Difference between revisions of "2024 AIME II Problems/Problem 11"
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==Solution 4== | ==Solution 4== | ||
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We will use Vieta's formulas to solve this problem. We assume <math>a + b + c = 300</math>, <math>ab + bc + ca = m</math>, and <math>abc = n</math>. Thus <math>a</math>, <math>b</math>, <math>c</math> are the three roots of a cubic polynomial <math>f(x)</math>. | We will use Vieta's formulas to solve this problem. We assume <math>a + b + c = 300</math>, <math>ab + bc + ca = m</math>, and <math>abc = n</math>. Thus <math>a</math>, <math>b</math>, <math>c</math> are the three roots of a cubic polynomial <math>f(x)</math>. | ||
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~GaloisTorrent <3 | ~GaloisTorrent <3 | ||
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+ | ==Solution 5== | ||
+ | Let's define <math>a=100+x</math>, <math>b=100+y</math>, <math>c=100+z</math>. Then we have <math>x+y+z=0</math> and | ||
+ | <math>6000000 = \sum a^2(b+c) = \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) </math> | ||
+ | <math>= \sum (20000 - 10000 x + x(40000-x^2)) = \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3</math>, so we get <math>x^3 + y^3 + z^3 = 0</math>. Then from <math>x+y+z = 0</math>, we can find <math>0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz</math>, which means that one of <math>a</math>, <math>b</math>,<math>c</math> must be 0. There are 201 solutions for each of <math>a=0</math>, <math>b=0</math> and <math>c=0</math>, and subtract the overcounting of 2 for solution <math>(200, 200, 200)</math>, the final result is <math>201 \times 3 - 2 = \boxed{601}</math>. | ||
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+ | Dan Li | ||
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+ | dan | ||
==Video Solution== | ==Video Solution== |
Revision as of 09:30, 13 February 2024
Contents
[hide]Problem
Find the number of triples of nonnegative integers
solution 1
Note . Thus,
. There are
cases for each but we need to subtract
for
. The answer is
~Bluesoul
Solution 2
, thus
. Complete the cube to get
, which so happens to be 0. Then we have
. We can use Fermat's last theorem here to note that one of a, b, c has to be 100. We have 200+200+200+1 = 601.
Solution 3
We have
.
Therefore, \[ \left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 . \]
Case 1: Exactly one out of ,
,
is equal to 0.
Step 1: We choose which term is equal to 0. The number ways is 3.
Step 2: For the other two terms that are not 0, we count the number of feasible solutions.
W.L.O.G, we assume we choose in Step 1. In this step, we determine
and
.
Recall . Thus,
.
Because
and
are nonnegative integers and
and
, the number of solutions is 200.
Following from the rule of product, the number of solutions in this case is .
Case 2: At least two out of ,
,
are equal to 0.
Because , we must have
.
Therefore, the number of solutions in this case is 1.
Putting all cases together, the total number of solutions is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4
We will use Vieta's formulas to solve this problem. We assume ,
, and
. Thus
,
,
are the three roots of a cubic polynomial
.
We note that , which simplifies to
.
Our polynomial is therefore equal to
. Note that
, and by polynomial division we obtain
.
We now notice that the solutions to the quadratic equation above are , and that by changing the value of
we can let the roots of the equation be any pair of two integers which sum to
. Thus any triple in the form
where
is an integer between
and
satisfies the conditions.
Now to count the possible solutions, we note that when , the three roots are distinct; thus there are
ways to order the three roots. As we can choose
from
to
, there are
triples in this case. When
, all three roots are equal to
, and there is only one triple in this case.
In total, there are thus distinct triples.
~GaloisTorrent <3
Solution 5
Let's define ,
,
. Then we have
and
, so we get
. Then from
, we can find
, which means that one of
,
,
must be 0. There are 201 solutions for each of
,
and
, and subtract the overcounting of 2 for solution
, the final result is
.
Dan Li
dan
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.