A three digit number is 9. The new number obtained by inverting ten digits and one digit is 9 larger than the original number. How many are the original two digits?

A three digit number is 9. The new number obtained by inverting ten digits and one digit is 9 larger than the original number. How many are the original two digits?

Let two digits be: ab
a+b=9
10b+a-(10a+b)=9
b-a=1
therefore
b=5,a=4
therefore
The original double digits were 45

The sum of a two digit number and a ten digit number is 9. If the new number obtained by transposing the two digit number and a ten digit number is 9 larger than the original number, the original two digit number is ()
A. 54B. 27C. 72D. 45

Let X be the one digit number of the original number, then the ten digit number is 9-x. according to the meaning of the question: 10x + (9-x) = 10 (9-x) + X + 9, the solution is: x = 5, 9-x = 4, then the original two digit number is 45

Insert a 0 between the ten and the single digits of two numbers, and the two digits become three digits, which is exactly 9 times of the original two digits

Let the original number be 10A + B, then the three digits after inserting 0 are 100A + B
100a+b=9（10a+b）
10a=8b
a：b=4：5
So B can only be 5 and a can only be 4
So this double-digit number is 45