Thursday, September 6, 2018

SECOND EXAMPLE OF CRYPTARITHMETIC PROBLEM

Here we are provided with the following information
           BASE
         +BALL
       ------------
       GAMES
There are seven distinct digits from 10 preliminary digits that are from [0-9]:A,B,S,E,LG,M

As we are just adding 2 numbers so possible carry overs are either 1 or 0
But, when we are adding B+B it gives us some carry that is greater than zero as both 4 digit numbers add up to form a five digit number 
Hence, G can't be zero
So, G=1

Now our question will look somewhat like this:
           BASE
         +BALL
       ------------
        1 AMES
Then,
if 2 B's are added and they are giving us value greater than equal to 10
Then,B must be greater than equal to 5


Consider B=9 first


if B=9 then A=8 which is not possible as then B+B+1 is not equal to A
if B=8 then A=6 which is not possible for same reason
Take B=7 then A=4 which is possible as it is not leaving any carry over.

So,

M=A+A+x where x is carry over from S+L+y where y is carry over from E+L
knowing the fact that x and y can take maximum value of 1 and minimum value of 0
We can say


M should be either 8 or 9
8 when x=0
9 when x=1


Assuming M=8 
that is x=0
S+L+y=E
E+L= 10*y+S
S+L+y+L=10*y+S
2*L=9*y
where y can take either 0 or 1


if y=0 then L=0 then S and E are not distinct which is not possible according to question.
if y=1
then L=4.5 which is again not possible as L must be a single digit among 0 to 9
So, M=8 is not possible
So,our question further reduces to
             1
           74SE
         +74LL
       ------------
        1 49ES

Now, 
S+L=10*1+E=10+E,and
E+L=10*y+S
So,
here y can take maximum and minimum values of 1 and 0 respectively.
So,
if y=0
Then, 
E+L=S,or S-L=E,and
S+L=10+E
Adding both we get,
2*S=2*E+10
or 
S=E+5
Subtracting 1st equation from 2nd we get
2*L=10
or L=5

So,our question further reduces to
             1 0
           74SE
         +7455
       ------------
        1 49ES

So,
E+5=S,and
S+5+0=E+10*1
S+5=E+10
S=E+5
Now we have to choose digits from 0,1,2,3,4,5,6,7,8,and 9 other than 1,4,7,9,and 5.
So,we can take values of E and S from 0,2,3,6,8 such that S=E+5
0 is there 5 can't be chosen
2 is there 7 can't be chosen
6 is there 11 can't be chosen
8 is there 13 can't be chosen 
So, only possible pair left is 3 and 8 which satisfy our constraints.
Hence,
S=8 and E=3

             1 0
           7483
         +7455
       ------------
        1 4938
So,
A=4,
B=7,
S=8,
E=3,
L=5,
G=1,
M=9

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