How many ones are used in writing all of the numbers from 0 to 2018 (inclusive) in binary form?
2048 = 2^11 so the numbers below this all have 11 or fewer binary digits. Allow leading zeros so that all have exactly 11 bits. From 0 through 2047, there are 2048*11=22528 binary digits, of which half are 1 bits as every ordered set of binary digits is present. That's 11264 1bits.
But we want to go only through 2018 and so must subtract out the 1 bits from 2019 through 2047. That's
11111100011 through
11111111111
There are 20472018 = 29 of these numbers, so
The highorder 6 bits account for 6*29=174 1bits.
If the range had included 32 numbers there'd be 32*5/2 1bits accounted for by the loworder five bits, or 80 of them, but two of these are not in the range: in 00001 and 00010. That makes 78 to be subtracted out on this basis.
So 11264  (174+78) = 11012 1bits are in the range asked for.
then
DefDbl AZ
Dim crlf$
Private Sub Form_Load()
Form1.Visible = True
Text1.Text = ""
crlf = Chr(13) + Chr(10)
For i = 1 To 2018
b$ = base$(i, 2)
For j = 1 To Len(b)
If Mid(b, j, 1) = "1" Then tot = tot + 1
Next
Next
Text1.Text = Text1.Text & crlf & tot & " done"
End Sub
Function base$(n, b)
v$ = ""
n2 = n
Do
q = Int(n2 / b)
d = n2  q * b
n2 = q
v$ = Mid("0123456789abcdefghijklmnopqrstuvwxyz", d + 1, 1) + v$
Loop Until n2 = 0
base$ = v$
End Function
finds 11012 also.

Posted by Charlie
on 20180618 11:13:43 