2024 AIME II Problems/Problem 11
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)
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 | ||
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