How to prove that $7^{15} + 7^{16} + 7^{17} - 1$ is divisible by $10$?

Question

This was a question on my math exam. We weren't able to use calculators so proving by manually calculating the exact value would take too long. In the end I ignored this question to save time but I'm still curious.

Answer 1

You can first make some observations:

$$ 7^1 \equiv 7 \mod 10$$ $$ 7^2 \equiv -1 \mod 10$$ $$ 7^3 \equiv 3 \mod 10$$ $$ 7^4 \equiv 1 \mod 10$$ $$ 7^5 \equiv 7 \mod 10$$ $$ 7^6 \equiv -1 \mod 10$$ $$ 7^7 \equiv 3 \mod 10$$ $$ 7^8 \equiv 1 \mod 10$$

You can conclude (or prove by induction) that

$$ 7^{4k+1} \equiv 7 \mod 10$$ $$ 7^{4k+2} \equiv -1 \mod 10$$ $$ 7^{4k+3} \equiv 3 \mod 10$$ $$ 7^{4k+4} \equiv 1 \mod 10$$

So

$$ 7^{15} + 7^{16}+ 7^{17} - 1\equiv 3 + 1 + 7 - 1 \equiv 0 \mod 10.$$

Answer 2

Try modular arithmetic: $$7^2=49\equiv-1\mod 10,$$ so $$7^{15}+7^{16}+7^{17}-1\equiv (-1)^{7}7+(-1)^{8}+(-1)^{8}7-1\equiv -7+1+7-1\equiv0\mod 10$$ and $7^{15}+7^{16}+7^{17}-1$ is divisible by $10$.

Answer 3

$$7^{15}+7^{16}+7^{17}=7^{15}(1+7+7^2)\equiv7^{15}\cdot7\pmod{50}$$

Now $7^2=49\equiv-1\pmod{50}$

$\implies7^{16}=(7^2)^8\equiv(-1)^8\pmod{50}$

Answer 4

$$7^{15}+7^{16}+7^{17}-1=57\times 7^{15}-1$$

$7^2\equiv -1\pmod {10}$ so we have $7^{14}\equiv (-1)^7\equiv -1\pmod {10}$ and therefore $7^{15}\equiv -7$ and

$$57\times 7^{15}\equiv (-3)\times (-7)\equiv 21\equiv 1\pmod {10}$$

Answer 5

7^15(1+7+7^2)-1 =7^15(57)-1

(10-3)^15*(3(19))-1

(10*k - 3^15) *3 *19 -1 (10*K-3^16)*19-1

19*3^16-1 must be divisible by 10 if this is true

3^0=1 3^1=3 3^2=9 3^3=27 last digit of 3^4=1 ^5=3 9 7... repeats -3^16 will have last digit of 1 19*-xxxxx1*-1

(10+9)*(-xxxxx1)-1 =-10*xxxx1 + 9*xxxxx1-1 =10*K -xxxxxxx9 -1 =10*k-xxxxxxxx0 so 10 divides