I can't figure out how to crack this problem. It's as if I don't have enough information. Walk me through please
Jogging is a popular sport in Brilliantia. According to a recent survey, between 80% and 84% of Brillantians jog at least once a week. This conclusion was reached with 90% confidence.
How many Brilliantians were in the surveyed random sample?
The $z$-score for $90\%$ confidence is $1.645$ and the standard error estimate for a sample proportion with size $n$ is $$ SE = \sqrt{\frac{\hat{p} (1-\hat{p})}{n}} $$
Sounds like you got to the right place: $$ \begin{split} \hat{p} &- 1.645 E\ &= 0.8\\ \hat{p} &+ 1.645 E\ &= 0.84 \end{split} $$ Now add the equations to get the value of $\hat{p}$ (which you could also get from the symmetry of the confidence interval) and then solve for $E$.