One lily doubling daily covers the pond on day thirty, so eight lilies must finish in 30/8 = 3.75 days. They finish on day twenty-seven, because eight is two cubed and that slides the whole schedule exactly three days earlier. The article carries the general rule that k lilies save the floor of log base two of k, the case where five lilies save only two, and the non-overlap assumption the answer quietly rests on.
Three feet up each day, one foot back each night, ten feet to climb. Dividing ten by the net two feet a day gives five days, and it charges the snail for a night it never spends. The dawn heights settle the day, the last climb settles the moment, and the closed form the video had no room to voice is a single ceiling function.
Fifty white marbles, fifty black, two jars, and a fair coin choosing which jar gets drawn from. The even split gives exactly one half, and so does every other split where the jars are the same size. Isolating a single white marble reaches 74/99, an exchange argument proves nothing beats it, and three quarters turns out to be a ceiling no arrangement ever touches.
On the unit interval the indicator of the rationals has upper sum 1 and lower sum 0 for every partition ever written, so the two never meet and Riemann returns nothing at all. Give the k-th rational an interval of width ε/2^k and the whole countable set sits inside a total length of ε, which puts its measure at 0 and its Lebesgue integral at 0. Lebesgue's criterion turns that into the general law — a bounded function on a compact interval is Riemann integrable exactly when its discontinuities have measure zero — which is why Thomae's function, discontinuous on the same dense set, is integrable and this one is not. The trade is not free: sin x over x on the half line has an improper Riemann value of π/2 and no Lebesgue integral at all.