Difference between revisions of "2019 Mock AMC 10B Problems/Problem 16"
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Thus, <math> n = 14 + 1 = \boxed{(E) 15}.</math> | Thus, <math> n = 14 + 1 = \boxed{(E) 15}.</math> | ||
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+ | Note: <math>g</math> does not need to be <math>\geq 1</math> because the inequality <math>a+b+c+d+e+f\leq n</math> is not strict. So this solution is incorrect. |
Latest revision as of 14:06, 11 August 2024
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,
Note: does not need to be
because the inequality
is not strict. So this solution is incorrect.