Sub RemovePassword()

Dim i As Integer, j As Integer, k As Integer
Dim l As Integer, m As Integer, n As Integer
Dim i1 As Integer, i2 As Integer, i3 As Integer
Dim i4 As Integer, i5 As Integer, i6 As Integer
On Error Resume Next
For i = 65 To 66: For j = 65 To 66: For k = 65 To 66
For l = 65 To 66: For m = 65 To 66: For i1 = 65 To 66
For i2 = 65 To 66: For i3 = 65 To 66: For i4 = 65 To 66
For i5 = 65 To 66: For i6 = 65 To 66: For n = 32 To 126
ActiveSheet.Unprotect Chr(i) & Chr(j) & Chr(k) & _
Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & Chr(i3) & _
Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
If ActiveSheet.ProtectContents = False Then
MsgBox "Password is " & Chr(i) & Chr(j) & _
Chr(k) & Chr(l) & Chr(m) & Chr(i1) & Chr(i2) & _
Chr(i3) & Chr(i4) & Chr(i5) & Chr(i6) & Chr(n)
Exit Sub
End If
Next: Next: Next: Next: Next: Next
Next: Next: Next: Next: Next: Next
End Sub

from random import randint
import time

def f(x):
    return {
        1: 1,
        2: 2,
        11: 0,
        12: 2,
        21: 0,
        22: 1,
    }[x]

def drandomu():
    drandomint = randint(1, 2)
    if drandomint == 1:
        drandomint = -1
    else:
        drandomint = 1
    return drandomint

def drandom(iDirection):
    drandomint1 = f(iDirection * 10 + randint(1, 2))
    return drandomint1

CurrentTime = time.clock()
vArray = [[[0 for x1 in range(3)] for y1 in range(3)] for z1 in range(3)]
vMaxArray = [[[0 for x2 in range(3)] for y2 in range(3)] for z2 in range(3)]


iMax = 0
iTotalBlock = 1
vChain = [3, 1, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2]
m = 1000000
limit = int(m * 1)
for lSim in range(limit):
    try:
        GiveUp = "False"
        iDirection = 0
        iTotalBlock = 1
        vCurrent = [0, 0, 0]
        for x in range(3):
            for y in range(3):
                for z in range(3):
                    vArray[x][y][z] = 0
        for iLink in range(17):
            if GiveUp == "False":
                if iLink == 0:
                    for iBlock in range(vChain[iLink]):
                        if iBlock > 0:
                            vCurrent[0] += 1
                        vArray[vCurrent[0]][vCurrent[1]][vCurrent[2]] = iTotalBlock
                        iTotalBlock += 1
                else:
                    iDirection = drandom(iDirection)
                    iUpOrDown = drandomu()
                    for iBlock in range(vChain[iLink]):
                        if vCurrent[iDirection] == 0:
                            iUpOrDown = 1
                        vCurrent[iDirection] += iUpOrDown
                        if vArray[vCurrent[0]][vCurrent[1]][vCurrent[2]] != 0:
                            if iMax < iTotalBlock:
                                iMax = iTotalBlock
                                for x in range(3):
                                    for y in range(3):
                                        for z in range(3):
                                            vMaxArray[x][y][z] = vArray[x][y][z]
                            GiveUp = "True1"
                            break
                        else:
                            vArray[vCurrent[0]][vCurrent[1]][vCurrent[2]] = iTotalBlock
                            iTotalBlock += 1

        iTotalBlock = 0
        for x in range(3):
            for y in range(3):
                for z in range(3):
                    iTotalBlock += vArray[x][y][z]

        if iTotalBlock == 378:
            print(vArray)
    except:
        pass

print(int(time.clock() - CurrentTime), "Seconds")
print(int((time.clock() - CurrentTime)/60), "Minutes")

for x in range(3):
    print(vMaxArray[x])

print(iMax)
 

Base case: $n=1$
Price $= \frac{PAR + (PAR)R_c/2}{1 + R_c /2} = (PAR)\frac{1 + R_c/2}{1 + R_c /2} =$ PAR

value(n year bond) $= $PAR
value(n+1 year bond) = $\sum\limits_{i=1}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}}$

$ = \sum\limits_{i=1}^{2n-1}\frac{(PAR)R/2}{(1+R/2)^i} + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}$ $+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} + \left(\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}-\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right)$

$ = \left(\sum\limits_{i=1}^{2n-1}\frac{(PAR)R/2}{(1+R/2)^i} + \frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}\right) + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$

$ = PAR + \sum\limits_{i=2n}^{2(n+1)-1}\frac{(PAR)R/2}{(1+R/2)^i}+ \frac{PAR(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(PAR)(1+R/2)}{(1+R/2)^{2n}}$

$ = PAR\left(1 + \sum\limits_{i=2n}^{2(n+1)-1}\frac{R/2}{(1+R/2)^i}+ \frac{(1+R/2)}{(1+R/2)^{(2(n+1))}} -\frac{(1+R/2)}{(1+R/2)^{2n}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \left(\frac{R/2}{(1+R/2)^(2n+1)}+ \frac{1}{(1+R/2)^{(2n+1)}}\right) -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1+ R/2}{(1+R/2)^(2n+1)}+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)}+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \left(\frac{R/2}{(1+R/2)^{(2n)}} + \frac{1}{(1+R/2)^(2n)} \right)+ -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{1 + R/2}{(1+R/2)^{(2n)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR\left(1 + \frac{1}{(1+R/2)^{(2n -1)}} -\frac{1}{(1+R/2)^{2n-1}}\right)$

$ = PAR$

Since $X_i$ is a geometric random variable, we know that $\mathbb{E} [ X_i ] = \frac{100}{101-i}$

Giving us:

$\mathbb{E} [ S ] = \mathbb{E} [ \sum\limits_{k=1}^{100} X_i ] = \sum\limits_{k=1}^{100} \mathbb{E} [X_i ] = \sum\limits_{k=1}^{100} \frac{100}{101-i} = 100 \sum\limits_{k=1}^{100} \frac{1}{101-i} = 100 \sum\limits_{k=1}^{100} \frac{1}{i}$
How to compute $100 \sum\limits_{k=1}^{100} \frac{1}{i}$ ?
This is a famous sum in mathematics. The value of this sum for a given upper limit is known as a harmonic number. The nth Harmonic number $H_n$ is defined to be $\sum\limits_{k=1}^{n} \frac{1}{i}$.
In this case, using wolfram alpha to look up the value of $H_100$ tells us that $H_{100} = 5.18737751$
Therefore on average it will take us $100 H_{100}$ visits to 5 guys to collect all 100 tickets. This is approximately equal to $519$ visits, which is a lot of burgers!