2011 AIME I Problems/Problem 15
Contents
Problem
For some integer , the polynomial has the three integer roots , , and . Find .
Solution 1
From Vieta's formulas, we know that , and . Thus . All three of , , and are non-zero: say, if , then (which is not an integer). , let . If , then and if , then . We have Thus . We know that , have the same sign. So .
Also, if we fix , is fixed, so is maximized when . Hence, So . Thus we have bounded as , i.e. since . Let's 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,
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 ~firebolt360
Solution 6
Note that , so , or . Also, , so . Substituting , we can obtain , or . If it is not known that is prime, it may be proved in minutes or so by checking all primes up to . If divided either of , then in order for to contain an extra copy of , both would need to be divisible by . But then would also be divisible by , and the sum would clearly be divisible by .
By LTE, if is divisible by and neither are divisible by . Thus, the only possibility remaining is if did not divide . Let . Then, we have . Rearranging gives . As in the above solutions, we may eliminate certain values of by using mods. Then, we may test values until we obtain , and . Thus, , , and our answer is .
Video Solution
https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
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