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# Is (n^5/5)+(n^3/3)+(7n/15) an integer? 1. n is odd 2. n is

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Is (n^5/5)+(n^3/3)+(7n/15) an integer? 1. n is odd 2. n is [#permalink]  12 Apr 2004, 11:47
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Is (n^5/5)+(n^3/3)+(7n/15) an integer?

1. n is odd
2. n is even
Senior Manager
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agree with D
the initial thing can be transformed into:

X=[(3n^5)+(5n^3)+7n]/15

For X to be an integer, its numerator has to be divisible by 15.
consider (3n^5)+(5n^3)+7n closely; it is always divisible by 15 if n is an integer

my way to prove it is too long and not elegant...
another way is to take some numbers 0, 1, -1, 2, -2,....
Senior Manager
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stolyar wrote:
agree with D
the initial thing can be transformed into:

X=[(3n^5)+(5n^3)+7n]/15

For X to be an integer, its numerator has to be divisible by 15.
consider (3n^5)+(5n^3)+7n closely; it is always divisible by 15 if n is an integer

my way to prove it is too long and not elegant...
another way is to take some numbers 0, 1, -1, 2, -2,....

How do you know that n is always devided by 15 if n is an integer?
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Re: DS-43 [#permalink]  14 May 2004, 04:14
hallelujah1234 wrote:
Is (n^5/5)+(n^3/3)+(7n/15) an integer?

1. n is odd
2. n is even

(3*n^5 + 5*n^3 + 7*n)/15 = n(3n^4+5*n^2 + 7)/15.

3n^4+5n^2+7 = (-2n^4+2)+5(n^4+n^2+1).

by the theorem of Ferma: n(n^4-1) :: 5, and by the theorem of Ferma n(n^2-1) :: 3. => first part is interger anyway.

Second part: n(n^4+n^2+1)=n([n^4-n^2]+[2n^2+1]). Again, by the theorem of Ferma, n(n^4-n^2)=n(n^2-1)*(n^2) :: 3. n(2n^2+1) = n(2n^2-2+3) :: 3 again be the theorem of Ferma!

Regardless of 1 or 2 the stem is true!

P.S. theorem of Ferma: n^p-n always :: p, where p is prime, n - integer.
Re: DS-43   [#permalink] 14 May 2004, 04:14
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