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I want to find the residue by dividing $$1^5+2^5+....+1080^5$$ by $14$. Im supossed to attack this problem by using congruences but Im not pretty sure how to do it. All I know is that if $14 | 1^5+2^5+....+1080^5$ then $$14k+r=1^5+2^5+....+1080^5,$$

where $k \in \mathbb{Z}$ and $0 \leq r < k$. Please I aprreciate any help of this proof using congruences.

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  • $\begingroup$ If $a \mid b$, then $b = ak$ (so that $r = 0$) for some $k \in \Bbb Z$ $\endgroup$ – JavaMan Oct 30 '18 at 2:12
  • $\begingroup$ Hint: $(14k+r)^n = 14[k(14k+r)^{n-1}] + r^n$ $\endgroup$ – eyeballfrog Oct 30 '18 at 2:12
  • $\begingroup$ Hint: What is $1^5 + 2^5 + \dots + 13^5 + 14^5$? Then what is $1^5 + \dots + 14^5 + 15^5 + \dots 28^5$? $\endgroup$ – JavaMan Oct 30 '18 at 2:14
  • $\begingroup$ I need to use $1 \equiv some number(mod 14) , ....., 14 \equiv some number (mod 14)$ ? @JavaMan $\endgroup$ – Cos Oct 30 '18 at 2:30
  • $\begingroup$ Hint: the sum is even, so you only need to figure out the sum modulo $7$ and use Chinese Remainder theorem. $\endgroup$ – Thomas Andrews Oct 30 '18 at 2:37
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I will give you 'help' as you you have asked.

This procedure might be a bit long but will be most rewarding as you are in the learning stage: (i) To calculate $a^5 \pmod {14}$, you do not need to calculate 5th power and then divide by 14 for remainder. AT all intermediate stages of calculation you can reduce the intermediate answers modulo 14.

Example with $3^5$. Stop at $3^3=27$. This is one less than 28, and $28=0\pmod {14}$. So $3^3\equiv -1\pmod{14}$. Now $3^5= 3^3\times 3^2$ can be replaced by $-3^2\pmod{14}$, and this is $-9$ or $5$.

This way calculate $a^5$ for $a<8$. F0r $a=8,9,\ldots 13$, use $14-a = -a$. So, for example $11^5=- (3^5)-+9\pmod{14}$.

Now you can add up all $a^5$ for $a=0$ to 13. Call this sum $S$.

Now from 14 onwards use periodicity. That $(14+a)^5 = a^5\pmod{14}$. Now to sum up to $1080^5$ is same as checking how many times $S$ has to be added repeatedly. If $1080$ is not a multiple of $14$ (find out!) then finally a manual calculation for any leftover partial sum will do it.

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Hint $\bmod 14\!:\ n^{5}$ cancels with $\,(\overbrace{1078}^{\large \equiv\ 0\ }\!-\!n)^{5}\!\equiv -n^5,\,$ except for the middle term $1078/2 = 539,\,$ leaving only the terms $\,539^5+1079^5+1080^5\equiv 7^5+1^5+2^5\equiv \,\ldots$

Remark $ $ This method of cancelling out terms by pairing up inverses in sums (and products) is frequently useful, e.g. see Wilson's Theorem and related problems and see Gauss's grade school trick. It is a special case of exploiting involution (reflection) symmetry, here inversion (negation).

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You may proceed as follows by splitting up the calculation into $\mod 2$ and $\mod 7$.

  • Let $s = \sum_{k=1}^{1080}k^5$.
  • $\mod 2$: $\sum_{k=1}^{1080}k^5 \equiv_2 \sum_{i=1}^{540}(2i-1)^5 \equiv_2 540 \equiv_2 0 (\mod 2)$ $$\boxed{s \equiv 0 \mod 2}$$
  • $\mod 7$: Note that $$\sum_{k=1}^{\color{blue}{6}}k^5 \equiv_7 1^5 + 2^5+ 3^5 + (-3)^5 + (-2)^5 + (-1)^5 \equiv_7 0 \mod 7 $$ $$\stackrel{1080 \equiv 2 \mod 7}{\Longrightarrow} \sum_{k=1}^{1080}k^5 \equiv_7 {1079}^5 + {1080}^5 \equiv_7 1^5 + 2^5 \equiv_7 5 \mod 7$$ $$\boxed{s \equiv 5 \mod 7}$$ Now, you may just note that $\mod 14$ you only need to find the first positive even number which gives remainder $5$ when divided by $7$. Or you may use Chinese remainder theorem. Anyways, you get $$\Rightarrow \boxed{s \equiv 12 \mod 14}$$
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