I’ve been forwarded a classic problem, in bold below.

**The Monkey Problem**

**A rope over the top of a fence has the same length on each side, and weighs one-third of a pound per foot. On one end hangs a monkey holding a banana, and on the other end a dead-weight equal to the weight of the monkey. The banana weighs 2 ounces per inch. The length of the rope in feet is the same as the age of the monkey, and the weight of the monkey in ounces is as much as the age of the monkey’s mother. The combined ages of the monkey and its mother are 30 years. One-half the weight of the monkey, plus the weight of the banana is one-fourth the sum of the weights of the rope and the dead-weight.**

**The monkey’s mother is one-half as old as the monkey will be in when it is three times as old as its mother was when she was one-half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one-third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one-fourth as old as it is now.**

**How long is the banana in inches?**

In fact, I’ve seen this problem back when I was kid, though I’ve never tried to solve it, mostly because that last paragraph looked tedious and as though it would require a deeply nested set of multiplied and offset variables. It turns out however, to be simpler than I imagined. Here we go….

*A rope over the top of a fence has the same length on each side, and weighs one-third of a pound per foot.*

Everything else in the problem uses ounces, so let’s make it consistent. We’ll make up variables as go along. First we need rope weight and rope length.

*On one end hangs a monkey holding a banana, and on the other end a dead-weight equal to the weight of the monkey. *

Add in variables for Dead Weight and Monkey Weight.

*The banana weighs 2 ounces per inch. *

Add in variables for banana weight and banana length.

*The length of the rope in feet is the same as the age of the monkey *

Rope Length and Monkey Age.

*… and the weight of the monkey in ounces is as much as the age of the monkey’s mother. *

Add in MMA for Monkey Mother Age.

*The combined ages of the monkey and its mother are 30 years. *

*One-half the weight of the monkey, plus the weight of the banana is one-fourth the sum of the weights of the rope and the dead-weight. *

*The monkey’s mother is one-half as old as the monkey will be in when it is three times as old as its mother was when she was one-half as old as the monkey will be when it is as old as its mother will be when she is four times as old as the monkey was when it was twice as old as its mother was when she was one-third as old as the monkey was when it was as old as its mother was when she was three times as old as the monkey was when it was one-fourth as old as it is now. *

How long is the banana in inches?

No doubt these equations are easy to solve by hand, but why settle? We can write them in a form that can be solved by computer. Right now, I prefer the mathematical software Sage . Formatted for Sage, the equations look like this:

var('RW, RL, DW, MW, BW, BL, MA, MMA')
equations = [
RW == RL * 1/3 * 16,
DW == MW,
BW == 2 * BL,
RL == MA,
MW == MMA,
MA + MMA == 30,
1/2 * MW + BW == 1/4 * (RW + DW),
MMA == 1/2 * ( 3 * ( 1/2 * ( 4 * ( 2 * ( 1/3 * ( 3 * ( 1/4 * MA)))))))]
s = solve(equations, RW, RL, DW, MW, BW, BL, MA, MMA, solution_dict = True)
print s

And the output is:

Notice the banana weighs 11.5 ounces and the monkey only weighs 18 oz., so something is little implausible about the problem, but as mathematicians, that’s not our problem…