Difference between revisions of "2024 AIME II Problems/Problem 11"
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We now notice that the solutions to the quadratic equation above are <math>x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}</math>, and that by changing the value of <math>m</math> we can let the roots of the equation be any pair of two integers which sum to <math>200</math>. Thus any triple in the form <math>(100, 100 - x, 100 + x)</math> where <math>x</math> is an integer between <math>0</math> and <math>100</math> satisfies the conditions. | We now notice that the solutions to the quadratic equation above are <math>x = 100 \pm \frac{\sqrt{200^2 - 4(m - 20000)}}{2} = 100 \pm \sqrt{90000 - 4m}</math>, and that by changing the value of <math>m</math> we can let the roots of the equation be any pair of two integers which sum to <math>200</math>. Thus any triple in the form <math>(100, 100 - x, 100 + x)</math> where <math>x</math> is an integer between <math>0</math> and <math>100</math> satisfies the conditions. | ||
− | Now to count the possible solutions, we note that when <math>x \ne | + | Now to count the possible solutions, we note that when <math>x \ne 0</math>, the three roots are distinct; thus there are <math>3! = 6</math> ways to order the three roots. As we can choose <math>x</math> from <math>0</math> to <math>99</math>, there are <math>100 \cdot 3! = 600</math> triples in this case. When <math>x = 0</math>, all three roots are equal to <math>100</math>, and there is only one triple in this case. |
In total, there are thus <math>\boxed{601}</math> distinct triples. | In total, there are thus <math>\boxed{601}</math> distinct triples. | ||
~GaloisTorrent <3 | ~GaloisTorrent <3 | ||
+ | - minor edit made by MEPSPSPSOEODODODO | ||
==Solution 5== | ==Solution 5== |
Revision as of 07:06, 29 July 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
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 \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 <3 - 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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