Paper 4, Section II, E

Numbers and Sets
Part IA, 2015

Let pp be a prime. A base pp expansion of an integer kk is an expression

k=k0+pk1+p2k2++pkk=k_{0}+p \cdot k_{1}+p^{2} \cdot k_{2}+\cdots+p^{\ell} \cdot k_{\ell}

for some natural number \ell, with 0ki<p0 \leqslant k_{i}<p for i=0,1,,i=0,1, \ldots, \ell.

(i) Show that the sequence of coefficients k0,k1,k2,,kk_{0}, k_{1}, k_{2}, \ldots, k_{\ell} appearing in a base pp expansion of kk is unique, up to extending the sequence by zeroes.

(ii) Show that

(pj)0(modp),0<j<p\left(\begin{array}{l} p \\ j \end{array}\right) \equiv 0 \quad(\bmod p), \quad 0<j<p

and hence, by considering the polynomial (1+x)p(1+x)^{p} or otherwise, deduce that

(pij)0(modp),0<j<pi\left(\begin{array}{c} p^{i} \\ j \end{array}\right) \equiv 0 \quad(\bmod p), \quad 0<j<p^{i}

(iii) If n0+pn1+p2n2++pnn_{0}+p \cdot n_{1}+p^{2} \cdot n_{2}+\cdots+p^{\ell} \cdot n_{\ell} is a base pp expansion of nn, then, by considering the polynomial (1+x)n(1+x)^{n} or otherwise, show that

(nk)(n0k0)(n1k1)(nk)(modp)\left(\begin{array}{l} n \\ k \end{array}\right) \equiv\left(\begin{array}{l} n_{0} \\ k_{0} \end{array}\right)\left(\begin{array}{l} n_{1} \\ k_{1} \end{array}\right) \cdots\left(\begin{array}{l} n_{\ell} \\ k_{\ell} \end{array}\right) \quad(\bmod p)