Question 10808: is 1^5+2^5+3^5+.....+100^5 congruent to 0(mod 4).
i know that all even numbers in this problem are divisible by 4. my problem lie in the odd numbers since some of them has a remainder 1 and some 3.is there a pattern for these odd numbers to know how many of them has a remainder 1 and how many has a remainder 3.
Answer by khwang(438)   (Show Source):
You can put this solution on YOUR website! Your observation is all right, but note the remainders of 1 & 3
appearing in pairs.
Consider the sum of odd terms
1^5+3^5+5^5+.....+99^5
Use the same trick as finding the sum of arithemic series:
1^5+ 99^5 + 3^5+ 97^5+ 5^5 + 95^5 ...+ 49^5 + 51^5 (50 pairs)
Since a^5 + b^5 = (a + b) (a^4 - a^3b+ a^2b^2 - ab^3 +b^4)
, a^5 + b^5 is divided by a+b.
So, k^5 + (100 - k)^5 is divisible by 100. for all odd k, 1<= k <= 49.
Therefore, 100 is a factor of
1^5+ 99^5 + 3^5+ 97^5+ 5^5 + 95^5 ...+ 49^5 + 51^5
and so it is 0 mod 4.
Another way,as 1^5+3^5 = 4k1 = 0 mod 4.
5^5+ 7^5 = 0 mod 4 (since it is divisible by 12)
...
97^5 + 99^5 = 0 mod 4 (since it is divisible by 97+99 = 196)
In fact, for the odd numbers in the form 2k+1,
(2k+1)^5 mod 4 = 1 and for another form 2k+3 of odd numbers, we have
(2k+3)^5 = 3^5 = 9*9*3 = 1*1*3 = 3 mod 4.
Since 1+3 = 4 = 0 mod 4, we see that the remainders of odd terms alternating
between 1 and 3 and appeariing in pairs. We obtain this sum should be
0 mod 4.
Kenny
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