Residue of Powers of 2

Number Theory Level 3

Find the smallest non-negative integer $a$ such that

$2^{1993} \equiv a \pmod{11}$

2 solutions

Aditya Raut
Aug 1, 2014

Fermat's little theorem says

$2^{10} \equiv 1 \pmod{11}$

gives us

$2^{1993} \equiv 2^3 \equiv 8 \pmod{11}$

Hence the answer is $\boxed{8}$

Fermat's little theorem , Can u explain it..???

Rahul Jain - 6 years, 10 months ago

$\begin{cases} p \text{ - prime}\\ \gcd(a,p)=1\end{cases}\implies a^{p-1}\equiv 1\pmod p$

mathh mathh - 6 years, 10 months ago

we can get the result we got from fermat's little theorem by euler's totient function also.

Adarsh Kumar - 6 years, 10 months ago

well, Euler's totient function is just a generalization of fermat's little theorem.

mathh mathh - 6 years, 10 months ago

(2^10)^199 * 2^3/11=(1)^199 8/11=8/11=rem. 8 (as 11 93=1023 & 2^10=1024 so rem. is 1 and 1 power any thing is 1)

rahul kumar - 6 years, 9 months ago

Rwit Panda
Dec 23, 2015

$2^{5}\equiv-1\pmod{11}$

$2^{1993}\equiv(-1)^{398}\times2^{3}\pmod{11}$

So, least non negative a is $\boxed{8}$