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The positive difference of the fourth powers of two consecut

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The positive difference of the fourth powers of two consecut  [#permalink]

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New post 08 May 2013, 21:54
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The positive difference of the fourth powers of two consecutive positive integers must be divisible by

(A) one less than twice the larger integer

(B) one more than twice the larger integer

(C) one less than four times the larger integer

(D) one more than four times the larger integer

(E) one more than eight times the larger integer

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Re: The positive difference of the fourth powers of two consecut  [#permalink]

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New post 08 May 2013, 22:24
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emmak wrote:
The positive difference of the fourth powers of two consecutive positive integers must be divisible by

(A) one less than twice the larger integer

(B) one more than twice the larger integer

(C) one less than four times the larger integer

(D) one more than four times the larger integer

(E) one more than eight times the larger integer


Let the consecutive integers be (x-1) and x. Thus, \(x^4-(x-1)^4 = [x^2-(x-1)^2][x^2+(x-1)^2] = (2x-1)*[x^2+(x-1)^2]\)

The larger integer is x. Thus it is always divisible by 2x-1.

A.
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Re: The positive difference of the fourth powers of two consecut  [#permalink]

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New post 08 May 2013, 22:33
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emmak wrote:
The positive difference of the fourth powers of two consecutive positive integers must be divisible by

(A) one less than twice the larger integer

(B) one more than twice the larger integer

(C) one less than four times the larger integer

(D) one more than four times the larger integer

(E) one more than eight times the larger integer


Using Picking Numbers for this, Let the two positive integers be 2, 3
\(3^4 - 2^4 = 81-16 = 65\)

(A) one less than twice the larger integer = 2(3) - 1 = 5 --> 65 is divisible by 5
(B) one more than twice the larger integer = 2(3) + 1 = 7 --> ruled out
(C) one less than four times the larger integer = 4(3) - 1 = 11 --> ruled out
(D) one more than four times the larger integer = 4(3) + 1 = 13 --> 65 is divisible by 13
(E) one more than eight times the larger integer = 8(3) + 1 = 25 --> ruled out

Another set of numbers, 3,4
\(4^4 - 3^4 = 256-81 = 175\)

(A) one less than twice the larger integer = 2(4) - 1 = 7 --> 175 is divisible by 7
(D) one more than four times the larger integer = 4(4) + 1 = 17 --> ruled out

Answer : A
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Re: The positive difference of the fourth powers of two consecut  [#permalink]

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New post 08 Aug 2018, 14:11
emmak wrote:
The positive difference of the fourth powers of two consecutive positive integers must be divisible by

(A) one less than twice the larger integer

(B) one more than twice the larger integer

(C) one less than four times the larger integer

(D) one more than four times the larger integer

(E) one more than eight times the larger integer


Can anyone help me with the solution? \((n+1)^4-n^4\) how to proceed after that?
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The positive difference of the fourth powers of two consecut  [#permalink]

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New post 08 Aug 2018, 20:35
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aghosh54 wrote:
emmak wrote:
The positive difference of the fourth powers of two consecutive positive integers must be divisible by

(A) one less than twice the larger integer

(B) one more than twice the larger integer

(C) one less than four times the larger integer

(D) one more than four times the larger integer

(E) one more than eight times the larger integer


Can anyone help me with the solution? \((n+1)^4-n^4\) how to proceed after that?


Hi aghosh54,

You know \(a^2-b^2=(a+b)*(a-b)\), Since \((n+1)^4-n^4\) is in a similar format. Let's apply it:-

\((n+1)^4-n^4\)=\(((n+1)^2)^2-(n^2)^2\), where a=n+1 and b= n

So, \(((n+1)^2)^2-(n^2)^2\)=\(((n+1)^2+n^2)((n+1)^2-n^2)=(2n^2+2n+1)(2n+1)\)=\((2n^2+2n+1)(2n+1+1-1)=(2n^2+2n+1)(2(n+1)-1)\)

As per your picked numbers, (n+1) is the larger number.

So, the given expression is divisible by 2(n+1)-1, which is one less than twice the larger integer.

Ans. (A)

Hope it helps.
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The positive difference of the fourth powers of two consecut &nbs [#permalink] 08 Aug 2018, 20:35
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