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2 bases (Posted on 2017-07-21) Difficulty: 3 of 5
The palindromic decimal number N=abccba displays an interesting feature:
The value of abc in base 9 equals the value of cba in base 7.

What is N's prime factorization?

See The Solution Submitted by Ady TZIDON    
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Solution computer solution (spoiler) Comment 2 of 2 |
DefDbl A-Z
Dim crlf$


Private Sub Form_Load()
 Form1.Visible = True
 
 
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 
 For n1 = 100 To 999
   n2 = 100 * (n1 Mod 10) + 10 * ((n1 \ 10) Mod 10) + n1 \ 100
   n = 1000 * n1 + n2
   n1s$ = LTrim(Str(n1))
   n2s$ = LTrim(Str(n2))
   If fromBase(n1s, 9) = fromBase(n2s, 7) Then
     Text1.Text = Text1.Text & n & " " & Str(fromBase(n1s, 9)) & crlf
   End If
   xx = xx
 Next
 
 Text1.Text = Text1.Text & crlf & " done"
  
End Sub

Function fromBase(n$, b)
  v = 0
  For i = 1 To Len(n$)
    c$ = LCase$(Mid(n$, i, 1))
    If c$ > " " Then
      v = v * b + InStr("0123456789abcdefghijklmnopqrstuvwxyz", c$) - 1
    End If
  Next
  fromBase = v
End Function


provides two pseudo-answers and the real answer:


182281  155
305503  248
487784  403

The pseudoanswers first:

When 182 is treated as if it were a base-9 number, its value is 155. Then, while 281 can't really be a base-7 number, we could still treat it as if it could: 2*7^2 + 8*7 + 1 = 155. A similar thing happens when 784 is converted from base 7, as if 8 were a valid base-7 digit; it comes out to the same 403 in decimal that 487 comes out when treated as a base-9 number.

The true sought answer is the 305 in base 9 is the same as 503 in base 7, as each is 248 in decimal notation.

But the puzzle asks for N's prime factorization.

N = 305503 = 11 * 27773

so N happens to be a semiprime.


  Posted by Charlie on 2017-07-21 10:47:40
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