Question

5.
If
ABC
+ CBA
then the number of possible values for A, B, C, D, E satisfying this
equation where A, B, C, D and E are distinct digits is
DE DD
A
6
B)
5
C)
4
D) 3​

Answer 1

Answer:

4

Step-by-step explanation:

  A B  C

  C  B  A

D E  D  D

D must be 1 as D is carryover of sum of two digits

C+ A = D

=> C + A = 11

A + C = E => E = 2 With a Carry over from B + B

Now B + B + 1 = D  = 11  ( as Carry over is also required)

=> B = 5

A    5    C

C    5    A

1  2   1    1

A & C Can be

3 , 8

4 , 7

7 , 4

8 , 3

number of possible Values = 4

358 + 853 = 1211

457 + 754 = 1211

754 + 457 = 1211

853 + 358 = 1211