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CRYPT arithmetic
- Thread starter Saumyojit
- Start date
D
Deleted member 4993
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To start,:FORTY + TEN + TEN = SIXTY, find the value of each letter
N + N = 10 or 0 → N = 5 or 0
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem
2N+Y=Y+10
N=5
2T+R+1= X+10 OR X+20
O+1 OR O+2= I +10
F+1=S
IN 4TH COLUMN,
SUMMATION OF 2E+1 HAS TO BE ZERO, THEN ONLY ADDITION OF T+ ZERO =T
2E+1= 10
E CANNOT BE 4.5 . SOMETHING WRONG
So, N=0
E=5
2T+R+1= X+10 OR X+20
O+1 OR O+2= I +10
F+1=S
N=5
2T+R+1= X+10 OR X+20
O+1 OR O+2= I +10
F+1=S
IN 4TH COLUMN,
SUMMATION OF 2E+1 HAS TO BE ZERO, THEN ONLY ADDITION OF T+ ZERO =T
2E+1= 10
E CANNOT BE 4.5 . SOMETHING WRONG
So, N=0
E=5
2T+R+1= X+10 OR X+20
O+1 OR O+2= I +10
F+1=S
Thank you for posting a new thread with a new problem. This one is rather elegant.
Based on your private message to me about this problem, I think you would be better served by attacking these problems by using formal equations and specifying possibilities that you can later reject one by one SYSTEMATICALLY.
[MATH]Y + N + N = Y + 10\alpha, \ 0 \le \alpha \le 2;\\ \alpha + T + E + E = T + 10\beta, \ 0 \le \beta \le 2;\\ \beta + R + T + T = X + 10 \gamma, \ 0 \le \gamma \le 2;\\ \gamma + O + 0 + 0 = I + 10\delta, \ 0 \le \delta \le 1; \text { and }\\ \delta + F + 0 + 0 = S.[/MATH]Arithmetic and the rules of the game allow us both to specify the possible values and to reject the possibilities one by one.
So we know instantly that [MATH]F \ne 0, \ S \ne 0, \text { and } T \ne 0.[/MATH] Why?
[MATH]Y + N + N = Y + 10 \alpha \implies 2N = 10\alpha \implies N = 5 \alpha \implies N = 0, \ 5, \text { or } 10.[/MATH]
But we can immediately reject N = 10 as inconsistent with the rules.
[MATH]\text {If } N = 5 \implies \alpha = 1 \implies 1 + T + E + E = T + 10\beta \implies 2E = 10\beta - 1.\\ \text {But it is impossible for an even number to equal an odd number} \implies N = 0 \implies \alpha = 0.[/MATH][MATH]\therefore 0 + T + E + E = T + 10 \beta \implies 2E = 10\beta \implies E = 0, \ 5, \text { or } 10.[/MATH]
But 10 is out because E is a digit, and 0 is out because N = 0. Therefore E is 5, and beta is 1.
There are only two possibilities for delta. Can we eliminate one easily? Yes.
[MATH]\text {If } \delta = 0 \implies 0 + F + 0 + 0 = S \implies F = S, \text { which is impossible under the rules} \implies \delta = 1. [/MATH]
Now I suggest rewriting our equations to keep track of what we have learned.
[MATH]N = 0;\\ E = 5;\\ 1 + R + T + T = X + 10\gamma, \ 0 \le \gamma \le 2;\\ \gamma + O = I + 10; \text { and }\\ 1 + F = S.[/MATH]What can we deduce about gamma, O, and I?
Based on your private message to me about this problem, I think you would be better served by attacking these problems by using formal equations and specifying possibilities that you can later reject one by one SYSTEMATICALLY.
[MATH]Y + N + N = Y + 10\alpha, \ 0 \le \alpha \le 2;\\ \alpha + T + E + E = T + 10\beta, \ 0 \le \beta \le 2;\\ \beta + R + T + T = X + 10 \gamma, \ 0 \le \gamma \le 2;\\ \gamma + O + 0 + 0 = I + 10\delta, \ 0 \le \delta \le 1; \text { and }\\ \delta + F + 0 + 0 = S.[/MATH]Arithmetic and the rules of the game allow us both to specify the possible values and to reject the possibilities one by one.
So we know instantly that [MATH]F \ne 0, \ S \ne 0, \text { and } T \ne 0.[/MATH] Why?
[MATH]Y + N + N = Y + 10 \alpha \implies 2N = 10\alpha \implies N = 5 \alpha \implies N = 0, \ 5, \text { or } 10.[/MATH]
But we can immediately reject N = 10 as inconsistent with the rules.
[MATH]\text {If } N = 5 \implies \alpha = 1 \implies 1 + T + E + E = T + 10\beta \implies 2E = 10\beta - 1.\\ \text {But it is impossible for an even number to equal an odd number} \implies N = 0 \implies \alpha = 0.[/MATH][MATH]\therefore 0 + T + E + E = T + 10 \beta \implies 2E = 10\beta \implies E = 0, \ 5, \text { or } 10.[/MATH]
But 10 is out because E is a digit, and 0 is out because N = 0. Therefore E is 5, and beta is 1.
There are only two possibilities for delta. Can we eliminate one easily? Yes.
[MATH]\text {If } \delta = 0 \implies 0 + F + 0 + 0 = S \implies F = S, \text { which is impossible under the rules} \implies \delta = 1. [/MATH]
Now I suggest rewriting our equations to keep track of what we have learned.
[MATH]N = 0;\\ E = 5;\\ 1 + R + T + T = X + 10\gamma, \ 0 \le \gamma \le 2;\\ \gamma + O = I + 10; \text { and }\\ 1 + F = S.[/MATH]What can we deduce about gamma, O, and I?
