Question

the sum of 2 - digit number is 9 when we subtract 63 from it, its digits get interchange. find the sum and product of the original number and the number obtained by subtracting 63 from it.​

Answer 1

Hi....heres your answer....>

Let the ones place digit number be x.

Let the tens place digut number be y.

Now,

The original number= 10y+x.

When the digits are interchanged,

the numbe= 10x+y.

A/Q,

In 1st Case

x+y=9-----------eqn.1

In 2nd Case

(10y+x)-63=10x+y

=>10y+x -63=10x+y

=>9y-9x=63

=>y-x=7------------eqn.2

Now,

eqn.1 + eqn.2, we get,

2y=16

y=8

Now,

Putting y=8 in eqn.1,we get,

x+8=9

=>x=1

So,

The original two digit no.=10*8+1=81

The reversed two digit no.=10*1+8=18

Now,

Product of the digits of the original number=8

Sum of the digits of the original number

=9

And,

The Number obtained by subtracting 63 from the original no.=81-63=18