# eLitmus Cryptarithmetic Questions and Answers

## eLitmus Cryptography Questions and Answers

eLitmus Cryptarithmetic questions and answers are discussed in this page. You will find every possible eLitmus Cryptarithmetic problems with solutions that is asked in eLitmus online test.

The page will vary from low difficulty to high difficulty Cryptarithmetic eLitmus Questions with set of rules and step wise solutions which will help you understand Cryptarithmetic eLitmus problems more easily and efficiently.

1. All Rules to solve Cryptarithmetic
2. Tricks and Shortcuts
3. Most asked Questions from latest Drive Name of Exam eLitmus (pH Test) No. of Questiuons 4 Difficulty Medium – Hard Importance High ### Rules to Solve Infosys Cryptarithmetic Questions

1. Every Character/letter must have a unique and distinct value
2. The values of a character/letter can not be changed, and should remain same throughout.
3. Starting character of number can not be zero example – 0341 should be simply 341.
4. The problem will have only and only one solution
5. Addition of two similar numbers is always even
6. In case of addition of two numbers, if there is carry then, the carry can only be 1
7. Once all the characters/letters are replaced with numbers, arithmetic operations must be correct

#### Explaining Rules Stepwise

Rule 1 : Every Character/letter must have a unique and distinct value i.e.,

`A = 0; B = 1; C = 2; D = 3; E = 4; F = 5; G = 6 etc`

In the above line numeric value is not fixed for, these alphabets these are just examples.

Rule 2 : The values of a character/letter can not be changed, and should remain same throughout the solution.

```            X Y Z
+ X Z Y
--------
Y Z X```

Here if we find the value of Z = 9 or something else then the value of Z will be 9 in the entire process.

Rule 3 : Starting character of number can not be zero.

```            X Y Z
+ X Z Y
--------
Y Z X```

Here the value of X can not be zero because then the first number i.e. X Y Z will be just Y Z.

Rule 4 : Addition of two similar numbers is always even.

Here if we take Z = 1 then 1 + 1 = 2.

Or Z = 2 then 2 + 2 = 4

i.e in every case that weather the number is odd or even their sum is always even.

```            Z
+ Z
--------
EVEN```

Here the value of Z + Z will always be even number.

Rule 5 : Z + Y = Z then the possible value for Y will be either 0 (Zero) or 9.

For example

when Y = 0,

2 + 0 = 2

when Y = 10

2 + 10 = 12

Here also the unit place represent Z whose value is 2 and the tenth place is 1 which will be either a number or it will act as carry for the next number.

```             Z
+  Y
--------
Z```

Here either Y will be 0 (Zero) or 9.

Rule 6 : In case of addition of two numbers, if there is carry then, the carry can only be 1.

Now if we assign the maximum possible value i.e.  9 to Z then also by adding

Z + Z will be 9 + 9 = 18 then also it will give the maximum carryover of 1 only.

```             Z
+  Z
--------
X Y```

Here the maximum value for Z can be 9 acc. to rule 1.

## eLitmus Cryptoarithmetic Question 1 Question – X Y Z  + X Z Y = Y Z X. Find the value of X + Y + Z?

1. 15
2. 11
3. 7
4. 18

Given information we have :

• X, Y, Z have unique and distinct value (From Rule 1)
• Value of X cannot be zero because then the first number will be just Y Z (From Rule 3)
```            X Y Z
+ X Z Y
--------
Y Z X```

Step 1 :

For least significant bit i.e

Z
+ Y
——
X

Now there is 2 possible case i.e.,

Case 1 :

Z + Y = X

(which means that there is no carry over is getting generated)

Case 2 :

Z + Y = 10 + X

(which means 1 as a carry is getting generated)

```            X Y Z
+ X Z Y
--------
Y Z X```

Step 2 :

Case 1 :

Z + Y = X

If we follow Case 1 then The value of Y in Y + Z = Z will be Zero

0
+ Z
——
Z

```            X Y Z
+ X Z Y
--------
Y Z X```

Step 3 :

