Given a positive integer k, call n good, if among \binom{n}{0},\binom{n}{1},\binom{n}{2},...,\binom{n}{n} at least 0.99n of them are divisible by k. Show that exists some positive integer N such that among 1,2,...,N, there are at least 0.99N good numbers.
some try: for all prime p and nonnegative integers m,n,
\nu_p \binom{m+n}{n} = \frac{s_p(m)+s_p(n)-s_p(m+n)}{p-1}
which is equal to the number of carries when adding m and n in mod p.
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