How many were left?

Question

The total population of group $\alpha$ and $\beta$ equals $2000$.

Do:

• Each member of group $\alpha$ removes three members of group $\beta$.

• Each remaining member of group $\beta$ removes two members of group $\alpha$.

• Then each remaining member of group $\alpha$ removes three members of group $\beta$

Until $Population\left(\alpha\right) = Population\left(\beta\right)$

Find the population at this point.

How I am approaching the problem: $$ \begin{align} -3 \times (2000 - \beta) &= \text{Removed}_{\beta} \\ (2000 - \alpha) - \text{Removed}_{\beta} &= \text{Population}_\beta \end{align} $$

The problem is that this clearly is not going to work since I have two unknowns and one equation. I am not sure what I am doing wrong or if my approach is incorrect all together. Hints are preferable to out right answers.

Answer 1

How I ended up doing it, starting from the end and working backward seems to be the way to do it™.

α does the last removal

    Round         Population α           Population β       Total
  ---------- ---------------------- ---------------------- -------
     Last              x                      x                2x
   2nd Last            x                  x + 3x = 4x            5x
   3rd Last     x + 2 × 4x = 9x                 4x               13x
   4th Last            9x             4x + 3 × 9x = 31x          40x
   5th Last    9x + 2 × 31x = 71x              31x              102x
   6th Last           71x            31x + 3 × 71x = 244x       315x
   7th Last   71 + 2 × 244x = 559x             244x             803x

β does the last removal

    Round         Population α           Population β        Total
  ---------- ---------------------- ----------------------- -------
     Last              x                       x               2x
   2nd Last         x + 2x = 5x                  x               4x
   3rd Last            3x               x + 3 × 3x = 10x         13x
   4th Last    3x + 2 × 10x = 23x               10x              33x
   5th Last           23x             10x + 3 × 23x = 79x       102x
   6th Last   23x + 2 × 79x = 181x              79x             206x
   7th Last           181x            79x + 2 × 181x = 441x     622x

From the problem it seems there are at least 3 rounds, looking at α doing that last removal we get $40x = 2000$ and $x = 50$.

Answer 2

Haven't thought this through, but have you started "in reverse"?

For instance, popA == popB for some population << 2000

then add 2 or three to each population per the rules above until popA + popB == 2000 ?

Seems to lend itself to brute forcing a solution in code...

Hope this helps!

Answer 3

Not sure what the proper method would be but I used simultaneous equations and got the final populations as

50

Answer 4

For a systematic approach, let $\alpha_k,\beta_k$ be the group sizes after $k$ steps (so in partizular $\alpha_0=\alpha$, $\beta_0=\beta$). For a certain $n$, we have $\alpha_n=\beta_n=:d$. Then we observe that $\gcd(\alpha_i,\beta_i)=d$ for all $i$ (because $\gcd(a,b-3a)=\gcd(a,b)=\gcd(a-2b,b)$). Thus $d\mid \alpha_0+\beta_0=2000$. We can find the sequence $(\alpha_i/d,\beta_i/d)$ by working backwards (assuming $n$ is even): $$(1,1)\leftarrow(3,1)\leftarrow(3,10)\leftarrow(23,10)\leftarrow(23,79)\leftarrow(181,79)\leftarrow(181,622)\leftarrow(1425,622) $$ and here can stop because the values start exceeding $2000$. Neither $3+10$ nor $23+79$ nor $181+622$ is a divisor of $2000$ (but $1+1$ is leading to the trivial solution where $\alpha_0=\beta_0=1000$). For odd $n$, we obtain a different result: $$(1,1)\leftarrow (1,4)\leftarrow (9,4)\leftarrow(9,31)\leftarrow(71,31)\leftarrow(71,244)\leftarrow(559,244)$$ and can stop again. Here, $1+4$ and $9+31$ are divisors of $2000$, meaning we could have started with $\alpha_0=400$, $\beta_0=1600$ (and end with $800$) or with $\alpha_0=450$, $\beta_0=1550$