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If x^3x=n and x is a positive integer greater than 1, is n
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31 Jan 2012, 16:25
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63% (02:07) correct 37% (02:31) wrong based on 730 sessions
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If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.
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Re: Is n divisible by 8?
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31 Jan 2012, 16:34
enigma123 wrote: If \(x^3\) – x = n and x is a positive integer greater than 1, is n divisible by 8?
(1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.
I will really appreciate if you can tell me whether I am right or wrong. I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. Yes, your reasoning is correct. If x^3  x = n and x is a positive integer greater than 1, is n divisible by 8? x^3x=x(x^21)=(x1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8 (basically if x=odd then (x1)x(x+1) will be divisible by 8*3=24). (1) When 3x is divided by 2, there is a remainder > 3x=odd > x=odd > (x1)x(x+1) is divisible by 8. Sufficient. (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient. Answer: D. Hope it helps.
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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03 Apr 2013, 09:33
Samirc2 wrote: Bunuel wrote: Samirc2 wrote: Hi,
Quick question: I do not understand why, when considering the product (x1)x(x+1), you conclude that one of the factor is divisible by 4.
Let me explain my line of reasoning here: we know x is odd, so both (x1) and (x+1) are even, hence my conclusion is that (x1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
Thanks
By the way awesome book, thanks for sharing
Sam
Please check here: ifx3xnandxisapositiveintegergreaterthan1isn126854.html#p1037233As you could see from my post, I did check the previous posts and quoted them but I cannot figure out what is wrong with my reasoning. (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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03 Apr 2013, 10:08
Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0 so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor. Hope this helps..
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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12 Oct 2013, 06:48
nt2010 wrote: Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0
so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.
Hope this helps.. I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4.



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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12 Oct 2013, 09:06
audiogal101 wrote: nt2010 wrote: Just list out all the even numbers  0, 2, 4, 6, 8, 10, 12, 14, 16 and for this problem x > 0
so consecutive even numbers are 2, 6, 8, 10, 12, 14, 16, ... = 2 x ( 1, 2, 3, 4, 5, 6, 7, 8,..).. every even number has 2 as a factor and every even number in the listed ( 1, 2, 3, 4, 5, 6, 7, 8,..) series gives another 2 as a factor. Hence, in a sequence of even numbers one of the numbers always has 4 as a factor.
Hope this helps.. I believe the last statement would be correct only if the sequence starts with even integer 2. For instance, if there are two consecutive even numbers, 0 and 2, then 4 is not a factor. However, if the two consecutive numbers are 2 and 4, then 4 is a factor of 2*4. That's not true. Zero is divisible by EVERY integer except zero itself, since 0/integer=integer (or, which is the same, zero is a multiple of every integer except zero itself). Thus if (x1)x(x+1)=0*1*2=0, then the product is still divisible by 8. Hope it's clear.
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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04 Dec 2013, 12:47
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. Statement 1: 3x/2 gives remainder . This means x is odd. odd^3 = odd and odd odd = even. x >1 and an odd integer . Lets take x = 3 n = x^3  x = 273 n= 24. 24/8 >Remainder = 0 Lets take x = 5 n = x^3  x = 1255 n= 120. 120/8 >Remainder = 0 Sufficient Statement 2: x = 4y+1 . y>integer Put x in equation n = x^3  x n = (4y+1)^3  (4y+1) apply formula(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 (4y+1)^3 = 64y^3 + 3*16x^2 + 3*4y*1 + 1 = 64y^3 + 48x^2 + 12y + 1 n = 64y^3 + 48x^2 + 12y + 1  4y  1 n = 64y^3 + 48x^2 + 8y n = 8(8y^3 + 6x^2+1) Hence n is divisible by 8. > Sufficient Hence D



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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25 Dec 2013, 12:56
Is there a general pattern for what the remainder is when the square of an odd number is divisible by even numbers?
For instance, is the remainder always 1 when divided by 2, 4 and 8?



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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20 Jun 2014, 09:10
(2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so.



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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20 Jun 2014, 09:14
sagnik242 wrote: (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so. \(x = 4y + 1\). Now, \(4y\) is even, because of 4, and 1 is odd, thus \(x=even+odd=odd\). Does this make sense?
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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20 Jun 2014, 10:18
Bunuel wrote: sagnik242 wrote: (2) x = 4y + 1, where y is an integer > x=even+odd=odd > (x1)x(x+1) is divisible by 8. Sufficient.
Question : x = even + odd + odd, where does this come from? I think it comes from (2) but I don't get how if that's so. \(x = 4y + 1\). Now, \(4y\) is even, because of 4, and 1 is odd, thus \(x=even+odd=odd\). Does this make sense? Yes thanks so much:)



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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10 Jul 2014, 23:47
if x is odd, how is x(x+1)(x1) divisible by 8. I can't understand. Help?



