Difference between revisions of "2011 AIME I Problems/Problem 15"
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== Solution 3 == | == Solution 3 == | ||
− | Let us first note the obvious that is derived from Vieta's formulas: <math>a+b+c=0, ab+bc+ac=-2011</math>. Now, due to the first equation, let us say that <math>a+b=-c</math>, meaning that <math>a,b>0</math> and <math>c<0</math>. Now, since both <math>a</math> and <math>b</math> are greater than 0, their absolute values are both equal to <math>a</math> and <math>b</math>, respectively. Since <math>c</math> is less than 0, it equals <math>-a-b</math>. Therefore, <math>|c|=|-a-b|=a+b</math>, meaning <math>|a|+|b|+|c|=2(a+b)</math>. We now apply Newton's sums to get that <math>a^2+b^2+ab=2011</math>,or <math>(a+b)^2-ab=2011</math>. Solving, we find that <math>49^2-390</math> satisfies this, meaning <math>a+b=49</math>, so <math>2(a+b)=\boxed{098}</math>. | + | Let us first note the obvious that is derived from Vieta's formulas: <math>a+b+c=0, ab+bc+ac=-2011</math>. Now, due to the first equation, let us say that <math>a+b=-c</math>, meaning that <math>a,b>0</math> and <math>c<0</math>. Now, since both <math>a</math> and <math>b</math> are greater than 0, their absolute values are both equal to <math>a</math> and <math>b</math>, respectively. Since <math>c</math> is less than 0, it equals <math>-a-b</math>. Therefore, <math>|c|=|-a-b|=a+b</math>, meaning <math>|a|+|b|+|c|=2(a+b)</math>. We now apply Newton's sums to get that <math>a^2+b^2+ab=2011</math>,or <math>(a+b)^2-ab=2011</math>. Solving, we find that <math>49^2-390</math> satisfies this, meaning <math>a+b=49</math>, so <math>2(a+b)=\boxed{098}</math>. |
==Solution 4== | ==Solution 4== |
Revision as of 21:03, 20 January 2019
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, maximize when if we fixed . Hence, .
So .
so .
Now we have limited to .
Let's us analyze .
Here is a table:
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,
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 |
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