How do I solve this congruency?

Question

I have this congruence relation:

2013 ≡ 1012 (mod m)

I am supposed to find all m in the natural system.

score 5 · Accepted Answer · answered Mar 05 '13 at 18:47

5

$2013\equiv 1012\pmod m\implies m|(2013-1012)\implies m|1001$.

Thus all factors of $1001$ satisfy this.

answered Mar 05 '13 at 18:47

Aang

14,672

1

I see, I was able to get this point. I was confused as to what to do next. Thanks. – 0xFF Mar 05 '13 at 18:50
That's why we are all here: to help and to be helped :) – Aang Mar 05 '13 at 18:51

score 3 · Answer 2 · answered Mar 05 '13 at 18:47

3

HINT: By definition $2013\equiv 1012\pmod m$ if and only if $m\mid 2013-1012$.

answered Mar 05 '13 at 18:47

Brian M. Scott

616,228

score 2 · Answer 3 · answered Mar 05 '13 at 18:54

2

$\begin{eqnarray}{\bf Hint}\ \ \ \rm mod\ m\!:\ 2013&\equiv& 1012\\ \rm \iff m\mid 2013&-&1012 = 1001 = 10^3\!+\!1 = (10\!+\!1)(10^2\!-\!10\!+\!1) = 11\cdot 91 = 11\cdot 7\cdot 13\end{eqnarray} $

answered Mar 05 '13 at 18:54

Math Gems

19,574

How do I solve this congruency?

3 Answers3