The digits of a positive integer, having three digits, are in A.P and their sum is $15$. The number .. The digits of a positive integer, having three 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 original number.
My Attempt:
Let the three digits number be $100x+10y+z$ where $x$, $y$ and $z$ are in A.P.
Then, 
$y=\frac {x+z}{2}$
$2y=x+z$.
Then, what should I do??
 A: So,
$$x-2y+z = 0$$
$$x+y+z = 15$$
$$100z+10y+x +594= 100x+10y+z \implies 99x-99z=594$$
i.e.
$$\left[ \begin{array}{ccc}
1 & -2 & 1 \\\
1 & 1 & 1 \\\
99 & 0 & -99 \end{array} \right] \left[ \begin{array}{ccc}
x \\\
y \\\
z \end{array} \right] = \left[ \begin{array}{ccc}
0 \\\
15 \\\
594 \end{array} \right]$$
Solving for $x,y,z$, our number is 852. 
A: I wouldn't bother using algebra on such an easy problem. The three digits are in arithmetic progression and add up to $15$ so the middle digit is $5.$ Since the number is bigger than its reversal, the only possibilities are $654,753,852,$ and $951.$ Let's see, $654-456=198,$ nope. $753-357=396,$ nope. $852-258=594,$ we have a winner. The answer is $852.$
A: Suppose the number is $abc$, notice that $594=abc-cba=99(a-c)$, so $a-c=6$.
Since the digits are in arithmetic progression they must be $a,a-3,a-6$. Since they add $15$ we have $a=8$.
So the number is $852$
A: You have used the fact that the digits are in arithmetic progression.  You have not used the fact that the difference between the original number and the reversed number is $594$.  You need to do so.  Now you have $(100x+10y+z)-(100z+10y+x)=594$.  Does this help?
A: You are off to a good start, but you need to use all of the given information. 
The fact that $x,y,z$ are an arithmetic progression means that $y = \dfrac{x+z}{2}$. 
The problem also gives that the sum of the digits is $15$, i.e. $x+y+z = 15$.
Finally, reversing the digits gives the number $100z+10y+x$. So we have $100z+10y+x = (100x+10y+z)-594$. 
Now, you have $3$ linear equations and $3$ unknowns. Is this enough to solve the problem?
A: You know that $x+y+z=15$, so if you replace $x+z$ with $2y$ in this equation, you'll find $y$. Also, using the fact that this is an arithmetic progression, you can say that $x=y+d$ and $z=y-d$, where $d$ is the difference of this progression (and $y$ already has a known numerical value). That leaves you with only one unknown quantity $d$. Set the difference of the original and reversed numbers to be equal $594$, and you'll be able to find $d$.
