Question

the sum of a two digit 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

Let the number be 10x+y
Then reversed no is 10y+x
According to question ;
10x+y+10y+x=66
11x +11y =66
x+y=6 [1]
Also
x-y=2 [2]
Adding [1] and[2]
2x=8
x=4
From [1]
y=2
So the required no is 42
There are two such number that is 24 and 42

Answer 2

Let the numbers be( x+10y) and on reversing (y+10x)

(x+10y)+(y+10x) = 66

11x+11y = 66

11(x+y) = 66

x+y = 66/11

x+y = 6 (be the equation 1)

Digits differ by 2, so x-y = 2 (be the equation 2)

Adding 1 and 2 equations we get , x+y = 6 + x-y = 2

2x = 8

x = 8/2

so, x = 4

Put value of x in equation 1 .

x + y = 6

4 + y = 6

y = 6 - 4

y = 2

∴ number is :- ( 4+10*2) = 24 or 42