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Question about arithmetic progressions

  1. Oct 31, 2014 #1
    1. The problem statement, all variables and given/known data
    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
     
  2. jcsd
  3. Oct 31, 2014 #2

    jedishrfu

    Staff: Mentor

    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...
     
  4. Oct 31, 2014 #3
    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)
     
  5. Oct 31, 2014 #4

    jedishrfu

    Staff: Mentor

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