Answer:
[tex]p_r = 65\ 536 \div 2^r = 65\ 536 \div 2^-^r = 2^{16-r}; r\ \in \{0,1,2,3...16\}[/tex]
Step-by-step explanation:
Pet peeve of mine: let's call it [tex]p_r[/tex] (read p sub r)since it's a representation over the natural numbers and not over [tex]\mathbb{R}[/tex].
Old man yelling at clouds moment gone, we know that the number of people gets halved every time so
At round 0, we do not halve, so [tex]p_0 = 65\ 536\div 1[/tex]
At round 1, we halve once, so [tex]p_1 = 65\ 536\div 2[/tex]
At round 2, we halve twice, so [tex]p_2 = 65\ 536\div 4 = 65\ 536\div 2^2[/tex]
At round 3, we halve three times, so [tex]p_3 = 65\ 536\div 8 = 65\ 536\div 2^3[/tex]
...
At round k, we halved k times, so [tex]p_k = 65\ 536 \div 2^k[/tex]
...
At round 16 we [tex]2^1^6 =65\ 536[/tex] so we have a winner.
That allows us to write a formula. I personally find the last one to be the neater, but any of the three is formally correct.
[tex]p_r = 65\ 536 \div 2^r = 65\ 536 \div 2^-^r = 2^{16-r}; r\ \in \{0,1,2,3...16\}[/tex]
