A two-digit number is 18 more than the number formed by reversing the digits.If the sum of the digits is10.find the number??

Let the two-digit number be represented as 10a + b, where a is the digit in the tens place and b is the digit in the ones place.

According to the given conditions:
10a + b = 10b + a + 18
9a - 9b = 18
a - b = 2

Also, it is given that the sum of the digits is 10:
a + b = 10

Solving the above two equations simultaneously:
a - b = 2
a + b = 10

Adding the two equations:
2a = 12
a = 6

Substitute the value of a back into the equation a + b = 10:
6 + b = 10
b = 4

Therefore, the two-digit number is 64.

Let the ten's digit be x and unit's digit be y .
Then the number is 10x + y .
Reversing the digits we get 10y + x
Now 10x + y= 10y + x +18
= 9x -9y=18
= X-y=2
Also x+y=10
Adding we get 2x=12
= x = 6
= y=4
Therefore the number is 10x+y =10 (6)+4=64