2024 AIME II Problems/Problem 11
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[hide]Problem
Find the number of triples of nonnegative integers satisfying and
Solution 1
Note that . 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
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
Solution 6 : Author - Shiva Kumar Kannan (IN PROGRESS PLEASE DO NOT EDIT)
Since , . There is a well known algebraic identity known by those ineterested in Olympiad mathematics, which is :
If . Hence, as as mentioned above, .
Now expand the LHS of the equation : .
We are given in the problem that Notice that . This means that .
Simplify to get .
This means that .
We know that . We also know that .
Now the LHS can be written as . This simplifies to .
Now, we evaluate the right side. . Now we set the LHS and RHS equal to each other.
. Notice that the LHS is just times the RHS. If the RHS is equal to times itself, the only possible value the RHS can take is . The RHS was originally . This must equal . . This means one of or must be . The remaining two must sum up to as the three of them together sum to as indicated by the problem. WLOG Let us assume and . As and are nonnegative integers, we employ Stars and Bars to find that there are solutions to the equation. As or could in reality be , multiply by to get . However, the solution is counted thrice in total, but we only want it counted once, so subtract from to arrive at the final answer : The number of solutions is .
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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