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Crypt arithmetic2
- Thread starter Saumyojit
- Start date
Last edited:
You gave no explanation for how you deduced A = 1 or C = 5. Thus, I cannot identify what error or errors you may have made. Your logic may have been perfect, or you may be right by accident, or you may be wrong.ABCD + EBCB= AFGAG
A=1,C=5
2C+1=11
D+B=G+10
2B+1=G OR G+10
E+2 OR 1=F+10*A
I have shown you now several times a systematic method for solving these puzzles. Why do you refuse to use it?
[MATH]d + b = g + 10\alpha, \alpha = 0 \text { or } 1;[/MATH]
[MATH]\alpha + 2c = a + 10\beta, \beta = 0 \text { or } 1;[/MATH]
[MATH]\beta + 2b = g + 10\gamma, \gamma = 0 \text { or } 1;[/MATH]
[MATH]\gamma + a + e = f + 10\delta, \delta = 0 \text { or } 1; \text { and}[/MATH]
[MATH]\delta + 0 + 0 = a.[/MATH]
By the puzzle rules, a > 0 so delta must be > 0 so delta must be 1 so a must be 1.
If that was your logic, PERFECT.
[MATH]\therefore \alpha + 2c = a + 10\beta \implies \alpha + 2c = 1 + 10\beta.[/MATH]
[MATH]1 + 10\beta \text { is odd, but } 2c \text { is even} \implies \alpha = 1. [/MATH]
[MATH]1 + 2c = 1 + 10\beta \implies c = 5\beta \implies c = 0 = \beta \text { or } c = 5 \text { and } \beta = 1.[/MATH]
Now I point out two things. Whatever your logic was, it identified only one of two possibilities so it was flawed logic. Moreover, each possibility actually identifies two needed numbers rather than just one. So at least your presentation was also flawed.
This is part of the value of a systematic method: you will not overlook possibilities. My crticisms are meant to be constructive, not mean-spirited.
Now when we are faced with alternatives, we pick one arbitrarily and explore it until we reach an impossibility or an answer. But before we do that I think it is a good idea to write down what we know so as not to burden our memory.
[MATH]d + b = g + 10;[/MATH]
[MATH]c = 0 = \beta \text { or } c = 5 \text { and } \beta = 1;[/MATH]
[MATH]\beta + 2b = g + 10\gamma, \gamma = 0 \text { or } 1;[/MATH]
[MATH]\gamma + a + e = f + 10; \text { and}[/MATH]
[MATH]a = 1.[/MATH]
OK. Time to explore one of the alternatives. Which we explore is arbitrary, but I like to mark clearly that we are operating on an assumption, which may be false.
[MATH]\text {ASSUME temporarily that } c = 5 \text { and } \beta = 1.[/MATH]
Then as you say [MATH]1 + 2b = g + 10\gamma, \ \gamma = 0 \text { or } 1.[/MATH]
Just from this, we can conclude that g is 3, 7, or 9. Why?
Moreover gamma can be 0 or 1. With fewer alternatives on gamma, that is where to work. What do we do? We explore one of the alternatives by assuming temporarily that it is true.
[MATH]\text {ASSUME temporarily that } \gamma = 0.[/MATH]
We can conclude that g is not 3. Why?
So g is 7 or 9.
Now what?
You gave no explanation for how you deduced A = 1 or C = 5. Thus, I cannot identify what error or errors you may have made. Your logic may have been perfect, or you may be right by accident, or you may be wrong.
I have shown you now several times a systematic method for solving these puzzles. Why do you refuse to use it?
[MATH]d + b = g + 10\alpha, \alpha = 0 \text { or } 1;[/MATH]
[MATH]\alpha + 2c = a + 10\beta, \beta = 0 \text { or } 1;[/MATH]
[MATH]\beta + 2b = g + 10\gamma, \gamma = 0 \text { or } 1;[/MATH]
[MATH]\gamma + a + e = f + 10\delta, \delta = 0 \text { or } 1; \text { and}[/MATH]
[MATH]\delta + 0 + 0 = a.[/MATH]
By the puzzle rules, a > 0 so delta must be > 0 so delta must be 1 so a must be 1.
If that was your logic, PERFECT.
[MATH]\therefore \alpha + 2c = a + 10\beta \implies \alpha + 2c = 1 + 10\beta.[/MATH]
[MATH]1 + 10\beta \text { is odd, but } 2c \text { is even} \implies \alpha = 1. [/MATH]
[MATH]1 + 2c = 1 + 10\beta \implies c = 5\beta \implies c = 0 = \beta \text { or } c = 5 \text { and } \beta = 1.[/MATH]
Now I point out two things. Whatever your logic was, it identified only one of two possibilities so it was flawed logic. Moreover, each possibility actually identifies two needed numbers rather than just one. So at least your presentation was also flawed.
This is part of the value of a systematic method: you will not overlook possibilities. My crticisms are meant to be constructive, not mean-spirited.
Now when we are faced with alternatives, we pick one arbitrarily and explore it until we reach an impossibility or an answer. But before we do that I think it is a good idea to write down what we know so as not to burden our memory.
[MATH]d + b = g + 10;[/MATH]
[MATH]c = 0 = \beta \text { or } c = 5 \text { and } \beta = 1;[/MATH]
[MATH]\beta + 2b = g + 10\gamma, \gamma = 0 \text { or } 1;[/MATH]
[MATH]\gamma + a + e = f + 10; \text { and}[/MATH]
[MATH]a = 1.[/MATH]
OK. Time to explore one of the alternatives. Which we explore is arbitrary, but I like to mark clearly that we are operating on an assumption, which may be false.
[MATH]\text {ASSUME temporarily that } c = 5 \text { and } \beta = 1.[/MATH]
Then as you say [MATH]1 + 2b = g + 10\gamma, \ \gamma = 0 \text { or } 1.[/MATH]
Just from this, we can conclude that g is 3, 7, or 9. Why?
Moreover gamma can be 0 or 1. With fewer alternatives on gamma, that is where to work. What do we do? We explore one of the alternatives by assuming temporarily that it is true.
[MATH]\text {ASSUME temporarily that } \gamma = 0.[/MATH]
We can conclude that g is not 3. Why?
So g is 7 or 9.
Now what?
Yes I made a mistake C can be zero or five
But the Answer Will come considering 5
"Start a new thread"
Last edited by a moderator:
Perhaps. I am not working on this anymore because you said it was solved.Yes I made a mistake C can be zero or five
But the Answer Will come considering 5
"Start a new thread"
The new prob Is solvedPerhaps. I am not working on this anymore because you said it was solved.