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First Term Finding (Posted on 2016-05-04) Difficulty: 3 of 5
Consider two arithmetic sequences having the common first term as A, where A is an integer. All the terms of the two sequences are integers.

Determine the minimum absolute value of A such that:

The 20th term of the first sequence equals the 16th term of the second sequence, and:

The sum of the first 20 terms of the first sequence equals the sum of the first 16 terms of the second sequence.

No Solution Yet Submitted by K Sengupta    
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Solution computer solution Comment 2 of 2 |
Just using the finite first 20 and first 16 sequences:

The first terms and the last terms of each of these finite sequences (20 members and 16 members) are equal.  That would be impossible if both were positive. So A is negative.

The program tests the possibilities:

DefDbl A-Z
Dim crlf$

Private Sub Form_Load()
 Form1.Visible = True
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For a = -1 To -1000 Step -1
   For d1 = 1 To 2000
     fin = a + 19 * d1
     tot = 20 * (a + fin) / 2
     d2 = (fin - a) / 15
     If d2 = Int(d2) Then
     If tot = 16 * (a + fin) / 2 Then
       Text1.Text = Text1.Text & a & Str(fin) & "    " & d1 & Str(d2) & crlf
     End If
     End If
   Next
 Next a
  
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

The two totals are of course equal at zero, and the minimum absolute value of A is 285, when the sequences go from -285 to 285 with differences of 30 and 38 respectively.

-285 285    30 38
-570 570    60 76
-855 855    90 114

On hindsight of course, 285 if the LCM of 19 and 15, the number of differences between members of the sequences.

  Posted by Charlie on 2016-05-04 13:13:53
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