**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 ..........**

**Cryptal-Arithmetic -1[click to download]******Cryptal-Arithmetic-2[click to download]******Cryptal-Arithmetic-3[click to download]******Cryptal-Arithmetic-4[click to download]******Cryptal-Arithmetic-5[click to download]******Cryptal-Arithmetic-6[click to download]******Cryptal-Arithmetic-7[click to download]****

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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 :

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

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.

Thank You.

**N:B**

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