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        riddles >> medium >>  divisible by 10
        
(Message started by: tony123 on Sep 23^rd, 2007, 1:20am)

Title: divisible by 10
Post by tony123 on Sep 23^rd, 2007, 1:20am

a,b,c,d integers, a+b+c+d=0
prove

a^5+b^5+c^5+d^5 divisible by 10

Title: Re: divisible by 10
Post by BNC on Sep 23^rd, 2007, 3:35am

a+b+c+d=0 => d=-(a+b+c)
We need to prove:
a^5+b^5+c^5-(a+b+c)^5 = 0 mod(10)

Expanding (a+b+c)^5 gives us fifth order terms a^5+b^5+c^5 which cancel the other fifth order terms.
The other terms are mostly with coefficients of 10, 20, or 30. Given a, b, and c are integers, these terms are divisible by 10.

The remainder is: 5(a^4b+a^4c+b^4a+b^4c+c^4a+c^4b). Redistribute to 5ab(a^3+b^3) + 5ac(a^3+c^3)+abc(b^3+c^3).

Look, e.g., at 5ab(a^3+b^3): if at least one term (a or b) is even, then ab is even, and the expression is divisible by 10. If both odd, then both a^3 and b^3 are odd, hence (a^3+b^3) is even and the expression is divisible by 10. The same goes for the expressions with (b,c) and (a,c).

QED.

Title: Re: divisible by 10
Post by srn347 on Sep 23^rd, 2007, 12:15pm

a+b=-c-d
raise both sides to the fifth power
then move c^5 + d^5 to the left side
a^5+b^5+c^5 +d^5 =-10a^3 b^2-10a^2 b^3 -5ab^4 -5a^4 b -10c^3 d^2 -10c^2 d^3 -5cd^4 -5c^4 d
so a^5+b^5+c^5+d^5 is a multiple of 10 if everything else is. The variables with a coeffiecient of 10 drop out. All that is left is proving that 5ab^4 +5ba^4 +5cd^4 +5dc^4 is a multiple of 10. The 5's cancel out. ab^4 +ba^4 +cd^4 +dc^4 has to be even. Since 0 is even and a+b+c+d=0, a+b+c+d is even, so either they are all even, they are all odd, or two are odd and two are even. If they are all even, ab^4 +ba^4 +cd^4 +dc^4 must be even. If they are all odd, then since multiplying odds gets odds, it is the sum of 4 odd integers, which is even. If two are odd and two are even, if a and b are the odds, it is the sum of two odds and two evens(which is even). Same for c and d being the odds. If a and d are the odds, they are multiplied by even and it is the sum of two evens(which is even). Same for b and c being odd. Sorry about this answer being very long, but it is still the correct answer.

Title: Re: divisible by 10
Post by SWF on Sep 23^rd, 2007, 4:42pm

For any integer, n, n=n⁵ mod 10. Since a+b+c+d=0 mod 10, so does the sum of their fifth powers.

Title: Re: divisible by 10
Post by BNC on Sep 24^th, 2007, 4:41am

on 09/23/07 at 16:42:10, SWF wrote:

For any integer, n, n=n⁵ mod 10.

Interesting... I didn't know that.
Is there a name for this theorm?

Title: Re: divisible by 10
Post by Grimbal on Sep 24^th, 2007, 5:09am

~~Euler's Totient Theorem (http://mathworld.wolfram.com/EulersTotientTheorem.html)~~

[edit]Hm.... It is related but not quite that.[/edit]

Title: Re: divisible by 10
Post by towr on Sep 24^th, 2007, 8:14am

on 09/24/07 at 04:41:38, BNC wrote:

Interesting... I didn't know that.
Is there a name for this theorm?

I'm not sure it's worth a theorem, you only need to check 6 cases:
0 : 0⁵ % 10 = 0
1 : 1⁵ % 10 = 1
2 : 2⁵ % 10 = 32 % 10 = 2
3 : 3⁵ % 10 = 243 % 10 = 3
4 : 4⁵ % 10 = 1024 % 10 = 4
5 : 5⁵ % 10 = 3125 % 10 = 5
For 5 < n < 10 take 10 + [(n-10)⁵ % 10] = 10 - [(10-n)⁵ % 10], and all the [(10-n)⁵ % 10] are already known to be 10-n; 10 - (10-n) = n

Title: Re: divisible by 10
Post by SWF on Sep 25^th, 2007, 6:06pm

Not only is n^5-n a multiple of 10, but it is also evenly divisible by 3. Just factor it to see why. The sum of 5th powers is therefore a multiple of 30.

Title: Re: divisible by 10
Post by ecoist on Sep 25^th, 2007, 6:58pm

Ha, ha, SWF, I didn't notice that! Thanks to you I now see that n^p-n is divisible by 6p, for all primes p>3. And is also divisible by 10, for primes p>5 and congruent 1 modulo 4.

Title: Re: divisible by 10
Post by srn347 on Sep 25^th, 2007, 7:45pm

n⁵-n is divisable by 3, but you never asked why, so I shall answer the question with reverse psychology. n is either 3k+1, 3k-1, or 3k. Assuming it isn't 3k(because that ones given), if it's 3k+1, you get 27k⁵+252k³+84k²+15k+1-3k-1, which is a bunch of terms of k and coefficients of 3. 3k-1 is 243⁵-405k[sup]4/sup]+36k[sup]3/sup]-84k[sup]2+15k-1-3k+1, which is the same story as the former one.