Number theory question, based on the theory of congruences.
Asked
Active
Viewed 121 times
0
-
3^500 ≡ -5 mod 14 – Apr 01 '21 at 19:37
-
1How have you tried to attack the problem? – Alann Rosas Apr 01 '21 at 19:39
-
Clearly $,2\mid 3^{6k+2}+5 =: n,,$ and $7\mid n,$ by $!\bmod 7!:\ 3^{6k+2}\equiv (\color{#c00}{3^6})^k 3^2\equiv \color{#c00}1\cdot 9.,$ So $,2,7\mid n\Rightarrow 2\cdot 7\mid n,$ by lcm/CCRT – Bill Dubuque Apr 01 '21 at 20:18
-
Or use $,(a,14)=1\Rightarrow a^6\equiv 1\pmod{14},$ by the 4th dupe (my answer there shows this method generalizes widely). – Bill Dubuque Apr 01 '21 at 20:33
2 Answers
3
hint
$$3^3=27\equiv -1\mod 14$$
$$500=3\times 166+2$$
Remark
You can use Euler's Theorem $$gcd(3,14)=1\implies $$ $$3^{\phi(14)}\equiv 1 \mod 14$$
with $$\phi(14)=14(1-\frac 12)(1-\frac 17)=6$$
hamam_Abdallah
- 62,951
1
$2$ divides $3^{500}+5$, because $3^{500}+5\equiv1^{500}+5\equiv0\pmod 2$.
$7$ divides $3^{500}+5$, because $3^6\equiv1\pmod7$ (by Fermat's little theorem),
so $3^{498}\equiv1\pmod7 $, so $3^{500}+5\equiv3^2+5\equiv0\pmod7$.
J. W. Tanner
- 60,406