Why can the sum of the new number and the old number be divided by 11

Why can the sum of the new number and the old number be divided by 11

Let a be a in the tens and B in the ones,
Then the original number is 10A + B and the new number is 10B + a
The sum of them is: (10a + b) + (10b + a) = 11 (a + b)
So, sum can be divided by 11

Transpose a two digit single digit and a ten digit number to get a new two digit number. Try to explain that the sum of the new two digits and the original two digits must be divisible by 11!
I'm very polite!

Number 1: X number 2: y
Original number: 10x + y
New number: 10Y + X
（10y+x）+（10x+y）=（10x+x）+（10y+y）=11x+11y=11(x+y)

Insert a number from 0 to 9 between the ten and the single digits of two natural numbers, and the two digits will become three digits. The three digits after inserting a number between some two digits will be 9 times of the original two digits______ One

According to the meaning of the question, we can turn the original question into a number puzzle: from 5 × 9 = 45, we can get that B can only take 5, a × 9 + 4 must carry, so a = 1, 2, 3, 4, so the two digits are 15, 25, 35, 45 respectively. There are four such two digits. So the answer is: 4

For a natural number with two digits, the sum of the number in ten digits and the number in one digit is 9——————

10a+（9-a）=9a+9