Question

The sum of two digited number and the number obtained by reversing the digits is 66. If the digits

of the number differ by 2. Find the number. How many such numbers are there ?​

Answer 1

ANSWER:

• Original number = 42 and 24

GIVEN:

• Sum of two digited number and the number obtained by reversing the digits is 66.
• The digits differ by 2.

TO FIND:

• Original number

SOLUTION:

Let 'x' be the digit at tens place.

Let 'y' be the digit at once place.

Original number = 10x+y

Reversed number = 10y+x

Now

According to the question:

Case 1

=> 10x+y+10y+x = 66

=> 11x+11y = 66

=> 11(x+y) = 66

=> x+y = 66/11

=> x+y = 6. ....(i)

Case 2

When tens place is greater than once place.

=> x-y = 2 ....(ii)

Adding eq(i) and (ii) we get;

=> x+y+x-y = 6+2

=> 2x = 8

=> x = 8/2

=> x = 4

Putting x = 4 in eq (i) we get;

=> 4+y = 6

=> y = 2

In this case:

Original number = 10x+y

Original number = 10x+y= 10*4 + 2

Original number = 10x+y= 10*4 + 2= 42

Case 3

When digit of one's place is greater than tens place.

=> y-x = 2 .......(iii)

Adding eq (i) and (iii) we get;

=> 2y = 8

=> y = 4

Putting y = 4 in eq (I) we get;

=> x+4 = 6

=> x = 2

In this Case :

Original number = 10x+y

0x+y= 10*2+4

0x+y= 10*2+4= 24

• Original number = 42 and 24
• There are 2 such number satsfying the eqn.