Find the sum of all two-digit numbers that, when divided by 8, gives the remainder of 1.

A number that, when divided by 8, gives a remainder of 1 can be written as 8K + 1, K is a natural number The number is two-digit, so 9 <8K + 1 <101. Solve the inequality: 9 <8K + 1 <101. 8 <8K <100 1 k belongs to {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} => 8K + 1 belongs to {17, 25, 33, 41, 49, 57, 65, 73, 81, 89, 97} Adding these numbers, I get: 627.
Answer: 627