9 +9 = 18  ; 9 x9 =  81

Here the sum and the product pf two numbers give resuts where the  digits are revered. Can you find out more examples ?

Let m and n be the two numbers satisfying the required condition.

m+ n = 10 x + y and mn = 10y + x where x, y are any single digit number

m = 10x + y – n  = (10y +x )/n

which gives n²  = x(10n -1)- y(10-n)

To make n to be positive  x(10n-1)  > y(10-n)

n=1;  1 = 9(x-y) , there is now whole number solution

n = 2; 4 =  19x – 8y, which gives  x =4 and y =9  ,or m = 47

2 + 47 = 49 l 2 x 47 = 94

n=3 ; 9 = 29x – 7y, which provides x =2. Y =7 or m = 24

3 + 24 = 27  ; 3 x 24 = 72

For higher values of n , n = 4,5,6…., there is no such pair of numbers

If the sum and product are three digit numbers,then

 m+n = 100x + 10 y + z and mn = 100z + 10y + x

m = (100z +10y +x)/n = 110x + 10y +z – n

n² = x(100n -1) +10y (n-1) – z(100-n)

when n = 2 ; 4 = 199x + 10y – 98z

when x = 4 , z= 9  we have 786 + 19y – 882  which gives y = 9 0r m – 497

2 + 497 = 499 ; 2 x 497 = 0=994

This is true for any number of 9 in between 4 and 7.

2 + 4997 = 4999 ; 2 x 4997 = 9994

2+49997 = 40000 ; 2 x 49997 = 99994