Modular arithmetic to find the mod of a large number

Question

If $x \equiv 23 \bmod 317$ and $x \equiv 25 \bmod 331$, what is $x \bmod 104927$? What techniques are typically used to solve problems of this nature? It doesn't seem clear to me that it can be solved straightforwardly.

Chinese Remainder Theorem should ring the bell. – NasuSama Feb 06 '14 at 03:33 — NasuSama, Feb 06 '14 at 03:33
I'm having a difficult time applying it here. – user1038665 Feb 06 '14 at 03:46 — user1038665, Feb 06 '14 at 03:46

score 3 · Accepted Answer · edited Apr 13 '17 at 12:19

3

$x = 25+331n,\, $ so $\,{\rm mod}\ 317\!:\ 23\equiv x = 25+331n\,\Rightarrow\, 14n\equiv -2\,\Rightarrow\, 7n\equiv -1\,$ so, by Gauss's algorithm, $\, n\equiv -1/7\equiv -45/315 \equiv 272/(-2)\equiv -136.\,$ So $\,x\equiv 25-331(136)\pmod{317\cdot 331}$

edited Apr 13 '17 at 12:19

Community

1

answered Feb 06 '14 at 03:51

Bill Dubuque

272,048

Modular arithmetic to find the mod of a large number

1 Answers1