So it is! But what’s two orders of magnitude

I just wish Euler would accept 298,714,475,488.1710 as an answer. That’s what I get from the trendline coefficients. Ought to be good enough!

…mrt

I haven’t looked at #72, and I don’t have a lot of time spare at the moment, but if I do find time for it, and if I can solve it, I’ll definitely report back here.

By the way, 0.2*((10^6)^1.99) is much closer to 300 billion than 3 billion

Doug

If Prof Jarvis ever reads this thread, I apologize for misspelling his name above.

I’m not going to learn Python, I going to get a better handle on arrays. That’s quite (!) a speed jump.

Have you done #72 yet? My GCD method looks like it’d still be running into next week if I let it, and my recursive solution exhausts the stack somewhere north on 100,000. Using data points of D=10, 100, 1000, 10000, and 100,000, the answer is approximately 0.30*(D^1.99), or around 3 billion for D=10^6. Not quite good enough. I can’t noodle out the trick.

…mrt

Fairly straightforward with the help of Prof. Jarvis. Run time = about 0.3 sec

Function SqRtI(val As Long, Dig As Double) As Variant
Dim AA() As Long, BA() As Long, Digp As Long
Dim SumDig(1 To 1, 1 To 2) As Double
Dim i As Long, j As Long, k As Long, NumDig As Long
Dim BigVal As Long, dStep As Long, n As Long

SumDig(1, 2) = Timer
Digp = Dig + 10

For n = 2 To val
If n ^ 0.5 .NE. Int(n ^ 0.5) Then
ReDim AA(1 To Digp)
ReDim BA(1 To Digp)

AA(1) = n * 5
i = 1
‘ Split AA(1) into digits
Do While AA(i) .GT. 9
AA(i + 1) = AA(i)
AA(i) = AA(i) Mod 10
AA(i + 1) = Int(AA(i + 1) / 10)
i = i + 1
Loop

BA(1) = 5
j = 1

Do While j .LT=. Dig + 1

‘ Check if a .GT=. b
If i = j Then
k = i
Do While AA(k) = BA(k)
k = k – 1
Loop
If AA(k) .GT=. BA(k) Then BigVal = 1 Else BigVal = 2
Else
If i .GT. j Then BigVal = 1 Else BigVal = 2
End If

If BigVal = 1 Then
‘ a .GT=. b
For k = 1 To i
AA(k) = AA(k) – BA(k)
If AA(k) .LT. 0 Then
AA(k) = AA(k) + 10
AA(k + 1) = AA(k + 1) – 1
End If
Next k

Do While AA(i) = 0
i = i – 1
Loop

BA(2) = BA(2) + 1
If j = 1 Then j = 2
If BA(2) .GT. 9 Then
k = 2
Do While BA(k) .GT. 9
BA(k + 1) = BA(k)
BA(k) = BA(k) Mod 10
BA(k + 1) = Int(BA(k + 1) / 10)
k = k + 1
Loop
If k .GT. j Then j = k
End If
Else
‘ b .GT. a
For k = i + 2 To 3 Step -1
AA(k) = AA(k – 2)
Next k
AA(2) = 0
AA(1) = 0
i = i + 2

For k = j + 1 To 3 Step -1
BA(k) = BA(k – 1)
Next k
BA(2) = 0
j = j + 1
End If
Loop

‘ Find sum of digits
For j = Dig + 2 To Dig – 97 Step -1
SumDig(1, 1) = SumDig(1, 1) + BA(j)
Next j
End If

Next n
SumDig(1, 2) = Timer – SumDig(1, 2)
SqRtI = SumDig
End Function

I might have a go at a spreadsheet based solution as well.