Sharing Turnips.

Number Theory Level 3

There are 3 groups of people in a Green house, Blue house and Red house. All nearby Turnip patches.

They all use a strange currency of Golden Turnips and regular Turnips. With the conversion of $1 GT = N T$ for some unknown $N$ .

They all begin with 1 Golden Turnip each, which they all convert to $T$ at the local Turnip patch bank.

It's Christmas and everyone is feeling generous, so individually decide to divide ALL of their Turnips between ALL the occupants of the other houses (this is done perfectly evenly). (Someone from the Blue house gives an equal amount to everyone in the Red and Green houses etc). They put the Turnips into a sack and place it in the recipient's room.

Unfortunately a thief is about, and sneaks into 1 person's room at random from each house and steals all of the sacks in those rooms. They then take the sacks to the Turnip patch bank to change as many dirty Turnips into Golden Turnips as possible. Surprisingly they get exactly $8 GT$ back, with no spare $T$ !

Each house has $1$ president.

Let $P$ be the probability that the president's room was stolen from every house.

Rounding $P$ to $1$ s.f., it is equal to $a*10^{-n}$ (where $a$ is an integer $a \in [1,9]$ ). By comparing all possible values of $P$ , what is the minimum value of $a+n-1$ ?

The answer is 240.

1 solution

Alex Burgess
Apr 3, 2019

Let $X,Y,Z$ be the number of occupants in the Green. Blue and Red houses respectively.

Someone in the Green house receives $\frac{1}{X+Y} GT$ from everyone in the Red house and $\frac{1}{Z+X} GT$ from everyone in the Blue house, totalling $\left( \frac{Z}{X+Y} + \frac{Y}{Z+X} \right) GT$ . Similarly for occupants of the Blue and Red houses.

Hence the thief steals a total of $\left( \frac{Z}{X+Y} + \frac{Y}{Z+X} \right) + \left( \frac{Z}{X+Y} + \frac{X}{Y+Z} \right) + \left( \frac{X}{Y+Z} + \frac{Y}{Z+X} \right) GT = 2\left( \frac{Z}{X+Y} + \frac{Y}{Z+X} + \frac{X}{Y+Z}\right) GT = 8 GT$ .

Hence $\frac{Z}{X+Y} + \frac{Y}{Z+X} + \frac{X}{Y+Z} = 4$ .

See here

By implementing elliptical curves, one can determine the solution with the smallest value of $x$ (where $x \leq y \leq z$ ) to this equation is:

$x = 4373612677928697257861252602371390152816537558161613618621437993378423467772036 = 4.374*10^{78}$

$y = 36875131794129999827197811565225474825492979968971970996283137471637224634055579 = 3.688*10^{79}$

$z = 154476802108746166441951315019919837485664325669565431700026634898253202035277999 = 1.545*10^{80}$

In this question, $P = \frac{1}{XYZ}$ .

Using the values above $xyz = 2.49*10^{238}$ , so $P = \bar P = 4.01*10^{-239}$ , with $a + n = 243$ .

This is the minimum value of $n$ , but $a$ can be minimised by noticing that $(kx, ky, kz)$ is also a solution. Resulting in $P = \frac{\bar P}{k^3}$ .

$k = 2$ leads to $P = 5*10^{-240}$ , with $a + n = 245$ .

$k = 3$ leads to $P = 1*10^{-240}$ , with $a + n = 241$ .

Increasing $k$ from here, or looking at solutions not in the form $(kx, ky, kz)$ , will result in a greater value of $n$ , and hence a greater value of $a + n$ .

Hence the minimum value of $a + n - 1$ is $240$ .

Note, I mistyped the answer, so changed the question from $a + n$ to $a + n - 1$ .

Sharing Turnips.

The answer is 240.

1 solution

0 pending reports