Difference between revisions of "2024 AIME II Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Find the number of triples of nonnegative integers | + | Find the number of triples of nonnegative integers <math>(a,b,c)</math> satisfying <math>a + b + c = 300</math> and |
<cmath>a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.</cmath> | <cmath>a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.</cmath> | ||
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<math>ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000</math> | <math>ab(a+b)+bc(b+c)+ac(a+c)=300(ab+bc+ac)-3abc=6000000, 100(ab+bc+ac)-abc=2000000</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> | + | Note that |
+ | <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,Shen Kislay Kai | ~Bluesoul,Shen Kislay Kai | ||
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+ | == Solution 6 : Author - Shiva Kumar Kannan (IN PROGRESS PLEASE DO NOT EDIT) == | ||
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+ | Since <math> a + b + c = 300 </math>, <math> (100 - a) + (100 - b) + (100 - c) = 300 - (a + b + c) = 0 </math>. There is a well known algebraic identity known by those ineterested in Olympiad mathematics, which is : | ||
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+ | If <math> a + b + c = 0, a^3 + b^3 + c^3 = 3abc </math>. Hence, as <math> (100 - a) + (100 - b) + (100 - c) = 0 </math> as mentioned above, <math> (100 - a)^3 + (100 - b)^3 + (100 - c)^3 = 3(100 - a)(100 - b)(100 - c)</math>. | ||
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+ | Now expand the LHS of the equation : <math> (100 - a)^3 + (100 - b)^3 + (100 - c)^3 = 3 * 100^3 - 3 * 100^2 * (a + b + c) + 3 * 100 * (a^2 + b^2 + c^2) - (a^3 + b^3 + c^3)</math>. | ||
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+ | We are given in the problem that <cmath>a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 6,000,000.</cmath> Notice that <math> (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 6abc</math>. This means that <math> 300^3 = a^3 + b^3 + c^3 + 6abc + 3 * 6 * 10^6</math>. | ||
+ | Simplify to get <math> a^3 + b^3 + c^3 + 6abc = 9 * 10^6</math>. | ||
+ | This means that <math> a^3 + b^3 + c^3 = 9 * 10^6 - 6abc </math>. | ||
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+ | We know that <math> a + b + c = 300 </math>. We also know that <math> a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) = 300^2 - 2(ab + bc + ac)</math>. | ||
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+ | Now the LHS can be written as <math> 3 * 100^3 - 3 * 100^2 * 300 + 3 * 100 * (300^2 - 2(ab + ac + bc)) - (9 * 10^6 - 6abc)</math>. This simplifies to <math> 12 * 10^6 - 600(ab + bc + ac) + 6abc</math>. | ||
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+ | Now, we evaluate the right side. <math> (100 - a)(100 - b)(100 - c) = 10^6 - 10^4(a + b + c) + 100(ab + bc + ac) - abc = -2*10^6 - 100(ab + bc + ac) - abc</math>. Now we set the LHS and RHS equal to each other. | ||
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+ | <cmath>12 * 10^6 - 600(ab + ac + bc) + 6abc = -2*10^6 + 100(ab + bc + ac) - abc</cmath>. Notice that the LHS is just <math>-6</math> times the RHS. If the RHS is equal to <math>-6</math> times itself, the only possible value the RHS can take is <math>0</math>. The RHS was originally <math>3(100 - a)(100 - b)(100 - c)</math>. This must equal <math>0</math>. <cmath> 3(100 - a)(100 - b)(100 - c) = 0</cmath>. This means one of <math>a, b,</math> or <math>c</math> must be <math>100</math>. The remaining two must sum up to <math>200</math> as the three of them together sum to <math>300</math> as indicated by the problem. WLOG Let us assume <math>a = 100</math> and <math> b + c = 200</math>. As <math>b</math> and <math>c</math> are nonnegative integers, we employ Stars and Bars to find that there are <math>\binom{200 + 2 - 1}{2 - 1} = 201 </math> solutions to the equation. As <math>a, b,</math> or <math>c</math> could in reality be <math>100</math>, multiply <math>201</math> by <math>3</math> to get <math>603</math>. However, the solution <math>(a, b, c) = (100, 100, 100)</math> is counted thrice in total, but we only want it counted once, so subtract <math>2</math> from <math>603</math> to arrive at the final answer : The number of solutions is <math>\boxed{601}</math>. | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 11:00, 15 October 2024
Contents
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 \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
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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