Number Theory - Remainders

Question

A number is of the form $13k_1+12$ and of the form $11k_2+7$

That is $N = 13k_1 + 12 = 11k_2 + 7$

Now why must N also equal $(13 \times 11)k_3 + 51$ ?

Thanks

@ArpitKansal I've tried getting k1 in terms of k2. 51 seems to be the smallest value of N that satisfies the above conditions — Hormigas, Sep 09 '15 at 03:49
Note that N=7(mod 11) and N=12(mod13) Can you say something from here? — Arpit Kansal, Sep 09 '15 at 03:53

score 2 · Accepted Answer · answered Sep 09 '15 at 04:12

2

Alternately, $N=13k_1+12 = 11k_2+7 \implies N-51 = 13(k_1-3)=11(k_2-4)$

Thus $N-51$ must be a multiple of both $13$ and $11$.

answered Sep 09 '15 at 04:12

Macavity

46,381

Elegant and beautiful – Hormigas Sep 09 '15 at 05:22
Usually the trouble is in finding $51$, which @robjohn has so meticulously explained in his post below... – Macavity Sep 09 '15 at 13:38

GAVD · Answer 2 · 2015-09-09T04:05:48.060

0

Let me give a trick. We have $N + 92 = 13k_1 + 104 = 13(k_1 + 8)$ and $N+92 = 11k_2 + 99 = 11(k_2 + 9)$.

Then, $N+92 = 11\times 13 k_3$ or $N = 143k_3 -92$ or $N=143k_4 + 51$.

See also link

edited Sep 09 '15 at 04:05

answered Sep 09 '15 at 03:58

GAVD

7,296
1
16
30

Don't re-use $k_3$ for another value. $N=143(k_3-1)+51$. – Thomas Andrews Sep 09 '15 at 04:04
Uhm, OK, I changed it to $k_4$. :) – GAVD Sep 09 '15 at 04:06

score 0 · Answer 3 · answered Sep 09 '15 at 05:07

$N$ as defined is $12$ mod $13$ by first definition and $7$ mod $11$ by second definition, since both $13$ and $11$ are relatively prime (in fact primes) thus $N$ should have same mod ($13 \times 11$) as well, rest follows by manipulating the $12$, $7$ remainders with $13$, $11$

score 0 · Answer 4 · edited Apr 13 '17 at 12:20

Finding a Solution

To find an $N$ so that $$ \begin{align} N&\equiv12&\pmod{13}\\ N&\equiv7&\pmod{11} \end{align}\tag{1} $$ we can start by solving $$ 13x+11y=1\tag{2} $$ using the Extended Euclidean Algorithm. The implementation described in this answer gives $$ \begin{array}{r} &&1&5&2\\\hline 1&0&1&-5&11\\ 0&1&-1&6&-13\\ 13&11&2&1&0\\ \end{array}\tag{3} $$ which implies $$ 13(-5)+11(6)=1\tag{4} $$ Using $(4)$, we can show $$ \begin{align} 66&\equiv1&\pmod{13}\\ 66&\equiv0&\pmod{11} \end{align}\tag{5} $$ and $$ \begin{align} -65&\equiv0&\pmod{13}\\ -65&\equiv1&\pmod{11} \end{align}\tag{6} $$ $12$ times $(5)$ plus $7$ times $(6)$ gives $$ \begin{align} 337&\equiv12&\pmod{13}\\ 337&\equiv7&\pmod{11} \end{align}\tag{7} $$ Subtracting $2\cdot11\cdot13=286$ from the left sides of $(7)$ gives $$ \begin{align} 51&\equiv12&\pmod{13}\\ 51&\equiv7&\pmod{11} \end{align}\tag{8} $$

Finding a General Solution

If $N_1$ and $N_2$ are any two solutions to $(1)$, then $$ \begin{align} N_1-N_2&\equiv0&\pmod{13}\\ N_1-N_2&\equiv0&\pmod{11} \end{align}\tag{9} $$ Therefore, $$ N_1-N_2\equiv0\pmod{11\cdot13}\tag{10} $$ Putting together $(8)$ and $(10)$, we get that all the solutions of $(1)$ are given by $$ N\equiv51\pmod{143}\tag{11} $$

Number Theory - Remainders

4 Answers4