Difference between revisions of "2011 AIME I Problems/Problem 15"
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As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | ||
+ | |||
+ | ==Solution 5 (mod to help bash)== | ||
+ | First, derive the equations <math>a=-b-c</math> and <math>ab+bc+ca=-2011\implies b^2+bc+c^2=2011</math>. Since the product is negative, <math>a</math> is negative, and <math>b</math> and <math>c</math> positive. Now, a simple mod 3 testing of all cases shows that <math>b\equiv \{1,2\} \pmod{3}</math>, and <math>c</math> has the repective value. We can choose <math>b</math> not congruent to 0, make sure you see why. Now, we bash on values of <math>b</math>, testing the quadratic function to see if <math>c</math> is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for <math>b=10</math>, <math>c=39, -49</math>. Choosing <math>c</math> positive we get <math>a=-49</math>, so <math>|a|+|b|+|c|=10+29+39=\boxed{098}</math> | ||
==Note== | ==Note== |
Revision as of 22:47, 16 February 2021
Contents
Problem
For some integer , the polynomial
has the three integer roots
,
, and
. Find
.
Solution
With Vieta's formulas, we know that , and
.
since any one being zero will make the other two
.
. WLOG, let
.
Then if , then
and if
, then
.
We know that ,
have the same sign. So
. (
and
)
Also, if we fix ,
is maximized when
. Hence,
.
So .
so
.
Now we have limited to
.
Let's analyze .
Here is a table:
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We can tell we don't need to bother with ,
, So
won't work.
,
is not divisible by
,
, which is too small to get
.
,
is not divisible by
or
or
, we can clearly tell that
is too much.
Hence, ,
.
,
.
Answer:
Solution 2
Starting off like the previous solution, we know that , and
.
Therefore, .
Substituting, .
Factoring the perfect square, we get: or
.
Therefore, a sum () squared minus a product (
) gives
..
We can guess and check different ’s starting with
since
.
therefore
.
Since no factors of can sum to
(
being the largest sum), a + b cannot equal
.
making
.
and
so
cannot work either.
We can continue to do this until we reach .
making
.
, so one root is
and another is
. The roots sum to zero, so the last root must be
.
.
Solution 3
Let us first note the obvious that is derived from Vieta's formulas: . Now, due to the first equation, let us say that
, meaning that
and
. Now, since both
and
are greater than 0, their absolute values are both equal to
and
, respectively. Since
is less than 0, it equals
. Therefore,
, meaning
. We now apply Newton's sums to get that
,or
. Solving, we find that
satisfies this, meaning
, so
.
Solution 4
We have
As a result, we have
So,
As a result,
Solve and
, where
is an integer
Cause
So, after we tried for times, we get
and
then ,
As a result,
Solution 5 (mod to help bash)
First, derive the equations and
. Since the product is negative,
is negative, and
and
positive. Now, a simple mod 3 testing of all cases shows that
, and
has the repective value. We can choose
not congruent to 0, make sure you see why. Now, we bash on values of
, testing the quadratic function to see if
is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for
,
. Choosing
positive we get
, so
Note
This is a misplaced number theory problem.Which requires a bit of trial and error strategy.
Video Solution
https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.