Question

Find three-digit numbers that are divisible by 5 as well as 9 and whose consecutive digits are in A.P

Answer 1

Step-by-step explanation:

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.(1)

But we also know number is divisible by 9 so ((a−d)+a) should be equal to 9 or 18

2a−d=9 or 18.(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.

Answer 2

135 990 17 are three numbers which are divisiblep by 5