There is a three digit number. Now move the leftmost digit to the rightmost digit, which is 45 times smaller than the original number. It is also known that 9 times of the number on the hundred digit is 3 times smaller than the two digit number composed of ten digit and one digit. Try to find the original number Solution of quadratic equation of two variables

There is a three digit number. Now move the leftmost digit to the rightmost digit, which is 45 times smaller than the original number. It is also known that 9 times of the number on the hundred digit is 3 times smaller than the two digit number composed of ten digit and one digit. Try to find the original number Solution of quadratic equation of two variables

Let X be the number on the hundred and y be the two digits of ten and one
10y+x=100x+y-45
9x=y-3
The solution of this system of linear equations is x = 4, y = 39
So the original number was 439

A two digit, ten digit is a, single digit is B, exchange ten digit and single digit, can be divided by 9

A two digit, ten digit is a, single digit is B, that is 10A + B
Exchange ten and one digits, that is 10B + a
…… Can it be divided by 9? Less typing?
10A + B - (10b-a) = 9a-9b = 9 (a-b), so the difference can be divided by 9

It is known that the sum of the number on the ten digit and the number on the one digit of a two digit number is 9. The two digit number obtained by transposing the two digit number is the same as the original two digit number
The difference is 27, find these two numbers

Let X be one bit, then 9-x be ten bits
x+10(9-x)-[(9-x)+10x}=27
x+90-10x-9-9x=27
-18x=-54
x=3
The two numbers are 63
Or (9-x) + 10x - [x + 10 (9-x)] = 27
9-x+10x-x-90+10x=27
18x=108
x=6
These two numbers are 36