2019 Mock AMC 10B Problems/Problem 16
Revision as of 14:52, 5 November 2023 by Excruciating (talk | contribs)
Solution by excruciating:
For each of , they all need to be at least
, so we have:
or
without the "
need to be
restriction." So now it's stars and bars:
which simplifies to
which we can write as
.
Expanding, we get .
Because we only have terms,
is to be around
the highest divisor. Thus, we must have
as a divisor too.
is one too. We have left is
and
, to make
terms, so we have
, and
. Thus we have:
Thus,