Answers to Mind Games on the Web

Vol 2, No 1 Answers Presented by Smart Science ™ and Bill Graham as a free service for entertainment and education
To subscribe to Mind Games on the Web, click here

ANSWER TO
THE EGG FARMER

by Bill Graham

egg-laying hens The best way to solve this problem is to begin at the end. Farmer Humphrey ends up with no eggs. Add 1/2 egg and double to get the number of eggs just before that. Continue this process. The table below shows the process and the answer: 127 eggs.

Step Eggs
left Eggs
bought

After 0 0

7 1 1

6 3 2

5 7 4

4 15 8

3 31 16

2 63 32

1 127 64

The further question of replacing 1/2 with a/b requires some analysis and is more difficult in many ways than the initial question. You must have a number of eggs, N, that is multiplied by a/b and then has a/b added to it.

Na/b + a/b = N' Here, N is the number of eggs the farmer has and N' the number sold in this round. On the last round, N must equal N'. A little rearrangement after the substitution results in the following equations. a = N(b-a)
N = a/(b-a) Because N is an integer, b-a must divide a evenly. That means that either b = a+1 or b and a have a common divisor. If a and b have a common divisor, then the fraction is not in simplest terms and is not of interest to the solution.

The only fractions, a/b, that are in simplest terms and that can be used by farmer Humphrey have the form a/(a+1). For example, 1/2, 2/3, 3/4, 4/5, etc.

To see what the actual number of eggs will be for each fraction, do a little math with the equations above thar represent the last round. Because b-a is one, the last equation above becomes N = a.

Finally, you must have a way to calculate the previous round from each known round (still working backward). Let N be the amount in the previous round and N" be the amount left after selling N' = Na/b + a/b.

N - N' = N"
N - (Na/b + a/b) = N"
N(1-a/b) = N" + a/b
N[1-a/(a+1)] = N" + a/(a+1)
N/(a+1) = N" + a/(a+1)
N = N"(a+1) + a It's not too hard to show from the above that the n^th member of this series is (a+1)ⁿ-1, where n=1 corresponds to the last sale.

Replace a+1 with b.

N = N"b + b - 1 The formula for N(n), the n^th term, becomes N(n) =? bⁿ - 1 Substituting appropriately, you'll get the following. N(n) = N(n-1)b + b - 1
bⁿ - 1 =? b(b^n-1 - 1) + b - 1
bⁿ - 1 =? bⁿ - b + b - 1
bⁿ - 1 =? bⁿ - 1 The equality shows that the guess is correct. Of course, you could also have had the fun of deriving this formula by induction.

Click here