Find all possible 3-digit positive perfect powers that are divisible by the sum of their respective digits.
The program checks that the GCD of the powers of the prime factors is greater than 1. That GCD is the power.
n as pwr s.o.d.
100 10^2 1
144 12^2 9
216 6^3 9
225 15^2 9
243 3^5 9
324 18^2 9
400 20^2 4
441 21^2 9
512 2^9 8
576 24^2 18
900 30^2 9
from
DefDbl A-Z
Dim fct(20, 1), crlf$
Private Sub Form_Load()
Text1.Text = ""
crlf = Chr(13) & Chr(10)
For n = 100 To 999
f = factor(n)
g = fct(1, 1)
For i = 2 To f
g = gcd(g, fct(i, 1))
Next
If g > 1 Then
If n Mod sod(n) = 0 Then
Text1.Text = Text1.Text & n & " " & n ^ (1 / g) & "^" & (g) & Str(sod(n)) & crlf
End If
End If
Next
Text1.Text = Text1.Text & "done"
End Sub
Function factor(num)
diffCt = 0: good = 1
n = Abs(num): If n > 0 Then limit = Sqr(n) Else limit = 0
If limit <> Int(limit) Then limit = Int(limit + 1)
dv = 2: GoSub DivideIt
dv = 3: GoSub DivideIt
dv = 5: GoSub DivideIt
dv = 7
Do Until dv > limit
GoSub DivideIt: dv = dv + 4 '11
GoSub DivideIt: dv = dv + 2 '13
GoSub DivideIt: dv = dv + 4 '17
GoSub DivideIt: dv = dv + 2 '19
GoSub DivideIt: dv = dv + 4 '23
GoSub DivideIt: dv = dv + 6 '29
GoSub DivideIt: dv = dv + 2 '31
GoSub DivideIt: dv = dv + 6 '37
If INKEY$ = Chr$(27) Then s$ = Chr$(27): Exit Function
Loop
If n > 1 Then diffCt = diffCt + 1: fct(diffCt, 0) = n: fct(diffCt, 1) = 1
factor = diffCt
Exit Function
DivideIt:
cnt = 0
Do
q = Int(n / dv)
If q * dv = n And n > 0 Then
n = q: cnt = cnt + 1: If n > 0 Then limit = Sqr(n) Else limit = 0
If limit <> Int(limit) Then limit = Int(limit + 1)
Else
Exit Do
End If
Loop
If cnt > 0 Then
diffCt = diffCt + 1
fct(diffCt, 0) = dv
fct(diffCt, 1) = cnt
End If
Return
End Function
Function gcd(a, b)
x = a: y = b
Do
q = Int(x / y)
z = x - q * y
x = y: y = z
Loop Until z = 0
gcd = x
End Function
Function sod(n)
s$ = LTrim(Str(n))
tot = 0
For i = 1 To Len(s$)
tot = tot + Val(Mid(s$, i, 1))
Next
sod = tot
End Function
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Posted by Charlie
on 2015-07-29 09:58:22 |
computer solution
