# How To Solve Cryptarithmetic Questions Quickly

## Solve Crptarithmetic Questions:

Crypt arithmetic is a mathematical genre, where all the digits are replaced by any other symbol or the letters of an alphabet, and if the same letter reappears in a word, then it must be allotted the similar digit or a number, each time when it is being used. Go through this page to get an idea about How to Solve Cryptarithmetic Questions Quickly.

Each letter will be assigned with an individual digit, and no two letters will have the same number.

Now the main questions arise when we have to find the unique digit assigned to each corresponding letter.

Cryptarithmetic is considered to be, both a science as well as an art. Hence apart from logic, one must use his/her presence of mind and a little bit of common sense to solve the problems. ## Solve Cryptarithmetic Questions Quickly

### Question 1

Decode and solve the below mentioned Crypt Arithmetic problem:

Solution: Since it is already mentioned that the carry value of resultant cannot be 0 then lets presume that the carry value of D is 1

We know that the sum of two similar values is even, hence R will have an even value
Hence S+S= R So R is an even number for sure.
So the value of R can b (0, 2, 4, 6, 8)

Value of R cannot be 0 as two different values cannot be allotted the same digit,(if S=10 then their sum = 20 carry forward 2, then the value of R= 0) which is not possible.
IF S= 1 :
Not possible since D has the same value.

IF S = 2

Then R= 4 which is possible Hence S= 2 and R= 4
C4+C+R= A+10
C4+C+4= A+10
C4+C>5 (Being the value of carry will be generated when the value of C is greater than 5

C= 9
C1+S+D= E
C1+2+1= E
Therefore E= 3
C4+C+R= A+10
C4+9+4= A+10
Therefore A= 3 but it cannot be possible as E= 3
Now lets Consider S+D+C1= E
2+1+0= 3
Therefore E= 3 making C2= 0 since 2+1=3
Now lets consider the equation again:
C+R+C4= A+10
9+4+0= A+10
13= A+10
Therefore A= 3 but E= 3
So A is not equal to 3
Again considering R= 6 So S= 3 C4= 0
C+R+C4= A+10
9+6+0= A+10
15= A+10
Therefore A= 5
And S+D+C1= E
3+1+0= E therefore E= 4 and C2= 0
Now considering the equation
R+O+C3= N
6+0+C3= N
So 6+0+C3< or equal to 3
Let C3= 1
Then O< or equal to 2
That is O= 0, 1, 2
Let O =2
Again considering R+O+C3= N
6+2+1= N
Hence N= 9 but C= 9 so N cannot be equal to 9.
Now let N= 8 and C3= 0
Let us consider equation
O+A+C2= G
Therefore G= 7
Hence D= 1 S= 3 A= 5 G= 7 C= 9 O= 2 E= 4 R= 6 N= 8
And C1= 0 C2= 0 C3= 0 C4= 0 C5= 1
Now verifying the above values in the equation we get:

C5C4C3C2C1 ### Question 2.

If N O + G U N + N O = H U N T, find the value of HUNT.

A. 1082
B. 1802
C. 1208
D. 1280

#### Solution: Here H = 1, from the NUNN column we must have “carry 1,” so G = 9, U = zero. Since we have “carry” zero or 1 or 2 from the ONOT column, correspondingly we have N + U = 10 or 9 or 8. But duplication is not allowed, so N = 8 with “carry 2” from ONOT. Hence, O + O = T + 20 – 8 = T + 12. Testing for T = 2, 4 or 6, we find only T = 2 acceptable, O = 7. So we have 87 + 908 + 87 = 1082. Hence option A is the correct one.

### Question. 3

If “EAT + THAT = APPLE”, what is the sum of A+P+P+L+E?

A. 13
B. 14
C. 12
D. 15

#### Solution: From the given data, the value of A will be 1 because it is the only carry-over possible from the sum of 2 single-digit number. T maximum it can take only 9 and there should a carryover for T to give sum as 2 digit number. So T =9, P = 0, A = 1. T + T = 18, the value of E is 8 and 1 will be a carry over to the next column. That is 1 + A + A= L = 3. And finally H = 2. Hence, 819 + 9219 = 10038. A+P+P+L+E = 1+0+0+3+8 = 12. 