N is a 3-digit positive integer. The sum of all 3 digits of

Instead of writing the equations, I find the following approach to be time-saving:

We know that if we exchange the hundred and the unit digit, the new number is 99 less than the original. With this info, let us try the options:

A. 944
New number: 449
944-449 > 99

B. 449
New number: 944
449-944 < 99

C. 548
New number: 845
548-845 < 99

D. 845
New number: 845
845-548> 99

E. 746
New number: 647
746-647 = 99
E is the answer.

=>

Assume N is a three-digit integer represented by xyz.
Then we have an algebraic expression N = 100x + 10y + z.
We have x + y + z = 17, x + z = 13 and 100z + 10y + x = 100x + 10y + z – 99.
If we subtract the first 2 equations we get (x + y + z) – (x + z) = 17 – 13 and y = 4.
When we substitute this into the last equation, we have 100z + 10*4 + x = 100x +10*4 + z - 99, 99z – 99x = -99. If we multiply by -1 we get 99x - 99z = 99 or x – z = 1.
Then we have x = 7 and z = 6, since x + z = 13 and x – z = 1.
So, the original number N is 746.

Therefore, E is the answer.
Answer: E