Congruences. $1^5 + 2^5 + 3^5 + \cdots + 11^5 \equiv i \pmod {3}$

Question

Find $i$ if $1^5 + 2^5 + 3^5 + \cdots + 11^5 \equiv i \pmod {3}$

I know that $1^5 + 2^5 + 3^5 + \cdots + 11^5 \equiv (1 + 2 + 3 \cdots + 11)^5 \pmod {5}$ but I don´t know any property that could make the module smaller or something.

note: $1 +2 +3 \cdots +11 = 66 = 2*3*11$

Doing this with a calculator yields $i = 0$, I would really appreciate a hint or something.

Answer 1

By Fermat's little theorem, $k^5\equiv k^3\equiv k\pmod 3$. And $1+2+\cdots+11=66$, as you wrote. So, you can take $i=0$.

Answer 2

I'm not seeing the confusion here, as it looked like you already answered it--almost.

$x^5 \equiv_3 x$, so

$$1^5+2^5+\ldots + 11^5 \equiv_3 1+2+3+\ldots + 11 \equiv_3 66 \equiv_3 0$$

So $i=0$.

Meanwhile for general $n$ note that

$$1^5+2^5+ \ldots + n^5 \equiv_3 1+2+3+\ldots +n \equiv_3 1+2+0+1+2+0 \ldots (n \mod 3)$$

This is 1 iff $n \equiv_3 1$, and 0 otherwise.

Answer 3

Here's a different solution.

It is known that $$\sum_{k=1}^n k^5=\dfrac16n^6+\dfrac12n^5+\dfrac5{12}n^4-\dfrac1{12}n^2=\dfrac{(n(n+1))^2(2n^2+2n-1)}{12}.$$

From here it is clear that $\sum_{k=1}^{12} k^5$ is divisible by $12,$

and therefore $\sum_{k=1}^{11} k^5$ is divisible by $12$ and therefore by $3$.

Answer 4

A quick solution here... \begin{eqnarray} 1^5+2^5+\cdots+11^5\equiv(1^5+4^5+7^5+10^5)+(2^5+5^5+8^5+11^5)+(3^5+6^5+9^5)\\\equiv(4)+(4\times2^5)\equiv 4(1^5+2^5)\equiv0\pmod{3} \end{eqnarray}

Answer 5

Natural numbers mod $3$ are $$1,-1,0,1,-1,0,1,-1,0,1,-1,0,...$$

Note that the fifth powers are the same numbers as the first powers mod $3$

Adding the first $11$ results in $0$

Answer 6

More is true: Let $n$ and $m$ be any positive integers. Then

$$\sum_{k=1}^{2n-1}k^{2m+1}=\sum_{k=-(n-1)}^{n-1}(n+k)^{2m+1}\equiv\sum_{k=-(n-1)}^{n-1}k^{2m+1}=0\mod n$$

For the problem at hand, take $n=6$ and $m=2$. Then

$$\sum_{k=1}^{11}k^5\equiv0\mod6$$

and thus, since $3\mid6$, the sum is congruent to $0$ mod $3$ as well. Note, Fermat's Little Theorem and the primality of $3$ play no role in this approach, only the (odd) parity of $5$.

Answer 7

Using lil' Fermat, we know that for any $a$, $a3\equiv a\mod 3$, whence $a^5\equiv a^3\equiv a$.

Now, as any integer is congruent to $0,1$ or $-1\bmod 3$, any sum of three consecutive integers is congruent to $0$, so that $$1^5 + 2^5 + 3^5 + \cdots + 11^5 \equiv1 + 2 + 3 + \cdots + 11 \equiv 10+11\equiv 0\mod 3.$$