Obtain the digits ## x ## and ## y ##

[Math100]

Homework Statement

Assuming that $495$ divides $273x49y5$, obtain the digits $x$ and $y$.

Relevant Equations

None.

Observe that $495=5\cdot 9\cdot 11$.
This means $9\mid 273x49y5$ and $11\mid 273x49y5$.
Then $9\mid (2+7+3+x+4+9+y+5)\implies 9\mid (x+y+30)\implies x+y=6, 15$ and $11\mid (2-7+3-x+4-9+y-5)\implies 11\mid (y-x-1)\implies y-x=1$.
Now we compute these two systems of equations shown below:
$\{x+y=6, y-x=1\}$ and $\{x+y=15, y-x=1\}$
Thus $x=7$ and $y=8$.
Therefore, the digits $x$ and $y$ are $7$ and $8$.

 

[SammyS]

Homework Statement:: Assuming that $495$ divides $273x49y5$, obtain the digits $x$ and $y$.
Relevant Equations:: None.

Observe that $495=5\cdot 9\cdot 11$.
This means $9\mid 273x49y5$ and $11\mid 273x49y5$.
Then $9\mid (2+7+3+x+4+9+y+5)\implies 9\mid (x+y+30)\implies x+y=6, 15$ and $11\mid (2-7+3-x+4-9+y-5)\implies 11\mid (y-x-1)\implies y-x=1$.
Now we compute these two systems of equations shown below:
$\left \{ \begin{align*} x+y=6 \\ y-x=1 \end{align*} \right \}$ and $\left \{ \begin{align*} x+y=15 \\ y-x=1 \end{align*} \right \}$
Thus $x=7$ and $y=8$.
Therefore, the digits $x$ and $y$ are $7$ and $8$.

Fixed some LaTeX in the quoted text above.

 

[Delta2]

Fixed some LaTeX in the quoted text above.

Needed some additional fixing in LaTeX, or my browser for some reasons bugs seriously on displaying this message.

 

[fresh_42]

Homework Statement:: Assuming that $495$ divides $273x49y5$, obtain the digits $x$ and $y$.
Relevant Equations:: None.

Observe that $495=5\cdot 9\cdot 11$.
This means $9\mid 273x49y5$ and $11\mid 273x49y5$.
Then $9\mid (2+7+3+x+4+9+y+5)\implies 9\mid (x+y+30)\implies x+y=6, 15$ and $11\mid (2-7+3-x+4-9+y-5)\implies 11\mid (y-x-1)\implies y-x=1$

... since $0\leq x,y \leq 9.$

.
Now we compute these two systems of equations shown below:
$\{x+y=6, y-x=1\}$ and $\{x+y=15, y-x=1\}$

The first system yields $2y=7$ which has no integer solution, and the second ...

Thus $x=7$ and $y=8$.
Therefore, the digits $x$ and $y$ are $7$ and $8$.

Correct.