CAT 2019 LRDI Questions with Answers 1

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:

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 = 4. G = 5. which is not possible because H = 5 and each letter has a distinct number.

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.