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It is given that $911$ is a prime number. Integers $x$ and $y$ are chosen such that $9x+11y$ is a multiple of $911$. For what positive integer value $N<1000$ will $11x+Ny$ definitely be a multiple of $911$?

Someone please give the solution for it with explanation.

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Are you familiar with congruences? – Berci Mar 20 '13 at 20:26
yes, I a=b (mod c). – Raushan Kumar Mar 20 '13 at 20:38
Perfect. $\,\,\!\!$ – Berci Mar 20 '13 at 20:55

Hint $\rm\ mod\ 911\!:\, 9x\!+\!11y\equiv 0 \Rightarrow x \equiv -11y/9\Rightarrow 11x\equiv -11^2 y/9 \equiv (2\cdot 911\! -\!121)\,y/9\equiv \color{blue}{189}y$

Remark $\ $ Replying to a comment, I don't compute $\rm\,1/9\,\ mod\,\ 911.\:$ Instead I use that $\rm\:mod\ 911\!:\, -11^2/9\,\equiv\, (911\,n\!-\!11^2)/9,\:$ so I compute $\rm\:n\:$ such that $\rm\:9\mid 911\,n-11^2,\ $ i.e. $\rm\: mod\ 9\!:\ 0\,\equiv\, 911\,n\!-\!11^2\!\equiv\, \color{#C00}2n\!-\!\color{#0A0}4\,$ $\Rightarrow$ $\,\rm n\equiv 2,\:$ so $\rm\: \frac{-11^2}9 \equiv \frac{911(2) - 11^2}9$ $\equiv 100(2)\!-\!11 \equiv \color{blue}{189}.\:$
(Note that casting nines allows one to quickly do the above calculation purely mentally, since $\rm\: mod\ 9\!:\ 911\equiv 9\!+\!1\!+\!1\equiv \color{#C00}2,\:$ and $\rm\:11^2\!\equiv (1\!+\!1)^2\!\equiv \color{#0A0}4).$

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+1 Made it much easier... Anyway, this $/9$ here means $\cdot 405$ [modulo $911$], as $405$ is the multiplicative inverse of $9$ as I calculated. – Berci Mar 20 '13 at 20:53
@Berci Thanks. But I don't need to need to calculate $\rm\ 1/9\ mod\ 911.\:$ Instead, the arithmetic is all very simply single-digit calculations that can be quickly mentally performed, as I explain in the remark added in an edit. – Math Gems Mar 20 '13 at 21:33

Introducing the notation $a\equiv b \pmod{911}$ for $911\,|\, b-a$, we have $$9x+11y\equiv 0 \pmod{911}$$ That is, $9x\equiv -11y$. Now, the multiplicative inverse of $3$ modulo $911$ can be easily found, as $3\cdot 304=912\equiv 1\pmod{911}$. It follows, that $304^2=92416$ is multiplicative inverse of $3^2$ modulo $911$, which is congruent to $$304^2=92416=911\cdot100 + 911 + 405 \equiv 405 \pmod{911}$$ So, we have that $405\cdot 9\equiv 1 \pmod{911}$, so $$x+405\cdot 11y\equiv 405\cdot(9x+11y)\equiv 0 \pmod{911}$$ Further, $11\cdot 405=4455\equiv 811 \equiv -100 \pmod{911}$, so we have $x-100y\equiv 0$, now multiplying by $11$ we will soon get an $N$: $$11x-1100y\equiv 0 \pmod{911}$$ And finally, calculate the remainder of $1100$ mod $911$, that is $189$, so $911-189=722$ will be good for $N$.

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Hint $\ $ Below is a very simple alternative approach suggested by the form of the numbers:

$$\begin{eqnarray}\rm mod\ \ 911\!: &&\rm\ \ {-}11\,x&\equiv&\rm\, N\,y \\ &&\rm -900\,x\,&\equiv&\,\rm 1100\,y\\ \rm adding\ \Rightarrow &&\rm -911\,x &\equiv&\rm (N+1100)\,y \\ \Rightarrow && \qquad \ 0 &\equiv&\rm\ \, N + 189 \end{eqnarray}$$

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$$ \begin{align} 9x+11y&\equiv0\pmod{911}\tag{1}\\ 11x+ny&\equiv0\pmod{911}\tag{2} \end{align} $$ $11\times(1)-9\times(2)$ is $$ (121-9n)y\equiv0\pmod{911}\tag{3} $$ Since we don't necessarily have $y\equiv0\pmod{911}$, we must have $$ 9n\equiv121\pmod{911}\tag{4} $$ which, using an algorithm similar to the Euclid-Wallis Algorithm yields $$ n\equiv722\pmod{911}\tag{5} $$

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