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-   -   Always an integer. (https://www.mersenneforum.org/showthread.php?t=8628)

mfgoode 2007-07-08 16:16

Always an integer.
 
:rolleyes:

Show that (x^5)/5 + (x^3)/3 +(7x/15) is always an integer for integral values of x ?

Its easy if you know the method.

Similar problems are welcome.

Mally :coffee:

VolMike 2007-07-08 17:24

Well, I don't know the simplest way to reduce this problem, but it can be easy solved with one qiute obviously notice: any integer value can be represented as element of the set {15*k,15*k+1,15*k+2,...,15*k+14} for integer k=0,1,... Thus we can substitute this values for x in out initial expression and get results . Mathematica code, returns values of initial expression by substituting for x: [code] (1/15) x (7 + 5 x^2 + 3 x^4) /. x -> 15*k + Range[0, 14] // Expand [/code] And results: [code] {7 k + 1125 k^3 + 151875 k^5, 1 + 37 k + 675 k^2 + 7875 k^3 + 50625 k^4 + 151875 k^5, 10 + 307 k + 4050 k^2 + 28125 k^3 + 101250 k^4 + 151875 k^5, 59 + 1357 k + 12825 k^2 + 61875 k^3 + 151875 k^4 + 151875 k^5, 228 + 4087 k + 29700 k^2 + 109125 k^3 + 202500 k^4 + 151875 k^5, 669 + 9757 k + 57375 k^2 + 169875 k^3 + 253125 k^4 + 151875 k^5, 1630 + 19987 k + 98550 k^2 + 244125 k^3 + 303750 k^4 + 151875 k^5, 3479 + 36757 k + 155925 k^2 + 331875 k^3 + 354375 k^4 + 151875 k^5, 6728 + 62407 k + 232200 k^2 + 433125 k^3 + 405000 k^4 + 151875 k^5, 12057 + 99637 k + 330075 k^2 + 547875 k^3 + 455625 k^4 + 151875 k^5, 20338 + 151507 k + 452250 k^2 + 676125 k^3 + 506250 k^4 + 151875 k^5, 32659 + 221437 k + 601425 k^2 + 817875 k^3 + 556875 k^4 + 151875 k^5, 50348 + 313207 k + 780300 k^2 + 973125 k^3 + 607500 k^4 + 151875 k^5, 74997 + 430957 k + 991575 k^2 + 1141875 k^3 + 658125 k^4 + 151875 k^5, 108486 + 579187 k + 1237950 k^2 + 1324125 k^3 + 708750 k^4 + 151875 k^5} [/code] As can see, all expression values are integers for integer k. I think there is an shorter solution which uses the same idea, but represents in a brief way (expectedly with usage of properties of modulo operation).

axn 2007-07-08 17:27

[QUOTE=mfgoode;109867]Show that (x^5)/5 + (x^3)/3 +(7x/15) is always an integer for integral values of x[/QUOTE]

[spoiler]
Alternately we need to show that 3x^5 + 5x^3 + 7x is a multiple of 15.

Working modulo 3, we have x^3=x (by Fermat's Little Theorem)
Thus 3x^5 + 5x^3 + 7x == 3x^5 + 5x + 7x == 3x^5 + 12x == 0 (mod 3)

Working modulo 5, we have x^5=x
Thus 3x^5 + 5x^3 + 7x == 3x + 5x^3 + 7x == 10x + 5x^3 == 0 (mod 5)

Thus our expression is divisible by 15, as required.

Q.E.D
[/spoiler]

davieddy 2007-07-08 18:35

I tried to show 3x^5+5x^3+7x was a multiple of 15
by induction and failed:sad:

fivemack 2007-07-08 19:14

Proofs by working to a single modulus M are, surely, precisely proofs by induction; just with M distinct base cases and the induction rule being n -> n+M.

But in fact it seems to work for me by straight n->n+1 induction:

3(x+1)^5 = 3x^5 + 15x^4+30x^3+30x^2+15x+3
5(x+1)^3 = 5x^3 + 15x^2 + 15x + 5
7(x+1) = 7x + 7

so the sum is 3x^5 + 5x^3 + 7x (ex hypothesi divisible by 15)
+ lots of things which are in form a multiple of 15
+ 3 + 5 + 7 which equals 15

davieddy 2007-07-08 19:50

Thanks FiveMack.
Actually I tried to prove that 3x^4 +5x^2 +7
was divisible by 15. This could indeed be false,
when x is a multiple of 3 and/or 5.
David

And THX Mally - a successful puzzle :)
Look forward to the next!

mfgoode 2007-07-09 15:52

Precise and concise!
 
[QUOTE=davieddy;109876]Thanks FiveMack.
Actually I tried to prove that 3x^4 +5x^2 +7
was divisible by 15. This could indeed be false,
when x is a multiple of 3 and/or 5.
David

And THX Mally - a successful puzzle :)
Look forward to the next![/QUOTE]

:rolleyes:

Hold your horses Davie. Here is a precise and concise solution you can ever get. Thats why I put this problem.

Its the type Alpertron and Maxal will appreciate so here it is.

original Expression = [(x^5-x)/5] + [(x^3-x)/3] + x.
But by Fermat's theorem x^5 - x == 0 mod 5 and x^3 -x ==Mod 3 hence the expression is an integer!

Q.E.D

Mally :coffee:

davieddy 2007-07-09 19:23

[quote=mfgoode;109867]
Similar problems are welcome.

Mally :coffee:[/quote]

Neat.
Now we can generalize the problem to contain
any number of terms x^p/p where p is prime.
e.g.
x^7/7 +x^2/2 + 5x/14

David

m_f_h 2007-07-09 23:49

[quote=mfgoode;109930]original Expression = [(x^5-x)/5] + [(x^3-x)/3] + x.[/quote]
For me [(x^n-x)/n] is an integer for any n and x...
since [.] means floor(.) to me...
especially when it is put where it is not needed for anything else like grouping factors of a product or function arguments etc...:confused:

m_f_h 2007-07-09 23:58

[quote=davieddy;109953]we can generalize the problem to contain any number of terms x^p/p where p is prime. e.g.
x^7/7 +x^2/2 + 5x/14[/quote]
or x^11/11+x^101/101+999x/1111...

mfgoode 2007-07-12 08:33

General Eqn.
 
[QUOTE=davieddy;109953]Neat.
Now we can generalize the problem to contain
any number of terms x^p/p where p is prime.
e.g.
x^7/7 +x^2/2 + 5x/14

David[/QUOTE]

:smile:

Excellent Davie!

I would like you to note that the general expression is true for all +ve integers when p is a prime It is also true if p divides x. This is a corollary which results from Fermat’s little theorem.

Thus from your very expression when x is any number e.g. x = 14
14^7/7 + 14^2/2 +5*14/7 an integer although 7 and 2 both divide 14

A more restricted expression can be derived from Fermat’s little theorem, thus
x^6/7 + x^2/3 – 10/21 is also true and always an integer provided the primes
(In denominator) do not divide x , and the constant is changed accordingly as it’s not a function of x but rather of the primes.

Having said that could you derive the general expression in terms of x, p_1, and p_2……?

Try it with x = 5, 11 and any prime.

Mally :coffee:.


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