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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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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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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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(When p≠5)
Why did the vector cross the road?
It wanted to be normal.
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