Difference between revisions of "2023 AIME I Problems/Problem 9"
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− | a=-10 and a=-8 give values for b outside the range and a=-6 results in m= | + | a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328? |
Revision as of 14:14, 8 February 2023
Contents
[hide]Problem (Unofficial, please update when official one comes out):
is a polynomial with integer coefficients in the range
, inclusive. There is exactly one integer
such that
. How many possible values are there for the ordered triple
?
Solution
Solution 1
Plugging into
, we get
. We can rewrite into
, where
can be any value in the range. Since
must be
. The problem also asks for unique integers, meaning
can only be one value for each polynomial, so the discriminant must be
.
, and
. Rewrite to be
.
must be even for
to be an integer.
because
. There are 11 pairs of
and 41 integers for
, giving
~chem1kall
Solution 2
a=-10 and a=-8 give values for b outside the range and a=-6 results in m=2. Therefore shouldn't the answer be 41*8=328?