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#1 2007-09-22 20:20:03

tony123
Member
Registered: 2007-08-03
Posts: 228

divisible by 10

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

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

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#2 2007-09-23 02:19:10

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: divisible by 10

This works because of the curious fact that x and x^5 always have matching last digits, and are either both positive or both negative (or both 0).

Therefore, if a+b+c+d=0, then the last digits of a^5, b^5, c^5 and d^5 will also add to 0, if you attach minuses where needed. The last digit of that sum is 0, which makes it divisible by 10.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-09-23 08:02:45

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: divisible by 10

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#4 2007-09-23 08:38:20

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: divisible by 10

Here is a somewhat related problem:

For any prime p, 5 divides p^4 - 1


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2007-09-23 09:00:42

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: divisible by 10

(When p≠5)


Why did the vector cross the road?
It wanted to be normal.

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