Question about arithmetic progressions

[Robert Houdart]

Homework Statement

Of a 4 digit positive integer, the four digits form an Arithmetic progression from left to right. How many such 4 digit integers exist?

2. The attempt at a solution

If d = 1, the integers are 1234, 2345, …, 6789. These 6 integers and their reverses satisfy the given criterion. In addition to this, 3210 also satisfies the given criterion. So, if d = 1, there are 13 such integers. If d = 2, the integers are 1357, 2468 and 3579. These 3 integers and their reverses also satisfy the given criterion. In addition to this, 6420 also satisfies the given criterion. So, if d = 2, there are 7 such integers. If d = 3, the only integer is 9630. Thus, there are 21 such integers.

However the book states that for d=0 their exist 9 numbers

My question here is that does an arithmetic progression exist with common difference =zero

 

[jedishrfu]

I don't understand what you're asking.

Doesn't the d=0 mean the common difference is zero? and the nine numbers are 1111, 2222...

 

[Robert Houdart]

Yes, if d=0 is considered, nine more numbers are added to the solution set (1111 , 2222..., 9999)
However, my question is, a series (x, x, x, x, x, x, x) be considered an AP (that is can an AP have common difference=0)

 

[jedishrfu]

I think the answer is yes again although that's a trivial case like 1,1,1,1,1,1,1,1,1...

http://en.wikipedia.org/wiki/Arithmetic_progression