Difference between revisions of "2011 AIME I Problems/Problem 15"
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For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>. Find <math>|a| + |b| + |c|</math>. | For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>. Find <math>|a| + |b| + |c|</math>. | ||
− | ==Solution== | + | ==Solution 1== |
− | + | From Vieta's formulas, we know that <math>a+b+c = 0</math>, and <math>ab+bc+ac = -2011</math>. Thus <math>a = -(b+c)</math>. All three of <math>a</math>, <math>b</math>, and <math>c</math> are non-zero: say, if <math>a=0</math>, then <math>b=-c=\pm\sqrt{2011}</math> (which is not an integer). <math>\textsc{wlog}</math>, let <math>|a| \ge |b| \ge |c|</math>. If <math>a > 0</math>, then <math>b,c < 0</math> and if <math>a < 0</math>, then <math>b,c > 0</math>. We have <cmath>-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc</cmath> | |
+ | Thus <math>a^2 = 2011 + bc</math>. We know that <math>b</math>, <math>c</math> have the same sign. So <math>|a| \ge 45 = \lceil \sqrt{2011} \rceil</math>. | ||
− | + | Also, if we fix <math>a</math>, <math>b+c</math> is fixed, so <math>bc</math> is maximized when <math>b = c</math> . Hence, <cmath>2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}</cmath> | |
− | + | So <math>|a| \le 51</math>. Thus we have bounded <math>a</math> as <math>45\le |a| \le 51</math>, i.e. <math>45\le |b+c| \le 51</math> since <math>a=-(b+c)</math>. Let's analyze <math>bc=(b+c)^2-2011</math>. Here is a table: | |
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− | Let's | ||
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− | Here is a table: | ||
<table border = 1 cellspacing = 0 cellpadding = 5 style = "text-align:center;"> | <table border = 1 cellspacing = 0 cellpadding = 5 style = "text-align:center;"> | ||
− | <tr><th><math>|a|</math></th><th><math>a^2 | + | <tr><th><math>|a|</math></th><th><math>bc=a^2-2011</math></th></tr> |
<tr><td><math>45</math></td><td><math>14</math></td></tr> | <tr><td><math>45</math></td><td><math>14</math></td></tr> | ||
− | <tr><td><math>46</math></td><td><math> | + | <tr><td><math>46</math></td><td><math>105</math></td></tr> |
− | <tr><td><math>47</math></td><td><math> | + | <tr><td><math>47</math></td><td><math>198</math></td></tr> |
− | <tr><td><math>48</math></td><td><math> | + | <tr><td><math>48</math></td><td><math>293</math></td></tr> |
− | <tr><td><math>49</math></td><td><math> | + | <tr><td><math>49</math></td><td><math>390</math></td></tr> |
</table> | </table> | ||
<br /> | <br /> | ||
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<math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>, | <math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>, | ||
− | <math>198</math> is not divisible by <math>5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math> | + | <math>198</math> is not divisible by <math>5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math>. |
− | <math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much | + | <math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much. |
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<math>|-49|+10+39 = \boxed{098}</math>. | <math>|-49|+10+39 = \boxed{098}</math>. | ||
+ | |||
+ | |||
+ | == 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>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | We have | ||
+ | |||
+ | <math>(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc</math> | ||
+ | |||
+ | As a result, we have | ||
+ | |||
+ | <math>a+b+c=0</math> | ||
+ | |||
+ | <math>ab+bc+ac=-2011</math> | ||
+ | |||
+ | <math>abc=-m</math> | ||
+ | |||
+ | So, <math>a=-b-c</math> | ||
+ | |||
+ | As a result, <math>ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011</math> | ||
+ | |||
+ | Solve <math>b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}</math> and | ||
+ | <math>\Delta =8044-3c^2=k^2</math>, where <math>k</math> is an integer | ||
+ | |||
+ | Cause <math>89<\sqrt{8044}<90</math> | ||
+ | |||
+ | So, after we tried for <math>2</math> times, we get <math>k=88</math> and <math>c=10</math> | ||
+ | |||
+ | then <math>b=39</math>, <math>a=-b-c=-49</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> | ||
+ | ~firebolt360 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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== | ||
+ | |||
+ | https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx | ||
== See also == | == See also == | ||
− | {{AIME box|year=2011|num-b=14|after= | + | {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} |
+ | |||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 04:07, 24 July 2023
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:
![]() | ![]() |
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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
~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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