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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11 Jul 2014, 11:45
hamzakb wrote: if x is odd, how is x(x+1)(x1) divisible by 8. I can't understand. Help? Please read the whole thread: ifx3xnandxisapositiveintegergreaterthan1isn126854.html#p1207542(x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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14 Jan 2017, 05:43
Great Question. Here is what i did in this one=> As n=x^3x=> (x1)*x*(x+1)=> If x is odd => both x1 and x+1 will be consecutive even integers.So one of them would be multiple of 4 and other would be a multiple of 2 for sure.Hence n would be a multiple of 8.
Statement 1> 3x is odd => x is odd => Sufficient. Statement 2> x=4y+1 => Even+odd=odd Hence Sufficient.
Hence D
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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21 Sep 2017, 03:23
Hi, I don't understand why if x is odd then (x1)x(x+1) will be divisible by 8*3=24. Can someone pls help?



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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11 Oct 2017, 21:24
JsZJ wrote: Hi, I don't understand why if x is odd then (x1)x(x+1) will be divisible by 8*3=24. Can someone pls help? x^3x=x(x^21)=(x1)x(x+1), notice that we have the product of three consecutive integers, so one of (x1), x or (x+1) is divisible by 3. Now, if x=odd, then (x1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x1)(x+1) divisible by 2*4=8 (basically if x=odd then (x1)x(x+1) will be divisible by 8*3=24).
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Re: If x^3x=n and x is a positive integer greater than 1, is n
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12 Oct 2017, 02:42
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8?
(1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer.
x & n are Positive integers, so zero is not considered here and x > 1 n = x^3 – x = x (x^2  1) 1) 3x/2 yields reminder.........means x is odd number Let x = 3 ..........3 (91) = 3 * 8...........Answer is yes Let x = 5 ..........5 (251) = 5 *24...........Answer is yes Let x = 7 ..........7 (491) = 7 *4 8...........Answer is yes Let x = 9 ..........9 (811) = 9 * 80...........Answer is yes Let x = 15 ..........15 (2251) = 15 * 224...........Answer is yes There is pattern here Sufficient (2) x = 4y + 1, where y is an integer. Here, no need to rush in plugging number...........the equation yields odd number as above Sufficient Answer: D



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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03 Jan 2018, 19:34
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question.  x3x =n & x(2,3,...) Is x divisible by 8 ? stmt1: 3x = 2q + 1 => x = odd and can be represented by 2k+1 x3x = x(x+1)(x1) =(2k+1)(2k+2)(2k) =4(2k+1)(k+1)(k) Now (k+1)(k) is a product of 2 consecutive integers, one of which HAS to be even thus this product is even so x3x (=n) is divisible by8 => Sufficient Stmt 2: x=4y+1 n= x(x+1)(x1) = (4y+1)(4y+2)(4y) 8(4y+1)(2y+1)(y) => n is divisbile by 8 => Sufficient Both stmts are sufficient by itself



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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10 Sep 2018, 19:13
enigma123 wrote: If x^3 – x = n and x is a positive integer greater than 1, is n divisible by 8? (1) When 3x is divided by 2, there is a remainder. (2) x = 4y + 1, where y is an integer. I will really appreciate if you can tell me whether I am right or wrong: I have gone for D as an answer.
Let's simply the question stem a bit:
\(x^3\)  x = n which will give x (\(x^2\)1) = n which will give (x1), x and (x+1) = n. Therefore n is a product of 3 consecutive integers.
Now moving on to the statements
Statement 1 implies that x is ODD. Because ODD/EVEN will always end with some remainder.
X is ODD which means that the product will always be divisible by 2 three times and therefore will be divisible by 8.
Statement 2 implies that when x is divided by 4 it leaves a remainder 1 i.e. x is ODD. which is same as above and therefore sufficient.
Therefore for me both statements alone are sufficient to answer this question. my 'equal' key is not working so using colon : as equal!! 3x:2p+2 p:2 > 3x:6 >x:2...... 1x2x3 ....not div by 8 p:3 > 3x:8 >x:8/3..... 5/3 x 8/3 x 11/3 .... div by 8 p:4 > 3x:10 > x:10/3 .... 7/3 x 10/3 x 13/3 .... not div by 8 then how come 1 is sufficient?



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Re: If x^3x=n and x is a positive integer greater than 1, is n
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11 Apr 2019, 07:10
Good question.
Stmt 1: 3x=2a+r ===> x=(2a+r)/3. Since x is an integer, then x is divisible by 3 and minimum x=3 ===> (2)*(3)*(4) is divisible by 8 (and any other sequence is). Stmt 2: x=4y+1. Since x cant be 1 then y>0. For minimum y=1, then x=5 ===> (4)*(5)*(6) is divisible by 8 (and any other sequence is).
Therefore (D).




Re: If x^3x=n and x is a positive integer greater than 1, is n
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