Putnam 2007 B5 Session B Let kkk be a positive integer. Prove that there exist polynomials P0(n),P1(n),…,Pk−1(n)P_0(n), P_1(n), \dots, P_{k-1}(n)P0(n),P1(n),…,Pk−1(n) (which may depend on kkk) such that for any integer nnn, ⌊nk⌋k=P0(n)+P1(n)⌊nk⌋+⋯+Pk−1(n)⌊nk⌋k−1.\left\lfloor \frac{n}{k} \right\rfloor^k = P_0(n) + P_1(n) \left\lfloor \frac{n}{k} \right\rfloor + \cdots + P_{k-1}(n) \left\lfloor \frac{n}{k} \right\rfloor^{k-1}.⌊kn⌋k=P0(n)+P1(n)⌊kn⌋+⋯+Pk−1(n)⌊kn⌋k−1. (⌊a⌋\lfloor a \rfloor⌊a⌋ means the largest integer ≤a\leq a≤a.) Curated text Kedlaya Putnam archive. Quoted with source. Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution PreviousB4 NextB6 Locator putnam/2007/b5 · Tag VACY9TMB8S · Edit this page on GitHub · Keys: j next, k previous