Author |
Topic: divisible by 10 (Read 690 times) |
|
tony123
Junior Member
Posts: 61
|
 |
divisible by 10
« on: Sep 23rd, 2007, 1:20am » |
 Quote  Modify
|
a,b,c,d integers, a+b+c+d=0
prove
a^5+b^5+c^5+d^5 divisible by 10
|
|
 IP Logged |
|
|
|
BNC
Uberpuzzler
Gender: 
Posts: 1732
|
|
Re: divisible by 10
« Reply #1 on: Sep 23rd, 2007, 3:35am » |
Quote Modify
|
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.
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
srn437
Newbie
the dark lord rises again....
Posts: 1
|
|
Re: divisible by 10
« Reply #2 on: Sep 23rd, 2007, 12:15pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: divisible by 10
« Reply #3 on: Sep 23rd, 2007, 4:42pm » |
Quote Modify
|
For any integer, n, n=n5 mod 10. Since a+b+c+d=0 mod 10, so does the sum of their fifth powers.
|
|
IP Logged |
|
|
|
BNC
Uberpuzzler
Gender:
Posts: 1732
|
|
Re: divisible by 10
« Reply #4 on: Sep 24th, 2007, 4:41am » |
Quote Modify
|
on Sep 23rd, 2007, 4:42pm, SWF wrote:| For any integer, n, n=n5 mod 10. |
|
Interesting... I didn't know that.
Is there a name for this theorm?
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: divisible by 10
« Reply #5 on: Sep 24th, 2007, 5:09am » |
Quote Modify
|
Euler's Totient Theorem
[edit]Hm.... It is related but not quite that.[/edit]
|
| « Last Edit: Sep 24th, 2007, 5:15am by Grimbal » |
IP Logged |
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: divisible by 10
« Reply #6 on: Sep 24th, 2007, 8:14am » |
Quote Modify
|
on Sep 24th, 2007, 4:41am, 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
|
| « Last Edit: Sep 24th, 2007, 8:17am by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
SWF
Uberpuzzler
Posts: 879
|
|
Re: divisible by 10
« Reply #7 on: Sep 25th, 2007, 6:06pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
ecoist
Senior Riddler
Gender:
Posts: 405
|
|
Re: divisible by 10
« Reply #8 on: Sep 25th, 2007, 6:58pm » |
Quote Modify
|
Ha, ha, SWF, I didn't notice that! Thanks to you I now see that np-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.
|
|
IP Logged |
|
|
|
srn437
Newbie
the dark lord rises again....
Posts: 1
|
|
Re: divisible by 10
« Reply #9 on: Sep 25th, 2007, 7:45pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print