How to find the remainder, if you divide $10^{10^7}$ + $10^{10^6}$ + $10^{10^5}$ + $10^{10^4}$ by 7.

Question

I have been trying to find the remainder, if you divide $10^{10^7}$ + $10^{10^6}$ + $10^{10^5}$ + $10^{10^4}$ by 7.

Considering the divisibility criteria by 7, i first looked at the division of the powers of 10 by 7.

10 = 1*7+3 remainder of 3, $10^2$ = 14*7+2 remainder of 2, $10^3$ = 142*7+6 reminder of -1, $10^4$ = has a reminder of 4, $10^5$ = has a reminder of 5, $10^6$ = has a reminder of 1, $10^7$ = has a reminder of 3

Is it sufficient to say that the reminders are 1, 2, 3, 4, and 5?

Answer 1

As $10^i\equiv 4$ modulo $6$ for any positive integer $i$, we have that $10^{10^7}+10^{10^6}+10^{10^5}+10^{10^4}\equiv 4*10^4 \equiv 4*4=16\equiv 2$ modulo $7$.

Answer 2

$$10\equiv3\pmod7$$

Now using Why is $a^n - b^n$ divisible by $a-b$? $$10^{n+1}-10=10(10^n-1)\equiv0\pmod{10(10-1)}$$ for $n\ge0$

$$10^{n+1}\equiv10\pmod{90}\implies10^{n+1}\equiv10\pmod6\equiv4$$

$$\implies10^{10^{n+1}}\equiv3^{10^{n+1}}\equiv3^4\pmod7\equiv4$$