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Re: Is n divisible by 8? [#permalink]
31 Jan 2012, 15:34
2
This post received KUDOS
Expert's post
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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 07:46
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 (x-1), 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.
Hi,
Quick question: I do not understand why, when considering the product (x-1)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 (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 07:49
Expert's post
Samirc2 wrote:
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 (x-1), 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.
Hi,
Quick question: I do not understand why, when considering the product (x-1)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 (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 08:31
Bunuel wrote:
Samirc2 wrote:
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 (x-1), 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.
Hi,
Quick question: I do not understand why, when considering the product (x-1)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 (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 08:33
Expert's post
Samirc2 wrote:
Bunuel wrote:
Samirc2 wrote:
Hi,
Quick question: I do not understand why, when considering the product (x-1)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 (x-1) and (x+1) are even, hence my conclusion is that (x-1) and (x+1) are each divisible by 2, not by 4. So the only thing I can conclude is that (x-1)x(x+1) is divisible by 4 and not by 8. Could you please clarify this point?
As 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.
(x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 _________________
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
03 Apr 2013, 09:08
2
This post was BOOKMARKED
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.
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
12 Oct 2013, 05:48
1
This post was BOOKMARKED
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.
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
12 Oct 2013, 08:06
Expert's post
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 (x-1)x(x+1)=0*1*2=0, then the product is still divisible by 8.
\(x^3\) - x = n which will give x (\(x^2\)-1) = n which will give (x-1), 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 = 27-3 n= 24. 24/8 ->Remainder = 0
Lets take x = 5 n = x^3 - x = 125-5 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
(x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 _________________
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
09 Jul 2015, 23:27
Bunuel wrote:
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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
Answer: D.
Hope it helps.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient. How did u deduce this - x=even+odd=odd --> (x-1)x(x+1) _________________
Re: If x^3-x=n and x is a positive integer greater than 1, is n [#permalink]
10 Jul 2015, 00:47
Expert's post
honchos wrote:
Bunuel wrote:
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 (x-1), 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^3-x=x(x^2-1)=(x-1)x(x+1), notice that we have the product of three consecutive integers. Now, if x=odd, then (x-1) and (x+1) are consecutive even integers, thus one of them will also be divisible by 4, which will make (x-1)(x+1) divisible by 2*4=8 (basically if x=odd then (x-1)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 --> (x-1)x(x+1) is divisible by 8. Sufficient.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient.
Answer: D.
Hope it helps.
(2) x = 4y + 1, where y is an integer --> x=even+odd=odd --> (x-1)x(x+1) is divisible by 8. Sufficient. How did u deduce this - x=even+odd=odd --> (x-1)x(x+1)
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