=x^{16}-x^{16}=0\). 0 is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.

Answer: B.

Notes for statement (1):

If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\) and \(b=9\), both divisible by 3 ---> \(a+b=15\) and \(a-b=-3\), again both divisible by 3.

If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)): Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 ---> \(a+b=11\) and \(a-b=1\), neither is divisible by 3.

If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)): Example: \(a=5\) and \(b=4\), neither is divisible by 3 ---> \(a+b=9\), is divisible by 3 and \(a-b=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 ---> \(a+b=9\) and \(a-b=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 ---> \(a+b=4\) and \(a-b=0\), both are divisible by 4.

So according to above info that \(x\) is a multiple of 25, and \(y\) is a multiple of 20 tells us nothing whether \(x^{16}-y^8\) is divisible by 15.

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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26 Nov 2012, 12:17

pleonasm wrote:

Actually I came up with a slightly different explanation. I may be biased because I saw the answer accidentally before solving the problem.

Is x^16 -y^8+345 y^2 divisible by 15 - This can be broken down to :

x^16 -y^8+345 y^2 is divisible by 5 and 3 .

Statement 1: x is a multiple of 25 and y is a multiple of 20. So lets say x=25k y =20l now we have :

is (25k)^16 - (20l)^8 + 345*(20l)^2 divisible by 5 and 3.

This term is definitely divisible by 5 because all the individual terms end up in a 0 in the units place. However it cannot be determined if they are divisible by 3. So this is not sufficient.

Statement 2; y=x^2. So the question now becomes

Is x^16 - x^16 + 345 * x^4 divisible by 5 and 3

Is 345 * x^4 divisible by 5 and 3

Yes, because 345 is divisible by both 3 and 5 and any multiple of 345 has to be divisible by 3 and 5. Hence sufficient.

Answer B.

Though the explanation looks long, it took me under 2 minutes to solve.

-pradeep

Hi Pradeep,

This is really a lovely explanation however couldn't understand how a multiple of 25 raised to any degree of power end up with ZERO at the units digit place?

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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26 Nov 2012, 12:33

Hi there,

To me there question is very straight forward. We've to find out whether each term of expression X^16 - Y^8+ 345y^2 is divisible by 15 or not i.e. X^16/15 then Y^8/15 and 345Y^2/15 With help of 1 we can see it's not possible because 25*25*..../15 having reminder then 20*20*.../15 is also having reminder however the last term is divisible by 15 With help of 2 we can see the first and second term got vanished and the third is divisible by 15 so the answer is B

Re: M02 02 : Not convinced with method used to explaing the ans [#permalink]

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20 Nov 2013, 11:36

Here is my approach:

For x^16 -y^8+345 y^2 to be divisible by 15 we need that at least every factor of this expression to be divisible by 15. Statement 1 does not guarantee this. Conclusion not sufficient. Discard A or D. We have B,C,E left Statement 2: y=x^2 Using statement 2 to simplify the expression x^16 -y^8+345 y^2 leads us to: = (x^2)^8 -y^8+345 y^2 = y^8 - y^8 + 345 y^2 = 345 y^2

Any number divisible by 15 will still by divisible when multiplied by any other number, so, since 345 is divisible by 15, then 345 y^2 will always be divisible by 15.

B sufficient.

gmatclubot

Re: M02 02 : Not convinced with method used to explaing the ans
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20 Nov 2013, 11:36