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Tuesday, 1 January 2013

E-litmus Crypt-Arithmetic problem solving notes


                              E-LITMUS CRYPT-ARITHMETIC PROBLEM SOLVING MATERIAL





Here I am uploading some important algorithm to solve Crypto Arithmetic problem asked in Elitmus. Hope This section will help you  secure a good pH score ..........



** N:B

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Step by Step Solution to Crypto-Arithmetic Problem


Crypt-Arithmetic Problems are substitution problems where digits representing a mathematical operation are replaced by unique digits.
Like :
 P L A Y S + W E L L = B E T T E R

Where each unique alphabet represents a unique digit from among 0 to 9. So, if the solution to this puzzle is to be found, it would be (after a long computation) : 9 7 4 2 6 + 8 0 7 7 = 1 0 5 5 0 3


The basic rules are :
·                     Each unique digit must be replaced by a unique character.
·                     The number so formed cannot start with a ZERO.

There is no efficient way to calculating the correct substitution except simple mathematical deductions utilizing additive identity (which is a really cryptic way of saying A + 0 = A), multiplicative identity (A * 1 = A), and hit-and trial.

The solution becomes unmanageable if a systematic approach is not followed. So, here's a step-by-step approach to solving a Crypt-Arithmetic Problem.


                                                                 C1 C2 C3
                                                                 S   E   N   D
                                                                M  O   R   E
                                                           +______________
                                                           M   O   N   E   Y




SOLUTION


Value of M = 1 [ S<=9, M<=9, C1<=1, so S + M + C1 <=19 ]

Now, S = {9, 8}
when C1 = {0, 1}

Also, E + O + C2  = 10 + N,  if C1 = 1
Else,  E + O + C2 =  N, if C1 = 0

Similarly, N + R  + C3 = E + 10 , if C2 = 1
 Else,  N + R + C3 = E, if C2 = 0

And,  D + E = 10 + Y, if C3 = 1
Else,   D + E =  Y, if C3 = 0

Now, analyzing and deducing values from Right to Left, we get

If C1 = 1, S = 9 Then C1 + S + M = 11, which makes O=1 [False].

if C1 = 1, S = 8, Then C1 + S + M = 10, which makes O = 0.


if C1 = 0, S = 9, then C1 + S + M = 10, which makes O = 0.

So, O = 0 is valid.

Now,  Both the above given alternatives look equally probable at the moment.

Since O = 0, E + O + C2 will give carry only when C2 = 1 and E = 9.
But, that will give  E + O + C2 = 10, and N=0. [False, O = 0 is already established]

So, the second alternative must be correct, i.e. C1=0, S=9, O=0, M=1.

Now, E + O  + C2 = E + C2 = N.
If C2 = 0, then E = N [Invalid]
so, C2 = 1 and N = E + 1
Also,

 N + R + C3 = E + 10
 E + 1 + R + C3 = E + 10
or, R + C3 = 9

Now if C3 = 0, R = 9 [Invalid, S = 9]
So, C3 = 1 and R = 8.

Now,
Choices For E = {2, 3, 4, 5, 6, 7 }  so N = {3, 4, 5, 6, 7, 8 }
But 8 is already taken.

So, E = {2, 3, 4, 5, 6} and N = {3, 4, 5, 6, 7}

Now,
D + E = 10 + Y  >= 12

so, (D,E) = { (5, 7), (6, 7), (7, 5), (7, 6) }  which means N =  { 6, 7, 8}
But 8 is already taken so  N = {6, 7}
so, (D, E) = { (7, 5), (7, 6)}  i.e D = {7} and E = {5, 6}

Now, if E = 6, N=7 [invalid, 7 is already taken, D = 7]
so , E = 5, N = 6 and Y = 2.
Hence the Solution becomes ,

S = 9, E = 5, N = 6, D = 7, M = 1, O = 0, R = 8, Y = 2.

i.e.     9  5  6  7
      + 1  0  8  5
         ________
       1 0  6  5  2




That's about it.
Thank You.



N:B

All the file are password protected . click here and register for password . You will  get a mail containing password after registration.



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