Insert a digit from 0 to 9 between a two digit ten digit and a single digit, and the two digit becomes a three digit, which has the sum of the three digit and the two digit It is 10 times of this two digit number. Find all the three digits that meet the condition

Insert a digit from 0 to 9 between a two digit ten digit and a single digit, and the two digit becomes a three digit, which has the sum of the three digit and the two digit It is 10 times of this two digit number. Find all the three digits that meet the condition

Let the original two digits be AC and the three digits be ABC
Then (100a + 10B + C) + (10a + C) = 10 (10a + C)
It is concluded that a + B = 4C / 5
Then C = 5
A+B=4
When a = 1, B = 3, the three digit number is 135
When a = 2, B = 2, the three digit number is 225
When a = 3, B = 1, the three digit number is 315
When a = 4, B = 0, three digits 405

All natural numbers can be divided into 10 categories according to their equal digits: those with one digit are called the first category, and those with nine digit are called the ninth category
(1) Take out six different natural numbers randomly. Is the sum of 2 numbers a multiple of 10?
(2) Take out 7 natural numbers which are different from each other. Is the sum of 2 numbers a multiple of 10?
If so, please give reasons; if not, please give counter examples

If you take 1, 2, 3, 4, 5, 10, you can't get the sum of two numbers, which is a multiple of 10
7 words because of 1 + 9, 2 + 8, 3 + 7, 4 + 6, 5, 10
In this way, there are six combinations, of which the sum of two numbers in 1-4 combinations is a multiple of 10
Taking 7 numbers, there must be at least one group in 1-4 combinations, and taking 2 numbers, so (2) must meet the requirement