Factoring into p minus one times p plus one stops the question being about p: the two neighbours are consecutive even numbers so their product carries eight, and one of the three consecutive integers around p is a multiple of three which cannot be p itself. Since eight and three are coprime, 24 divides, and 24 is exactly maximal.
Counting to fifty in steps of one to ten, the first player wins, and exactly one of the ten legal openings does it. The stations are 6, 17, 28, 39 and 50, spaced eleven apart because eleven is one more than the largest legal step. The article carries the residue argument that proves the opening is unique, and the target 55 where the advantage flips.
Every one of the nine conditions leaves a remainder one short of its divisor, so x plus one is divisible by all of 2 through 10 and the answer is 2519. Minimality comes free, and the whole solution set is 2520k minus 1. The article carries the coprimality caveat, the near miss 209 that satisfies six of the nine, and a variant where no shift exists.
Multiplying four rings a minute by five gives a quantity in inverse minutes squared, so a dimension check kills the twenty-second answer before any number theory. Converting to gaps makes the ring times two arithmetic progressions whose intersection is the least common multiple. Shift one bell by a second and the two never coincide at all.
Adding fives and threes never reaches four, and that observation is correct. It is also about the wrong set, because a pour is a subtraction and the reachable amounts are the integer combinations rather than the natural ones. An exhaustive state-graph search proves six pours is minimal, and the four missing sums turn out to be the gaps of a numerical semigroup.
Una recta con coeficientes racionales aplica ℚ sobre ℚ. Nada curvo lo logra jamás. Tres filtros — interpolación, forma, denominadores — dejan la clasificación completa.