-5
$\begingroup$

I know these things If $x$ is divided by $11$ it remainder of $7$. If $x$ is divided by $13$ it remainder of $7$. If $x$ is divide by $3$ it remainder $3$. $x$ is less than $500$.

I am trying to prove that $x$ even ...

  • 1
    Do you know the Chinese Remainder Theorem? – 2017-02-03
  • 0
    If $x$ is divided by $3$ it has a remainder of *what*? – 2017-02-03
  • 0
    $3$ so it divides perfectly into it $x$ – 2017-02-03
  • 1
    I would call that a remainder of $0$ – 2017-02-03
  • 0
    yes it is $$ $$ – 2017-02-03

2 Answers 2

4

Hint $\ $ If $\,x\,$ is a solution so too is $\,x + 3\cdot 11\cdot 13,$ and they have opposite parity.

Maybe you seek to prove that the least nonegative solution is even. Then

coprime $\,11,13\mid x-7\,\Rightarrow\, 11\cdot 13= 143\mid x-7\ $ so $\,x = 7 + 143k$.

Thus ${\rm mod}\ 3\!:\,\ 3 \equiv x \equiv 7+143k\equiv 1+2k\ $ so $\,2k\equiv 2,\,$ so $\,k\equiv 1\ $ so $\, k = 1+3n$

Thus $\,x = 7+143(1+3n) = 150+ 429n$

  • 0
    what does it mean opposite parity? – 2017-02-03
  • 1
    One is odd, the other is even, since adding an odd intger to $x$ changes its parity. So there are both odd and even solutions. So you cannot prove that $x$ is even. – 2017-02-03
  • 0
    ok .. so is it $3 \cdot 11 \cdot 13$?? so 429? that isnt even – 2017-02-03
  • 0
    I don't really understand why you are multiplying $3,11,13$ anyway – 2017-02-03
  • 0
    Because adding that number to $x$ does not affect the remainder $x$ leaves when dividing by $3,11\,$ or $13.\ $ So it is also a solution. Did you not yet learn the Chinese Remainder Theorem? – 2017-02-03
  • 0
    ok so whatever x is you can add 429 and that works too and its going to be odd, so i have to check everything 0,2,4,6 up to 71 then it becomes above 500? – 2017-02-03
  • 0
    I checked everything up to 70 and nothing works for all three remainder ................................. – 2017-02-03
  • 0
    I FOUND IT $$x = 150$$ easy after all – 2017-02-03
  • 0
    theres a pattern to make it easier to check fast you can skip a lot – 2017-02-03
  • 1
    @terrace See my edit. It appears that you omitted a crucial constraint in your question, that you seek the *least nonnegative* solution. – 2017-02-03
1

If you are familiar with modular arithmetic, you can set up your information into modular equivalences: $$\begin{cases}x\equiv7\mod11\\x\equiv7\mod13\\x\equiv0\mod3\end{cases}$$ You can then use the Chinese Remainder Theorem to solve this. Note that $11,13,3$ are pairwise coprime, and the solution modulo $(3)(13)(11)=429$ is less than $500$. Also note that the last equivalence shows a remainder of $0$ when divided by three, which is equivalent to a remainder of $3$ when divided by $3$ (since the remainder should be less than the divisor).

  • 0
    I got it its $$x=150$$ – 2017-02-03