Difference between revisions of "2011 AIME I Problems/Problem 15"
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+ | ==Solution 6== | ||
+ | Note that <math>-c=b+a</math>, so <math>c^2=a^2+2ab+b^2</math>, or <math>-c^2+ab=-a^2-ab-b^2</math>. Also, <math>ab+bc+ca=-2011</math>, so <math>(a+b)c+ab=-c^2+ab=-2011</math>. Substituting <math>-c^2+ab=-a^2-ab-b^2</math>, we can obtain <math>a^2+ab+b^2=2011</math>, or <math>\frac{a^3-b^3}{a-b}=2011</math>. If it is not known that <math>2011</math> is prime, it may be proved in <math>5</math> minutes or so by checking all primes up to <math>43</math>. If <math>2011</math> divided either of <math>a, b</math>, then in order for <math>a^3-b^3</math> to contain an extra copy of <math>2011</math>, both <math>a, b</math> would need to be divisible by <math>2021</math>. But then <math>c</math> would also be divisible by <math>2011</math>, and the sum <math>ab+bc+ca</math> would clearly be divisible by <math>2011^2</math>. | ||
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+ | By LTE, <math>v_2011(a^3-b^3)=v_2011(a-b)</math> if <math>a-b</math> is divisible by <math>2011</math> and neither <math>a,b</math> are divisible by <math>2011</math>. Thus, the only possibility remaining is if <math>a-b</math> did not divide <math>2011</math>. Let <math>a=k+b</math>. Then, we have <math>(b+k)^3-b^3=2011k</math>. Rearranging gives <math>3b(b+k)=2011-k^2</math>. As in the above solutions, we may eliminate certain values of <math>k</math> by using mods. Then, we may test values until we obtain <math>k=29</math>, and <math>a=10</math>. Thus, <math>b=39</math>, <math>c=-49</math>, and our answer is <math>49+39+10=098</math>. | ||
==Video Solution== | ==Video Solution== | ||
Revision as of 14:17, 18 January 2023
Contents
[hide]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) | ||
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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