This is a project made in python programing language to solve crypt arithmetic problems which is a part of logical reasoning and can be asked in interviews.
Decode and solve the below mentioned Crypt Arithmetic problem:
CROSS+ ROADS= DANGER
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
If “EAT + THAT = APPLE”, what is the sum of A+P+P+L+E?
A. 13
B. 14
C. 12
D. 15
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.
