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Ask The Wizard #400
Farmer Brown allows his flock of six sheep to graze in a grass-filled, fenced area of his field, and it takes them three days to eat all the grass.
After that, he lets the grass grow back to its original height.
Then, he introduces three of his sheep into the same area, and it takes them a week to eat through the grass.
Next, he allows the grass to regenerate to its full height once more and brings in just one sheep into the area. How long do you think it will require that single sheep to finish eating the grass?
We will assume sheep consume grass at a steady rate, while the grass itself also grows at a defined rate.
Let's denote i as the number of days needed for one sheep to eat the initial field of grass, assuming it stays constant without regrowth.
Let g = grass growth in one day.
It is given that six sheep take three days to finish the initial grass plus the grass that has grown back during those three days. We can represent this relationship mathematically as:
i + 3g = 3*6
Similarly, it is stated that three sheep require seven days to eat the original grass along with the grass that has grown in that period. This can also be expressed with a formula:
i + 7g = 7*3
We now have a system of two equations with two unknown variables:
i + 3g = 18
i + 7g = 21
It is easy to solve for i and g as:
i = 63/4 = 15.75
g = 3/4 = 0.75
The problem asks how much time it would take for one sheep to clean the field completely. Let's represent that time with x in our equation:
i + xg = x
(63/4) + (3/4)g = x
63/4 = x/4
x = 63.
Thus, it will take one sheep a total of 63 days to clear the field effectively.
Consider a cuboid that measures x by y by z, made up of xyz individual cubes. If someone paints all its outer surfaces, how can we determine the dimensions where the number of painted cubes equals the number of unpainted ones?
I have identified 20 unique sets of dimensions that meet the criteria. Here they are:
- 5 X 13 X 132
- 5 X 14 X 72
- 5 X 15 X 52
- 5 X 16 X 42
- 5 X 17 X 36
- 5 X 18 X 32
- 5 X 20 X 27
- 5 X 22 X 24
- 6 X 9 X 56
- 6 X 10 X 32
- 6 X 11 X 24
- 6 X 12 X 20
- 6 X 14 X 16
- 7 X 7 X 100
- 7 X 8 X 30
- 7 X 9 X 20
- 7 X 10 X 16
- 8 X 8 X 18
- 8 X 9 X 14
- 8 X 10 X 12
You've repeatedly mentioned that the average number of trials required for an event with probability p to occur is 1/p. I challenge you to provide proof of this statement.
Let's define x as the expected number of trials before an event occurs.
x = 1*p + (1-p)*(1+x)
x = p + 1 + x - p - px
Subtracting x from both sides:
0 = p + 1 - p - px
Canceling p and -p:
0 = 1 - px
px = 1
x = 1/p
Let’s introduce q = 1-p, meaning the probability that the event does not happen.
Let's define x as the expected number of trials before an event occurs.
The expression for x can be represented as follows: x = 1 * P(success in one trial) + 2 * P(success in two trials) + 3 * P(success in three trials) + …
= 1p + 2pq + 3pq^2 + 4pq^3 + ..
x/p = 1 + 2q + 3q^2 + 4q^3 + ..
x/p - 1 = 2q + 3q^2 + 4q^3 + ..
x/p - 1 = q * (2 + 3q + 4q^2 + 5q^3 + ..)
x/p - 1 = q * (1 + 2q + 3q^2 + 4q^3 + .. + 1 + q + q^2 + q^3 + ..)
x/p - 1 = q * (x/p + 1 + q + q^2 + q^3 + ..)
Let y = 1 + q + q^2 + q^3 + ..
y-1 = q + q^2 + q^3 + ..
y-1 = q * (1 + q + q^2 + q^3 + .. )
(y-1)/q = 1 + q + q^2 + q^3 + ..
(y-1)/q = y
y/q - y = 1/q
y*(1/q - 1) = 1/q
y*(1/q - q/q) = 1/q
y*[(1-q)/q] = 1/q
y*(1-q) = 1
y = 1/(1-q)
x/p - 1 = q * (x/p + 1/(1-q))
x/p - 1 = q * (x/p + 1/p)
x/p - 1 = q * (1+x)/p
x/p - q * (1+x)/p = 1
x/p - qx/p = 1 + q/p
x*(1/p - q/p) = 1+q/p
x*(1-q)/p = 1+q/p
x*p/p = 1+q/p
x = 1+q/p
x = 1 + (1-p)/p
x = p/p + (1-p)/p
x = 1/p