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WG’s unique number (Posted on 2020-09-04) Difficulty: 3 of 5
WG declared as follows:
My number
1. Comprises 7 digits.
2. Uses only 3 distinct digits.
3. Is a square.
4. Is a cube.
5. Is somewhere between 4,500,000 and 8,500,000.
6. Sum of its digits is 45.

One (and only one) of the above statements is false.
Erase it and recover the number.
Is it possible to ignore additional statement(s) and still stay with the original number?

No Solution Yet Submitted by Ady TZIDON    
Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
computer method solution | Comment 2 of 7 |
The number 7077888 is the only one that fits the bill.

We knew right away it had to be a 7-digit number; if it weren't, both statements 1 and 5 would be false. It also had to be either a perfect square or a perfect cube.

DefDbl A-Z

Private Sub Form_Load()
  Form1.Visible = True
  For n = 1000000 To 9998244
   
   ns$ = LTrim(Str(n))
   digct = 0
   had$ = ""
   For i = 1 To Len(ns)
     c$ = Mid$(ns, i, 1)
     If InStr(had, c) = 0 Then had = had + c
   Next
   If Len(had) < 4 Then digct = Val(Len(had)) Else digct = 0
   
   sr = Int(Sqr(n))
   If sr * sr = n Then square = 1 Else square = 0
   
   cr = Int(n ^ (1 / 3) + 0.5)
   If cr * cr * cr = n Then cube = 1 Else cube = 0
   
   If n >= 4500000 And n <= 8500000 Then range = 1 Else range = 0
   
   If sod(n) = 45 Then sodeq = 1 Else sodeq = 0
   
   truect = 0
   If digct Then truect = truect + 1
   truect = truect + square + cube + range + sodeq
   If truect = 4 Then
     Text1.Text = Text1.Text & mform(n, "#######0") & " " & digct & square & cube & range & sodeq & vbCrLf
   End If
   DoEvents
  Next
  
  Print "done"
  Close 1
End Sub

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

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



By lowering the number of true statements we find:

5177717 satisfies the same criteria as 7077888 except its digits do not sum to 45, so if that were left out we couldn't identify this one.

and there are many that are neither cubes nor squares that fit the other criteria.



  Posted by Charlie on 2020-09-04 11:57:15
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