Difference between revisions of "2024 AIME II Problems/Problem 11"
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Note <math>(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0</math>. Thus, <math>a/b/c=100</math>. There are <math>201</math> cases for each but we need to subtract <math>2</math> for <math>(100,100,100)</math>. The answer is <math>\boxed{601}</math> | Note <math>(100-a)(100-b)(100-c)=1000000-10000(a+b+c)+100(ab+bc+ac)-abc=0</math>. Thus, <math>a/b/c=100</math>. There are <math>201</math> cases for each but we need to subtract <math>2</math> for <math>(100,100,100)</math>. The answer is <math>\boxed{601}</math> | ||
− | ~Bluesoul | + | ~Bluesoul,Shen Kislay Kai |
==Solution 2== | ==Solution 2== |
Latest revision as of 13:22, 3 September 2024
Contents
Problem
Find the number of triples of nonnegative integers \((a,b,c)\) satisfying \(a + b + c = 300\) and \begin{equation*} a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000. \end{equation*}
Solution 1
Note . Thus,
. There are
cases for each but we need to subtract
for
. The answer is
~Bluesoul,Shen Kislay Kai
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
has to be 100. We have
Solution 3
We have
\begin{align*}
& a^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\
& = ab \left( a + b \right) + bc \left( b + c \right) + ca \left( c + a \right) \\
& = ab \left( 300 - c \right) + bc \left( 300 - a \right) + ca \left( 300 - b \right) \\
& = 300 \left( ab + bc + ca \right) - 3 abc \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 10^4 \left( a + b + c \right) + 10^6
\right) \\
& = -3 \left(
\left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right)
- 2 \cdot 10^6
\right) \\
& = 6 \cdot 10^6 .
\end{align*}
The first and the fifth equalities follow from the condition that .
Therefore,
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 <Shen Kislay Kai>
- minor edit made by MEPSPSPSOEODODODO
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 |
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