Possible Numbers

A handy way to deal with digits is by expressing the number algebraically. Here we have $\overline{abc}=9\overline{ac}+4c$ $100a+10b+c=9(10a+c)+4c$ $10a+10b=12c$ $5(a+b)=6c$ Since 5 is a prime number, and 6 is not divisible by 5, $c$ must be a multiple of 5. However, $c$ can only be a single digit number as it is a digit in the number $\overline{abc}$ . Therefore we must have $c=0$ or $c=5$

If $c=0$ , then $a+b=0$ , which is impossible as $a$ is the first digit of the three digit number $\overline{abc}$ .

If $c=5$ , then $a+b=6$ , and hence we have 6 solutions $(1,5), (2,4), (3,3), (4,2), (5,1), (6,0)$ . Keep in mind that $a \neq 0$ .

Therefore, there are 6 numbers satisfying the equation: $\boxed{155, 245, 335, 425, 515, 605}$

Let the three digit number $\overline{abc}$ be $100a + 10b + c$ and similarly ; Let $\overline{ac}$ be $10a + c$ .

Now according to the given equation , $100a + 10b + c = 9(10a + c) + 4c \\ \implies 5(a+b) = 6c$

Now as $a,b,c$ are integers between 0 to 9 , and $a\neq 0$ , the only possibility is $a+b = 6$ and $c = 5$ .

Therefore , by taking $a= 1,2,3,4,5,6$ and using $b=6 - a$ , we get the total such numbers as $\large \color{#3D99F6}{\boxed{6}}$ .