Is 7($11^{106}$)+290 divisible by 13?

Question

I know I can check this using congruence, which is how I should be approaching this problem.

There are so many ways I can go about the problem, but I keep getting stuck on the $11^{106}$ term.

To reduce the exponent I have tried many things: I know $11^{106}\equiv -2^{106}\pmod {13}$ Which I can then make into $-2^{106}\equiv 4^{53}\pmod {13}$, but because 106 only has factors of 1, 2, 53 and 106 I have no idea how to reduce to make the number feasible to work with.

Obviously doing without calculator, any help is appreciated.

$(-2)^{6}=64\equiv-1\mod{13}$. I think the rest is obvious. – acat3 Mar 01 '20 at 05:41 — acat3, Mar 01 '20 at 05:41

score 0 · Answer 1 · answered Mar 01 '20 at 05:40

0

Use Fermat's little theorem to say $11^{12} \equiv 1 \pmod {13}$ and you can reduce the exponent a lot.

answered Mar 01 '20 at 05:40

Ross Millikan

383,099

score 0 · Answer 2 · answered Mar 01 '20 at 05:40

0

We have $11^{12} \equiv 1\pmod{13}$

$106=8(12)+10$

Hence the problem reduces to

$7(11^{10})+290 \equiv 7(-2)^{10}+290\equiv 7(2)^4(2^4)2^2+30 \equiv 7 (3)(3)(4)+30 \equiv -21+30 \equiv 9 \pmod{13}$

answered Mar 01 '20 at 05:40

Siong Thye Goh

153,832

Is 7($11^{106}$)+290 divisible by 13?

2 Answers2