Find the missing digits in this problem about an invoice

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The problem involves finding the missing digits in the number x679y, where 72 divides this number. To satisfy the divisibility by 9, the equation (x + y + 4) must equal 0 modulo 9, leading to the conclusion that x + y = 5. For divisibility by 8, the last three digits 79y must be divisible by 8, which determines y as 2. Subsequently, substituting y into the previous equation reveals x as 3. Thus, the missing digits are x=3 and y=2.
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Homework Statement
An old and some what illegible invoice shows that ## 72 ## canned hams were purchased for ## $x67.9y ##. Find the missing digits.
Relevant Equations
None.
Consider the missing digits in ## x679y ##.
Then ## 72\mid x679y ##.
Note that ## 72=8\cdot 9 ##.
This means ## 8\mid x679y ## and ## 9\mid x679y ##.
Now we have
\begin{align*}
&x679y\equiv 0\pmod {9}\\
&\implies (x+6+7+9+y)\equiv 0\pmod {9}\\
&\implies (x+y+22)\equiv 0\pmod {9}\\
&\implies (x+y+4)\equiv 0\pmod {9}.\\
\end{align*}
Thus ## x679y=x\cdot 10^{4}+6\cdot 10^{3}+7\cdot 10^{2}+9\cdot 10+y ##.
Observe that ## x679y\equiv 0\pmod {8}\implies x679y\equiv (4+2+y)\pmod {8}\implies (6+y)\equiv 0\pmod {8} ##
because ## 10^{3}\equiv 0\pmod {8} ## and ## 10^{4}\equiv 0\pmod {8} ##.
Since ## (6+y)\equiv 0\pmod {8}\implies y=2 ##, it follows that ## (x+6)\equiv 0\pmod {9}\implies x=3 ##.
Therefore, the missing digits are ## x=3 ## and ## y=2 ##.
 
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Math100 said:
Homework Statement:: An old and some what illegible invoice shows that ## 72 ## canned hams were purchased for ## $x67.9y ##. Find the missing digits.
Relevant Equations:: None.

Consider the missing digits in ## x679y ##.
Then ## 72\mid x679y ##.
Note that ## 72=8\cdot 9 ##.
This means ## 8\mid x679y ## and ## 9\mid x679y ##.
Now we have
\begin{align*}
&x679y\equiv 0\pmod {9}\\
&\implies (x+6+7+9+y)\equiv 0\pmod {9}\\
&\implies (x+y+22)\equiv 0\pmod {9}\\
&\implies (x+y+4)\equiv 0\pmod {9}.\\
\end{align*}
Thus ## x679y=x\cdot 10^{4}+6\cdot 10^{3}+7\cdot 10^{2}+9\cdot 10+y ##.
Observe that ## x679y\equiv 0\pmod {8}\implies x679y\equiv (4+2+y)\pmod {8}\implies (6+y)\equiv 0\pmod {8} ##
because ## 10^{3}\equiv 0\pmod {8} ## and ## 10^{4}\equiv 0\pmod {8} ##.
Since ## (6+y)\equiv 0\pmod {8}\implies y=2 ##, it follows that ## (x+6)\equiv 0\pmod {9}\implies x=3 ##.
Therefore, the missing digits are ## x=3 ## and ## y=2 ##.
Correct.

I would determine ##y## first. We must have that ##8## divides ##79y##, the right most 3 digits, That will give you that ##y=2## .

After that, it's almost trivial to find that ##x=3## .
 
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