Cutting a hundred million of a thirty-year bond down to fifty takes five hundred futures, and the usual arithmetic of fifty million over a hundred thousand lands there only because the contract's duration per dollar of face happens to match the bond's. What a hedge matches is dollars per basis point: 56,288.92 against 112.5778. Hold a thirty-year zero instead and the same job needs 1,234 contracts, while five hundred three-month contracts would cover 22.2 per cent of it.
Fourteen billion fill-ups a year divided by what a pump could do at full tilt gives 25,000 stations; divided by what a pump actually does it gives 149,829, inside the published range. The gap is exactly six, and the article proves that six is the ratio of the two throughput guesses alone, because the fleet, the fill-up frequency, the opening hours and the pumps per station all cancel. The utilisation of one sixth is Little's law read as 2.67 busy hours in a sixteen-hour day.
Two outlets in a town of fifty thousand is one per twenty-five thousand people, which scaled to the United States gives 13,600 against a published count near 13,500. That 0.74 per cent is luck, and the article shows why: the answer is exactly inversely proportional to the one density guess, and sweeping it across every defensible value spans 8,500 to 22,667. A second chain built from revenue, sharing no input at all, lands at 13,615.
A fair coin cuts probabilities into halves and quarters, and a short argument about the prime factorisation of two shows it can never reach one third in a bounded number of flips. Dropping the bound fixes it: flip twice, bin the tail-tail, and each child holds exactly a third for 8/3 flips on average. That naive scheme turns out to be the best any coin-flipping procedure can do for three outcomes, which stops being true at five.
Cut two diagonally opposite corners off a chessboard and 62 = 2 x 31 stays true, yet nothing fits. Writing the colour of a square as the sign (-1)^(i+j) turns the argument into arithmetic: every domino sums to zero, the two lost corners both carried +1, and the board left over is 30 against 32. The converse, Gomory's theorem, is the harder half and it goes the other way.
Person k flips every bulb that is a multiple of k, and after a hundred passes exactly the ten perfect squares are lit. Bulb n is flipped once per divisor, and the pairing d against n/d is fixed-point free unless n is a square, so the parity is decided by algebra rather than by accumulation. The lit fraction is one over the square root of the row, and stopping the process at person 50 inverts the answer to 54 bulbs.
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.