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Palindromes from Digit Sum Series (Posted on 2015-02-03) Difficulty: 3 of 5
Form a series, starting from 1,

Add to the last number the sum of its own digits, and then repeat.

Example: 1, 2, 4, 8, 16, 23, 28, 38 . . .

The 595th term is 10001, both of which are numerical palindromes.

How many such palindromic pairs does the series contain?

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution open ended solution/discussion Comment 1 of 1
I had to make the starting number, 1, the 0th term of the series in order for 10001 to be the 595th term. The palindrome pairs found, up through the thousandth term are:

0 1
1 2
2 4
3 8
11 77
151 1991
191 2552
595 10001
747 12221

These are the nine such palindromic pairs found by the program, which goes up to term number 1000, as mentioned:

DefDbl A-Z
Dim crlf$

Function mform$(x, t$)
  a$ = Format$(x, t$)
  If Len(a$) < Len(t$) Then a$ = Space$(Len(t$) - Len(a$)) & a$
  mform$ = a$
End Function

Private Sub Form_Load()
 Text1.Text = ""
 crlf$ = Chr(13) + Chr(10)
 Form1.Visible = True
 DoEvents
 
 termno = 0: member = 1: pals = 1
 Text1.Text = Text1.Text & termno & Str(member) & crlf
 ct = 1
 For termno = 1 To 1000
   member = member + sod(member)
  ' Text1.Text = Text1.Text & termno & Str(member) & crlf
  ' If termno < 11 Or termno = 595 Then
  '     Text1.Text = Text1.Text & termno & Str(member) & crlf
  ' End If
   If isPalin(termno) Then
     If isPalin(member) Then
       Text1.Text = Text1.Text & termno & Str(member) & crlf
       DoEvents
       pals = pals + 1
     End If
   End If
 Next termno
  
 Text1.Text = Text1.Text & pals & " done"
 
End Sub

Function isPalin(x)
  xs$ = LTrim(Str(x))
  good = 1
  For i = 1 To Len(xs) / 2
    If Mid(xs, i, 1) <> Mid(xs, Len(xs) + 1 - i, 1) Then good = 0: Exit For
  Next
  isPalin = good
End Function

Function sod(x)
  tot = 0
  xs = LTrim(Str(x))
  For i = 1 To Len(xs)
    tot = tot + Val(Mid(xs, i, 1))
  Next
  sod = tot
End Function

However, there are more, if we look farther:

22522 521125
50305 1254521
62726 1583851
1339331 42288224

(checked through the hundred millionth term)

I don't imagine there's any limit, but none were found after the 1,339,331th element through the 100,000,000th.


  Posted by Charlie on 2015-02-03 14:31:41
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