Find an integer a which satisfies $a\equiv5\pmod{8}$ and $a\equiv3\pmod{7}$

Question

We are learning congurence class part for an intro course. This is a sample exam question without a solution.

I have poor idea about this problem.

My idea is : suppose there is an a satisfies it.

then I have

$7a\equiv35\pmod{56}$ and $8a\equiv24\pmod{56}$

then by doing the substraction I get

$a\equiv-11\pmod{56}$

then

$a\equiv45\pmod{56}$

That is all the steps I can do up to now.

Any hint plz?

Thank you in advance.

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Addition: actually I found it is the correct anwser that $a = 45$. But I am not really sure why it is the answer. I got this answer suprisingly..

Well you got that $a\equiv 45\pmod{56}$ hence any $a=56k+45$ is solution to your problem,and you got it on a fine way. — kingW3, Feb 26 '15 at 09:39
@kingW3 So I can think [$a$] is a class in $a$ + $56Z$ where $a = 45$, is that right? Thank you! — SonicFancy, Feb 26 '15 at 09:41

score 3 · Answer 1 · answered Feb 26 '15 at 09:37

3

You can use the method of adding the modulus to solve a problem like this:

$\pmod{8}: a\equiv 5\equiv 13\equiv 21\equiv 29\equiv 37\equiv 45$.

We note that $45$ also satisfies the $\mod{7}$ congruence.

So the solution is $a\equiv 45\pmod{56}$.

answered Feb 26 '15 at 09:37

paw88789

Thank you. This is an excellent and brief way for small numbers. – SonicFancy Feb 26 '15 at 09:39
1

@Son General method is just as easy: solve ${\rm mod}\ 7!:,\ \color{#0a0}3\equiv a\equiv \color{#90f}{5}!+!\color{blue}8n\equiv 5!+!n\iff n\equiv \color{#c00}{-2}\equiv 5,,$ as in my answer, i.e. $,\color{blue}{8\equiv 1},$ so we must subtract it $\color{#c00}{\rm twice}$ from $,\color{#90f}5,$ to get $,\color{#0a0}3,,$ Equivalently, we can add it $,5,$ times, by $,-2\equiv 5.,$ The above answer is simply solving this equation by brute force testing $,n \equiv 1,2,3.\ldots,$ It's true that in some cases checking for small solutions can be much easier than the general CRT. – Bill Dubuque Mar 01 '15 at 20:15

score 3 · Answer 2 · answered Feb 26 '15 at 21:16

${\rm mod}\ 8\!:\,\ a\equiv 5\iff a = \color{#0a0}{5+ 8n}$

${\rm mod}\ 7\!:\,\ 3\equiv a\equiv \color{#0a0}{5+8n}\equiv 5+n\iff n\equiv -2\equiv 5\iff \color{#c00}{n = 5+7k}$

So we conclude $\ a = 5+8\color{#c00}n = 5+8(\color{#c00}{5+7k}) = 45 + 56k$

score 0 · Answer 3 · answered Feb 26 '15 at 09:45

I'm not sure this is gonna help you because this solution may not be one you are looking for, however it just uses very simple idea.

Suppose that $$ a \equiv 5 \;\text{(mod 8)} \quad a\equiv 3 \;\text{(mod7)} $$

Then $a+11$ must be divisible by both $8$ and $7$. So $a+11 \equiv 0 \; \text{(mod $56$)}$. Which gives unique solution up to mod $56$.

3 Answers3