Find the remainder when $2^{561}$ is divided by $561$ using simple congruence properties.

Question

$2^{561}\equiv ? \pmod{561}$

Few observations :
$561 = 3\times 11\times 17$
So Fermat's little theorem is not useful here. Any hints ?

If possible, kindly avoid carmichael numbers/group theory/euler theorems as this is from elementary number theory and we have just started congruences chapter.

Fermat's little theorem is very useful here (and the Chinese remainder theorem). — Daniel Fischer, Sep 17 '14 at 13:55
Oh interesting, also 561 is a square free integer. but im bit unclear on how to put everything together — AgentS, Sep 17 '14 at 13:58
I have used Chinese remainder theorem for solving systems like : $$x \equiv 1 \pmod 3 \ x\equiv 2 \pmod 7 \\cdots$$ the exponent is throwing me off in present problem — AgentS, Sep 17 '14 at 14:01
Compute $2^{561}$ modulo $3,11$, and $17$, then put things together with the Chinese remainder theorem. — Daniel Fischer, Sep 17 '14 at 14:02
Fermat said $2^{11}\equiv 2\pmod{11}$. That would means $2^{561}\equiv 2^{3\times 17}\pmod{11}$. — Quang Hoang, Sep 17 '14 at 14:02
I see, ty :) and I have $2^{561}\equiv 1\pmod 3 $ and $2^{561}\equiv 2^{4*140+1}\equiv 2 \pmod {17}$ — AgentS, Sep 17 '14 at 14:04
do I get : $$x \equiv 1 \pmod 3 \ x \equiv 2 \pmod {11} \ x \equiv 2 \pmod {17} \
(1)(1117)(1) + (2)(317)(8) + 2(3*11)(8) $$ ? — AgentS, Sep 17 '14 at 14:20
@DanielFischer please help, this is not giving me the correct answer — AgentS, Sep 17 '14 at 14:27
$$x \equiv 2 \pmod 3 \ x \equiv 2 \pmod {11} \ x \equiv 2 \pmod {17} \ (2)(1117)(1) + (2)(317)(8) + 2(3*11)(16) $$ this look okay or still see any errors ? — AgentS, Sep 17 '14 at 14:34
Maybe you can have a look here: http://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand — Martin Sleziak, Sep 18 '14 at 09:31

score 2 · Answer 1 · answered Sep 17 '14 at 17:05

2

We have $2^2=4\equiv1\pmod3,$

$2^5=32\equiv-1\pmod{11}\implies2^{10}\equiv(-1)^2\equiv1\pmod{11}$

$2^4=16\equiv-1\pmod{17}\implies2^8\equiv{-1}^2\equiv1\pmod{17}$

$\displaystyle\implies2^{\text{lcm}(2,10,8)}\equiv1\pmod{3\cdot11\cdot17}$

i.e., $\displaystyle2^{40}\equiv1\pmod{561}$

$\displaystyle\implies2^{561}=2(2^{40})^{14}\equiv2(1)^{14}\pmod{561}$

Reference : Carmichael function

answered Sep 17 '14 at 17:05

lab bhattacharjee

274,582

@rsadhvika, How about this? – lab bhattacharjee Sep 17 '14 at 17:06

score 2 · Answer 2 · answered Sep 18 '14 at 08:23

Fermats or Euler is exactly what you need :) (or anything else you know )

what you need to know is this :-

if $n= p_1p_2...p_i $ such that $p_1,p_2,..., p_i$ are distinct primes which means $GCD(p_j,p_k)=1$ for $j,k \leq i$ , and $x=m \bmod p_1=m \bmod p_2 = ...=m \bmod p_i$ then $x=m \bmod (p_1p_2...p_i) \rightarrow x=m \bmod n$

now GCD(3,7 )=1 ,GCD(3,11)=1 , GCD(7,11)=1 :-

$2^{{\color{DarkBlue} {16}}}={\color{Green} 1} \bmod {\color{Red} {17} }\\ 2^{{\color{DarkBlue} 2}}={\color{Green} 1} \bmod {\color{Red} 3}\\ 2^{{\color{DarkBlue} {10}}}={\color{Green} 1} \bmod {\color{Red} {11} }\\ 2^{{\color{DarkBlue} {16\times35}}}=2^{{\color{Magenta} {560} }}={\color{Green} 1} \bmod {\color{Red} {17} }\\ 2^{{\color{DarkBlue} {2\times280}}}=2^{{\color{Magenta} {560 } }}= {\color{Green} 1} \bmod {\color{Red} 3}\\ 2^{{\color{DarkBlue} {10\times56 } }}=2^{{\color{Magenta} {560} }}= {\color{Green} 1} \bmod {\color{Red} {11} }$

$2^{\color{Magenta}{560} }= {\color{Red} 1} \bmod {\color{Blue} {3\times 11\times 17}}\\ 2^{\color{Magenta}{560} }= {\color{Red} 1} \bmod {\color{Blue} {561}}\\$

thus :-

$2^{\color{Magenta}{561} }= {\color{Red} 2} \bmod {\color{Blue} {561}} $

let me know if you dint get it yet :D

Find the remainder when $2^{561}$ is divided by $561$ using simple congruence properties.

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