Now if we take the value of Y = 0 then the value of X in X+ X = Y must also be Zero which is not possible because of Rule 1 i.e. Every Character/letter must have a unique and distinct value

```            X Y Z
+ X Z Y
--------
Y Z X```

Step 4 :

Now by taking Case 2 : i.e.,

Z + Y = 10 + X

(which means 1 as a carry is getting generated)

```            X Y Z
+ X Z Y
--------
Y Z X```

Step 5 :

Here the possible value of Y is between 0 to 9 but

Y + Z = Z (it means either Y has to be Zero or 10 then only it is possible)

For example

0 + 5 = 5

10 + 5 = 15

(from above example it is clear that that value of Y has to be 10 and is only possible when there is one carry over and the original value of Y is 9)

From above discussion we can conclude

Step 7 :

Y = 9

1 (carry over)
9 (Value of Y)
+ Z
—
Z

```            X Y Z
+ X Z Y
--------
Y Z X```

After solving for the value of Y we can conclude that :

```            X 9 Z
+ X Z 9
--------
9 Z X```

Step 8 :

Note : Now there will definitely be a carry of 1 because 9 + Z will be surely above 10 and it will transfer of carry 1 on X

Now by Rule 1 we can that the value of both the X in X + X = 9 is same and there is 1 as carry over.

Thus we have an equation :

1 + X + X = 9

2X = 9-1;

2X = 8;

X = 8/2;

X = 4 ;

```            1 1            X 9 Z
+ X Z 9
--------
9 Z X```

Now by evaluating the value of X we can conclude that :

```            1 1            4 9 Z
+ 4 Z 9
--------
9 Z 4```

Step 9 :

Now by evaluating the least significant digit of both the numbers i.e., X Y Z & X Z Y

we can have an equation :

(Note : But you should keep in mind that adding these two number will create a carry of 1 on the next number)

Z + 9 = 4 ;

Z = 4 – 9;

Z = -5 ;

Z = 5 (because 5 + 9 = 14 where X =4 and 1 is acting as carry over for the next number.)

```            1 1            4 9 Z
+ 4 Z 9
--------
9 Z 4```

Here we can say that the value of

X = 4;

Y = 9;

Z = 5;

So the value of  X + Y + Z = 4 + 9 + 5 = 18 (which is option D).

```            1 1            4 9 5
+ 4 5 9
--------
9 5 4```

## eLitmus Cryptoarithmetic Question 2 Question –   For the following Cryptarithmetic find the values of J, B , A and D to the below questions? And assume that E = 4.

1. J = 9, B = 1, A = 0, D = 3

2. J = 8, B = 2, A = 0, D = 3

3. J = 9, B = 2, A = 0, D = 4

4. J = 7, B = 1, A = 0, D = 3

Given information we have :

• J, E, B, A, D have unique and distinct value (From Rule 1)
• Value of both B’s in B B are same. As both the Alphabets are same.
• Value of E is 4. (from question)
```              J E
x   B B
--------
J E            J E A          --------          B A D E   ```

Step 1 :

• The value of both the B’s in B B can only be 1 because when B B is getting multiplied by J E the output we are getting is J E.
```              J E
x   B B
--------
J E
J E A
--------
B A D E```

Step 2 :

• The only possible value for A can be 0 because the the space of GREEN “A” is blank. And blank space can only be filled by Zero.
• Hence the value of RED “A” will also be 0 (by Rule 2 ).
```              J E
x   1 1
--------
J E
J E A
--------
1 A D E```

Step 3 :

• The value of E is already know i.e. 4
```              J 4
x   1 1
--------
J 4
J 4 0
--------
1 D 4```

Step 4 :

• The only possible value for GREEN “J”  can only be 9 because then only with the help of carry 1 it will become 10 and 0 will come as output and 1 for the next digit.
```              J 4
x   1 1
--------
J 4
J 4 0
--------
1 0 D 4```

Step 5 :

• Now by the help of equation i.e.

9 + 4 = D

We get

D = 13

in which 3 will be the original value of D and 1 will act as carry over.

So the original value of D is :

D = 3

```              9 4
x   1 1
--------
9 4
9 4 0
--------
1 0 D 4```

From above analysis we can say that :

J = 9

B = 1

A = 0

D = 3

```              9 4
x   1 1
--------
9 4
9 4 0
--------
1 0 3 4```