A driver who eats one apple per loaded mile delivers 833 of 3000 across a thousand miles, because the price of a mile is the ceiling of the stock over the truck's capacity: three apples, then two, then one. The continuous optimum is 2500/3, but rounding the switch point down to mile 333 leaves 2001 apples needing three passes and produces a fake 834. A lower bound on loaded traversals proves 833 optimal without a dynamic programme, and the free return legs are the clause that separates 833 from 533.
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Read the full explanation — 833 Apples, and the Surplus Apple That Rides for Free
The midpoint of p and q sits strictly between them, and consecutive means precisely that no prime lives in that interval, so the answer is never and the proof is two lines with no arithmetic in it. The pair 2 and 3 survives for a different reason, since five halves is not an integer, and it is the only such pair.
Multiplying by the conjugate turns the difference into x over the sum of the same two terms, exactly, with no series and no error term, and dividing through by x reads off the limit one half. Completing the square gives a constant excess of a quarter, so the gap climbs toward a half strictly from below and never reaches it.
The factor pi over four cancels off both sides, so the comparison is 324 against 244, about a third more pizza. Written that way it is the law of cosines: lay the three widths out as a triangle and the corner between the two smaller sides opens to 109.47 degrees, wider than square, which is the answer with no arithmetic at all.