CAT 2019 Shift 1 Paper

F+F=F.
This implies that F has to be 0 . Since all the digits are less than 10 , the maximum sum of any two letters will be 17 (i.e. 9+8).
Therefore, a maximum of 1 can be carried over to the digits place on the left. In the ten thousands place, either the units digit of H+H=F or the units digit of H+H+1=F.
It cannot be H+H+1 because 2H+1 is an odd number and the units digit of that cannot be 0. From this, we can infer that the units digit of 2H must be 0.
This implies that H=5. Sum of both the 6 -digit numbers is a 7 -digit number. This implies that the leftmost digit in the 7 -digit number has to be 1. Therefore. A=1
Since H=5,H+H=10. This means that B+A+1=AA=10A+A.A=1. Therefore. B=9
In the hundreds place the units place of the sum A+F is C. Now A=1 and F=0.
But C cannot be 1. Hence, C has to be A+1, i.e. C=2. Therefore, G+K should be greater than 10 .
The units digit of the sum of G+K is 1. This implies that G and K are either 3 and 8 or 4 and 7. not necessarily in that order.
We now have the following: 

	9	5	1	1	G	0
	1	5	J	0	K	0
1	1	0	G	2	1	0

Since G is a single digit, J+1 is less than 9.
If J=3.G=4 and K=7. In that case. D and E will be 6 and 8 . not necessarily in that order.
If J=6.G=7 and K=4. In that case. D and E will be
3 and 8 . not necessarily in that order.
If j=7.G=8 and K=3. In that case, D and E will be
4 and 6 not necessarily in that order.
If J=8,G=9, which is not possible because B=9 and each letter has a distinct number.
A represents 1.