E=5 N=0 that I have written in the above post only .N=0;E=5;1+R+T+T=X+10γ, 0≤γ≤2;γ+O=I+10; and 1+F=S.N=0;E=5;1+R+T+T=X+10γ, 0≤γ≤2;γ+O=I+10; and 1+F=S.\displaystyle N = 0;\\ E = 5;\\ 1 + R + T + T = X + 10\gamma, \ 0 \le \gamma \le 2;\\ \gamma + O = I + 10; \text { and }\\ 1 + F = S.
1st column: F+1=S
2T+ R+1=X+ 20 or X+10 . (2 possibilties)
So 0≤γ≤2 . But it should be 1≤γ≤2 .
COZ ,
2T+ R+1=X (if γ=zero) cannot happen as this equation will lead to no carry in 2nd column.
2nd column: O+1 or O+2=I +10 (2 possibilties)
Okay I will apply HIT and trial . But before that What assumption will i consider from the above 4 possibilities?Hint: We have ten letters, namely E, F, I, N, O, R, S, T, X, and Y.
I wish you would not use notation that is hard to understand. Who knows how you have numbered your columns.E=5 N=0 that I have written in the above post only .
1st column: F+1=S
2T+ R+1=X+ 20 or X+10 . (2 possibilties)
So 0≤γ≤2 . But it should be 1≤γ≤2 .
COZ ,
2T+ R+1=X (if γ=zero) cannot happen as this equation will lead to no carry in 2nd column.
2nd column: O+1 or O+2=I +10 (2 possibilties)
Okay I will apply HIT and trial . But before that What assumption will i consider from the above 4 possibilities?
Yes I grasp that you had already deduced that N = 0, E = 5, and F + 1 = S. I was trying to show you an orderly and systematic way to do these puzzles. I was talking about method, WHICH YOU REFUSE TO USE.
And yes, you reason correctly that
[MATH]\text {If } \gamma = 0 \implies 0 + O = I + 10, \text { which is impossible for the digit } O \implies 1 \le \gamma \le 2. [/MATH]
Very good.
Note that I > 0 because N = 0. Therefore 10 < I + 10 = gamma + O.
[MATH]\text {If } O < 9 \implies \gamma + O \le 2 + 8 = 10, \text { which is impossible because } 10 < I + 10 = \gamma + O \implies O = 9.[/MATH]
[MATH]\text {If } \gamma = 1 \implies 1 + 9 = I + 10 \implies I = 0, \text { which is impossible because } N = 0 \implies \gamma = 2 \text { and } I = 1.[/MATH]
It is time to restate our equations because now the problem gets hard (I have not yet proved there is a unique solution).
[MATH]N = 0;\\ I = 1;\\ E = 5;\\ O = 9;\\ 1 + R + 2T = X + 20; \text { and}\\ 1 + F = S.[/MATH]See what you can do with that (remembering the rules as you go). In the mean time, I shall look at the issue of uniqueness.
Nice deduction of value of O and I .
Seriously this level of thinking did not came to my mind.
I didn't found any logic.
I am absolutely bad in terms of Deep thinking.
Seriously this level of thinking did not came to my mind.
Why so?I shall give a second hint: X cannot equal 3.
I didn't found any logic.
I am absolutely bad in terms of Deep thinking.
@Dr.Peterson please help
First, good work on getting here.6<_ T <_ 8
6<_ R <_ 8
2<_ X <_ 4 .
Don't know about F,S, Y
Second, I do not know what “deep thinking” is, but I am pretty sure that these puzzles do not require it. What they do require is perseverence.
Third, although you have solved the problem, you have not told us how. Probably you used trial and error, which is a valid method.
However, you can keep using my method of reductio ad absurdum. Here is how.
From your point
[MATH]6 \le T \le 8, \ 6 \le R \le 8, \text { and } 2 \le X \le 4[/MATH]
you can fairly easily deduce that in fact 6 < T.
If T = 6, 2T = 12 so R is either 7 or 8, making R + 2T equal to 19 or 20. Adding 1 makes X equal to 0 or 1, both of which are impossible.So T = 7 or 8.
Now a little indirect thinking helps. If T and R use two of the three digits 6, 7, and 8, then neither F nor S can be 6, 7, or 8 because 1 + F = S. So either F = 2 and S = 3 or F = 3 and S = 4. Therefore X cannot equal 3.
If T = 7 and R = 6, then 1 + 6 + 14 = 21 so X = 1. Impossible.
If T = 7 and R = 8, then 1 + 8 + 14 = 23 so X = 3. Impossible
If T = 8 and R = 6, then 1 + 6 + 16 = 23 so X = 3. Impossible.
Therefore X = 8 and R = 7 so 1 + 7 + 16 = 24. Therefore X = 4, F = 2, and S = 3. The only unused digit is 6 so Y = 6.
The reductio is a very powerful method.
@Subhotosh Khan
It's impossible to create a new thread for every CRYPT ARITHMETIC problemn each time .
If I would have asked a new question on a new topic then I Will post it as new thread.
Don't be so Strict
It's impossible to create a new thread for every CRYPT ARITHMETIC problemn each time .
If I would have asked a new question on a new topic then I Will post it as new thread.
Don't be so Strict
D
Deleted member 4993
Guest
You say - IMPOSSIBLE !@Subhotosh Khan
It's impossible to create a new thread for every CRYPT ARITHMETIC problemn each time .
If I would have asked a new question on a new topic then I Will post it as new thread.
Don't be so Strict
Why?
Please help me in my new post.You say - IMPOSSIBLE !
Why?