Remainder in this case using modular arithmetic

Question

How do you find the remainder when $870^{479}$ is divided by 65? My approach:

$870\equiv 25\pmod {65}$

=>$174^{479}\equiv 5^{479}\pmod {13}$

=>$174^{479}\equiv 25^{234.5}\pmod {13}$

I am stuck here. Please help.

See http://math.stackexchange.com/questions/1952810/finding-the-last-two-digits-of-5312442 OR http://math.stackexchange.com/questions/1844558/how-to-find-last-two-digits-of-22016 — lab bhattacharjee, Oct 22 '16 at 03:53

score 2 · Answer 1 · answered Oct 20 '16 at 16:54

2

$65=13\times5$

$870^{479}\equiv0 \pmod{5}$

$\color{red}{87}0^{479}\equiv\color{red}{9}0^{479}\equiv(-1)^{479}\equiv-1\pmod{13}$

So $x=870^{479}=-1+13k\equiv0 \pmod 5 \rightarrow k\equiv\frac{1}{13}\equiv\frac 13\equiv\frac63\equiv2 \pmod 5$

and therefore, $x=-1+13\times(2+5k')\equiv-1+26\equiv25 \pmod {65}$

answered Oct 20 '16 at 16:54

Evariste

2,511

I could not understand the second and third step. – MrAP Oct 20 '16 at 17:09
@MrAP He is applying CRT = Chinese Remainder Theorem. I posted an answer that avoids that. – Bill Dubuque Dec 17 '16 at 03:04

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

Note $\ ca\bmod cn\,=\, c\,(a\bmod n)\ $ as we explained here, $ $ hence

$\ \, \begin{align} 870^{\large 479}\!\bmod 65\, &=\, 5\,(\color{#0a0}{174}\cdot \color{#c00}{870}^{\large 478}\bmod 13)\\ &=\, 5\,(\ \color{#0a0}5\cdot (\color{#c00}{\bf -1})^{\large 478}\!\bmod 13)\\ &=\, 5\,(\,\color{#0a0}5\,) \end{align}\ \ \ $ by $\ \ \ \begin{align} \color{#0a0}{174}\,&\equiv\,\ \color{#0a0} 5\pmod{13}\\ \color{#c00}{870}\,&\equiv \color{#c00}{\bf -1}\!\!\!\pmod{13}\\ \phantom{.}\end{align}$

score 0 · Answer 3 · answered Oct 20 '16 at 16:51

0

HINT: $870^1\equiv870^3\pmod{65}$

answered Oct 20 '16 at 16:51

barak manos

43,109

Remainder in this case using modular arithmetic

3 Answers3