Let E=7and A=4find the other digits in the sum
BASE
+BALL
=GAMES
Answers
Answer:
Step-by-step explanation:
Solving problems like these involves understanding some basic principles and rules of addition and a lot of trial and error.
Take our example,
BASE
+BALL
------------
GAMES
------------
Since the result is one digit more than the numbers, it is quite obvious that there is a carry over and therefore G must be equal to 1. Now consider the part,
SE
+LL
-------
ES
-------
This suggests that S+L=E or 10+E (with 1 as a carry) and E+L=S or 10+S. Now this happens only when L=5 and S~E (difference b/w S and E) = 5
This gives us paired values of (0,5), (1,6), (2,7), (3,8) and (4,9) as possible values for (E,S) or (S,E). But out of these (0,5) and (1,6) cannot be accepted as G=1 and L=5.
The first rule of cryptic equations like this is that different letters cannot have the same numerical value. So we are left with the possibilities of (2,7) , (3,8) and (4,9). We can also infer that of S and E , E is the smaller value and S is the larger, because if E were larger , we would have a carry and then S+L=E would not be valid. This means that S+L=E has a carry over of 1.
Let these values remain for now, we shall come back to them later.
Now coming to the part
BA
+BA
--------
GAM
--------
Since there is a carry obtained here, we can infer that B must be greater than or equal to 5. But L=5, therefore B>5. Now we have two cases - If the sum A+A produces a carry, then A is odd, else A is even. Since there is a carry over from S+L=E, M has to be an odd number.
Now we have gathered all the information we can and there is nothing else to do.
To proceed we shall have to use the trial and error method substituting values for the letters keeping all the above points in mind.
Let us assume E=2 and S=7 and B=6. So we have,
1
6A72
+6A55
-----------
1AM27
------------
Now A can be either 2 or 3 depending on whether we have a carry from A+A or not. But since E=2, that means A must be 3 and therefore there is a carry. But replacing the other A's in the equation with 2's gives us two contradictions. Firstly M shall become equal to 7 (S is already equal to 7) and A+A does not produce a carry. Therefore our assumptions were wrong and we will have to try again for a different values.
(I shall skip to the combination which yields the solution, but you shall have to try for all possible values in between)
Now let us try for E=3 and S=8 and B=7. We have,
1
7A83
+7A55
----------
1AM38
-----------
This gives us A=4 or 5 based on whether there is a carry or not, but since already L=5, A must be equal to 4, therefore M=9. We have obtained values for all unknowns without any contradictions and hence this is the solution)
So finally we have
1
7483
+7455
---------
14938
---------
Therefore,
G=1
E=3
A=4
L=5
B=7
S=8
M=9
Some other useful observations would be.
If,
AB
+CD
-------
AE
--------
We can conclude that C=0
and if
AB
+CD
-------
EAF
-------
Then E =1 and C=9