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 Find the last member (Posted on 2015-03-05)
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 No Rating

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 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

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
Next

End Sub

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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