The digits of a positive

Question

The digits of positive integer having $3$ digits are in A.P and their sum is $15$. The number obtained by reversing the digits is $594$ less than the original number. Find the number.

My Attempt ;

Let the $3$ digit number be $100x+10y+z$ where $x$,$y$ and $z$ are at hundredth, tenth and ones places respectively.

According to question;

$100x+10y+z=15$

Now, what should I do next?

Answer 1

If $x,y,z$ are in $AP$ then $\frac{x+z}{2}=y$

$AP$ means arithmetic progression. The differences between $x$ and $y$ and $y$ and $z$ are equal. For more information see here.

The equation for the sum of the digits is $x+y+z=15$.

And finally we can take up your idea for the last condition:

$100x+10y+z=100z+10y+x+594$

The reversed number is 594 less than the origin number. Thus we have to add 594 to get an equality.

Answer 2

According to question: $x+y+z=15$. Can you go on from there?

Answer 3

You have $$\begin{cases}x+y+z=15\\100z+10y+x=100x+10y+z+594\end{cases}$$ so you get $$z-x=6\Rightarrow 2x+y=9$$ The possibilities come from $$\begin{cases}y\in\{1,3,5,7,9\}\\2x\in\{2,4,6,8\}\end{cases}$$ The three solutions are $$N\in \{177,258,339\}$$