The unit’s digit of a two-digit number is one more than the digit at ten’s place. If the number is more than five times of the sum of the digits of the number, then find the sum of all such possible numbers.

Let digit at tenʼs place be x.

Digit at unitʼs place = x + 1

5 times sum of digit = 5(x + x + 1)

=10x + 5

The number has to be greater than (10x + 5).

Also x can take values from 1 to 8 only.

x = 1, no. = 12

x = 2, no. = 23

x = 3, no. = 34

x = 4, no. = 45

For all these values of ‘x’, number is not more than 5 times the sum of digits.

x = 5, no. = 56

x = 6, no. = 67

x = 7, no. = 78

x = 8, no. = 89

For all these numbers the numbers are greater than 5 times the sum of digits.

So, sum of numbers = 56 + 67 + 78 + 89 = 290