2021 JMPSC Sprint Problems/Problem 1
Problem
Compute .
Solution
Solving the right side gives . Distributing into the left side gives
, so the answer is
.
Solution 2
and
, so the answer is
.
Solution 3
Therefore, the answer is
- kante314 -
Solution 4
Solution by pog
Recall that, for any nonzero ,
,
, we have that
. By Vieta's, we see that
,
, and
are the roots of the monic cubic polynomial
where
is an arbitrary constant equal to
.
In the given equation, we have that ,
, and
, so
and
, for
(Finding the former of the latter two coefficients would lead to bashy casework, which we avoid in this extremely elegant solution.)
Since our cubic has roots ,
, and
, the discriminant of our monic cubic polynomial
(where
,
, and
are the latter three coefficients of
, respectively--do not confuse them for the roots of our polynomial equal to
,
, and
, respectively) must not be equal to
, or our cubic will have a double root, which it clearly does not (proof here is left to the reader).
Thus, substitution of the coefficients into our discriminant gives that must not be equal to
. Letting
, expanding gives that
is not equal to
. Thus
must not be a root of our cubic, so from here we apply the cubic formula to find that
is not equal to (approximately)
,
, or
.
Recall that, from earlier, our answer is equal to . By bounding,
is between the aforementioned values of
and
, so since it must be a multiple of
(all answers in the Junior Mathematicians' Problem Solving Competition are integers), it is equal to
and the requested answer is equal to
. This corroborates with jasperE3's strict bounding.
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.