Here is the question to which I am referring:
Find three digit numbers that are divisible by 5 as well as 9 and whose consecutive digits are in AP.
My way to find answer:
- I first arbitrarily chosen three numbers $a-d$, $a$, $a+d$ in AP where $a-d$ is at hundreds place , $a$ is at tens place and $a+d$ is at ones place and here $d$ is the common difference.
- Then we know for a number to be divisible by $5$ it should have $0$ or $5$ at one's place so I first set $$a+d=0.\tag{1}$$ But we also know number is divisible by $9$ so $((a-d)+a)$ should be equal to $9$ or $18$ $$2a-d=9\text{ or }18. \tag{2}$$ Now I solved them simultaneously to get two values of $a$ and $d$ but one value was needed to be discarded. Same procedure I used for $$a+d=5$$ and finally I got three digit numbers whose consecutive digits are in AP Those numbers are $630$, $135$, $765$.
So my question is that is there a way to do this problem by another way which involves more AP concepts as I think this way requires very less use of AP concepts.
I think you can use the fact that $(a-d)+a+(a+d)=3a$ is divided by $9$. From here it is clear that $a=3$ or $a=6$. Now we will use the fact that the number is divided by $5$.
If $a=6$ then $d$ must be $-1$, and if $a=3$, then $d$ can be $2$ or $-3$.