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Find the last member (Posted on 2015-03-05) Difficulty: 2 of 5
Given a strictly increasing sequence of seven positive integers such that:

a. Each number (excluding the first) is a multiple of the one before it.
b. The sum of all seven numbers is 559.
c. Neither 0 nor 8 are used in any of the 7 numbers.

What is the last number in the sequence?

No Solution Yet Submitted by Ady TZIDON    
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Solution computer solution Comment 1 of 1
DefDbl A-Z
Dim h(6), crlf$, tot
Function mform$(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
  
 For s = 1 To 559 / 7
   m = 2
   Do
   rhs = (559 / s) * (m - 1) + 1
   pm = m
   m = rhs ^ (1 / 7)
   Loop Until m = pm
   Text1.Text = Text1.Text & mform(s, "##0") & "  " & mform(m, "##0.0000000") & mform(s * m ^ 6, "####0.0000000")
   tot = 0
   For i = 0 To 6
     tot = tot + s * m ^ i
   Next
   
   Text1.Text = Text1.Text & mform(tot, "####0.0000000") & crlf
   
 Next
 
 Text1.Text = Text1.Text & crlf & crlf
 
 For s = 1 To 50
   h(0) = s: tot = s
   addOn 1
 Next

End Sub

Sub addOn(wh)
  For m = 2 To 20
    h(wh) = h(wh - 1) * m
    tot = tot + h(wh)
     
    If tot <= 559 Then
      If tot < 559 Then
       If wh < 6 Then
         addOn wh + 1
       End If
      Else
        If wh = 6 Then
         For i = 0 To 6
          Text1.Text = Text1.Text & mform(h(i), "####0")
         Next
         Text1.Text = Text1.Text & crlf
        End If
      End If
    End If
    
    tot = tot - h(wh)
  Next m
End Sub

The first part of the program merely verifies there's no integral solution if the assumption is made that the same multiple is used each generation in the series.

The second part of the program gives the following series, based on rules a and b:

    1    2    4    8   16   48  480
    1    2    4    8   16  176  352
    1    2    4    8   32   64  448
    1    2    4    8   32  128  384
    1    2    4   12   36   72  432  **
    1    2    4   12   60  120  360
    1    2    4   24   48   96  384
    1    6   12   36   72  144  288

Only the sequence I've marked with double asterisk fits rule c.


  Posted by Charlie on 2015-03-05 10:48:21
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