CBSE Class 10 Answered

Let the x and y be the ten's and unit's digit respectively,

The number is 10x+y,

8(x+y) + 1 = 10x+y

13(x-y) + 2 = 10x+y

Equating the LHS of these equations since they have the same RHS.

8(x+y) + 1 = 13(x-y) + 2

21y = 5x + 1

x,y being positive integers,

Since RHS has to be at least 21, so that we can have y = 1.

x = 4, ie. 5x4 + 1 = 21

so 41 can be the number,

Also as we can see the RHS must be a multiple of 21,

and x can't be more than 9, since it's a two digit number.

5x9 + 1 = 46 which is 21x2 = 42 and 21x3 = 63.

Hence only one number is possible and it's 41.

Regards,

Team,

TopperLearning.