Taking one coin from each bag reads 9.9 ounces whichever bag is light, and the failure is blindness rather than imprecision. Loading i coins from bag i makes the dial an injective function of the culprit, at a cost of the tenth triangular number. A 45-coin variant is cheaper, and powers of two identify any subset of light bags from one reading.
Also asked ascounterfeit coins · light bag · one weighing
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.