Find the remainder when the number $52^{1989}$ is divided by 7.

Question

I tried arithmetic modular, but seems didn't work

you know that $52\equiv 3\mod 7$. Also, you can use the Euler-Fermat theorem, wich (in this case) says that $3^6\equiv 1\mod 7$, so in the exponent, you may work modulo $6$. — Ragnar, Dec 16 '13 at 14:41
@Ragnar If you make that an answer (and ping me), I'll upvote. — apnorton, Dec 16 '13 at 14:44

Asinomás · Answer 1 · 2013-12-16T14:54:21.950

3

$52\equiv 3\bmod 7$, by fermat $3^6\equiv 1 \bmod7$, therefore $52^{1989}\equiv 3^{1989}=3^{1986}*3^3=3^{331*6}*3^3\equiv1*3^3\equiv6$

edited Dec 16 '13 at 14:54

answered Dec 16 '13 at 14:45

Asinomás

105,651

score 2 · Answer 2 · answered Dec 16 '13 at 14:51

You know that $52\equiv3\mod7$. Also, you can use the Euler-Fermat theorem, wich (in this case) says that $36\equiv1\mod7$, so in the exponent, you may work modulo $6$. In general, Euler-Fermat says that $a^{\phi(m)}\equiv 1\mod m$, and here $m$ is prime, so $\phi(m)=p-1$. This is only true when $\gcd(a,m)=1$ (i.e. they are relative prime), because otherwise, a power of $a$ can never be $1\mod m$.

score 2 · Answer 3 · answered Dec 16 '13 at 15:19

2

Hint $\rm\ mod\ 7\!:\ 52^{3n} \equiv 3^{3n}\equiv (3^3)^n\equiv (-1)^{n} \equiv -1\ $ for odd $\rm\,n.\,$ (Here $\rm\,1989 = 3\cdot 663 = 3n).$

answered Dec 16 '13 at 15:19

Bill Dubuque

272,048

Find the remainder when the number $52^{1989}$ is divided by 7.

3 Answers3