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Codework
- Thread starter Ronanaron
- Start date
Hello, Ronanaron!
There is not enough information for a unique solution.
A slight modification will fix that . . .
\(\displaystyle \text{Suppose the 7-digit code is: }\:\underbrace{ABC}_x\underbrace{DEF}_yG\)
\(\displaystyle \text{Let the first 3-digit number be }x\text{; let the second 3-digit number be }y.\)
\(\displaystyle \text{We have: }\:\begin{Bmatrix}x - G &=& 447 \\ y - x &=& 447 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x &=& G + 447 \\ y &=& G + 894 \end{Bmatrix}\)
\(\displaystyle \text{If we let }G \,=\, 0,1,2, \cdots9,\,\text{ we get ten solutions.}\)
\(\displaystyle \text{The only solution which has seven }di\!f\!ferent\text{ digits is: }\:4508973\)
There is not enough information for a unique solution.
A slight modification will fix that . . .
A code has 7 different digits.
The first 3-digit number minus the last digit equals 447.
The second 3-digit number 3 minus the first 3-digit number also equals 447.
What's the code?
\(\displaystyle \text{Suppose the 7-digit code is: }\:\underbrace{ABC}_x\underbrace{DEF}_yG\)
\(\displaystyle \text{Let the first 3-digit number be }x\text{; let the second 3-digit number be }y.\)
\(\displaystyle \text{We have: }\:\begin{Bmatrix}x - G &=& 447 \\ y - x &=& 447 \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x &=& G + 447 \\ y &=& G + 894 \end{Bmatrix}\)
\(\displaystyle \text{If we let }G \,=\, 0,1,2, \cdots9,\,\text{ we get ten solutions.}\)
\(\displaystyle \text{The only solution which has seven }di\!f\!ferent\text{ digits is: }\:4508973\)
Hello, Denis!
I hope it is obvious how I read the problem.
I read "the second 3-digit number" to mean "the next 3-digit number"
. . . . . the number comprised of the 4th, 5th and 6th digits.
I found that your interpretation has no solutions.
\(\displaystyle \text{We have: }\:ABCDEFG\)
\(\displaystyle \text{Then: }\:\begin{array}{cccc}&A&B&C \\ -&&&G \\ \hline &4&4&7 \end{array}\)
\(\displaystyle \text{Since }0 \le G \le 9\text{, then: }\:447 \le ABC \le 456\)
\(\displaystyle \text{And we find that none of these cases have solutions.}\)
\(\displaystyle \text{We have: }\:\begin{array}{cccc} & A&B&C \\ -&B&C&D \\ \hline & 4&4&7 \end{array}\)
\(\displaystyle \text{Example: }\:ABC = 447:\quad\begin{array}{cccc}&4&4&7 \\ - & 4&7&D \\ \hline & 4&4&7 \end{array}\)
\(\displaystyle \text{Example: }\:ABC = 456:\quad\begin{array}{cccc}&4&5&6 \\ - & 5&6&D \\ \hline & 4&4&7 \end{array}\)
So I believe I had the correct interpretation.
I hope it is obvious how I read the problem.
I read "the second 3-digit number" to mean "the next 3-digit number"
. . . . . the number comprised of the 4th, 5th and 6th digits.
I found that your interpretation has no solutions.
\(\displaystyle \text{We have: }\:ABCDEFG\)
\(\displaystyle \text{Then: }\:\begin{array}{cccc}&A&B&C \\ -&&&G \\ \hline &4&4&7 \end{array}\)
\(\displaystyle \text{Since }0 \le G \le 9\text{, then: }\:447 \le ABC \le 456\)
\(\displaystyle \text{And we find that none of these cases have solutions.}\)
\(\displaystyle \text{We have: }\:\begin{array}{cccc} & A&B&C \\ -&B&C&D \\ \hline & 4&4&7 \end{array}\)
\(\displaystyle \text{Example: }\:ABC = 447:\quad\begin{array}{cccc}&4&4&7 \\ - & 4&7&D \\ \hline & 4&4&7 \end{array}\)
\(\displaystyle \text{Example: }\:ABC = 456:\quad\begin{array}{cccc}&4&5&6 \\ - & 5&6&D \\ \hline & 4&4&7 \end{array}\)
So I believe I had the correct interpretation.