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If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis

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If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis  [#permalink]

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New post 21 Oct 2016, 09:13
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Question Stats:

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If x is positive integer, is x^4 - 1 divisible by 5?

(1) x-1 is divisible by 5.

(2) When x^2 + 1 is divided by 5, the remainder is 2.

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Re: If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis  [#permalink]

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New post 21 Oct 2016, 09:45
idontknowwhy94 wrote:
If x is positive integer, is \(x^{4}\)-1 divisible by 5?

1) x-1 is divisible by 5.

2) When \(x^{2}\)+1 is divided by 5, the
remainder is 2.


Statement 1: This holds true for 6,11,16,.. all of which have powers ending with 1 or 6. so \(x^{4}\)-1 has to be divisible by 5
Statement 2: When \(x^{2}\)+1 is divided by 5, the remainder is 2. This indicates that the number can never be divisible by 5.

Hence, D must be the answer. A question like this, where the 2 statements give 2 contradictory answers is highly unlikely to come.
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Re: If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis  [#permalink]

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New post 21 Oct 2016, 10:11
1
mankodim wrote:
idontknowwhy94 wrote:
If x is positive integer, is \(x^{4}\)-1 divisible by 5?

1) x-1 is divisible by 5.

2) When \(x^{2}\)+1 is divided by 5, the
remainder is 2.


Statement 1: This holds true for 6,11,16,.. all of which have powers ending with 1 or 6. so \(x^{4}\)-1 has to be divisible by 5
Statement 2: When \(x^{2}\)+1 is divided by 5, the remainder is 2. This indicates that the number can never be divisible by 5.

Hence, D must be the answer. A question like this, where the 2 statements give 2 contradictory answers is highly unlikely to come.



Hi mankodim

rephrasing Question x^4-1=(x^2-1)(x^2+1)----->(x-1)(x+1)(x^2+1) is divisible by 5??

1) x-1 is divisible means (x-1)(x+1)(x^2+1) is div by 5----->Yes.....suff
2)X^2+1 is div by 5 means x=4, 6,9,11,14,16.....(series having two consecutive even then two consecutive odd pairs)
if we take x=4 then (x+1) is div. by 5.....suff..
if x=6 then x-1 is div by 5 .....suff.

both cases weare getting number divisible by 5
Ans D
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Re: If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis  [#permalink]

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New post 21 Oct 2016, 11:06
1
idontknowwhy94 wrote:
If x is positive integer, is \(x^{4}\)-1 divisible by 5?

1) x-1 is divisible by 5.

2) When \(x^{2}\)+1 is divided by 5, the
remainder is 2.



x^4 -1 = 5z??

(x^2+1)(x+1)(x-1)=5z??

Statement 1:

Sufficient

Statement 2:

x^2 +1 = 5y+2

On simplifying

x^2 - 1 = 5y

We know that

x^4 -1 = (x^2 +1)(x^2 - 1) = (5y+2)*5y


Therefore it is definitely divisible by 5.

D


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Re: If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis  [#permalink]

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New post 21 Mar 2017, 12:35
Great explanation acegmat123. It took me a while to understand what you wrote, but I got it.

The question: Is \((x^4 - 1)\) divisible by 5?
In other words, is this true: \((x^4 - 1)/5 = quotient + 0/5\)
The first thing to do is to factor \(x^4 - 1\) which is equal to: \((x^2+1)(x+1)(x-1)\)

Statement 1:
\((x-1)\) is divisible by 5
Well if (x-1) is divisible by 5, then the entire statement is divisible by 5, because (x-1) is part of \(x^4 - 1\). This makes statement 1 sufficient.

Statement 2:
When \(x^2 +1\) is divided by 5, the remainder is 2. Or in other words:
\((x^2 +1)/5 = quotient + 2/5\)
Simplify to:
\((x^2 +1) = (5*quotient) + 2\)
\((x^2 - 1) = (5*quotient) + 0\)

This proves that \((x^2 - 1)\) is divisible by 5, and \((x^2 - 1)\) is part of \((x^2+1)(x+1)(x-1)\), so then \(x^4 - 1\) is divisible by 5. Statement 2 is sufficient. Both statements are sufficient by themselves, so the answer is D.
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Re: If x is positive integer, is x^4 - 1 divisible by 5? (1) x-1 is divis &nbs [#permalink] 21 Mar 2017, 12:35
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