Difference between revisions of "2024 AIME II Problems/Problem 11"
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− | + | ==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== | ||
+ | <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> | ||
+ | |||
+ | ~Bluesoul | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | <math>a^2(b+c)+b^2(a+c)+c^2(a+b) = 6000000</math>, thus <math>a^2(300-a)+b^2(300-b)+c^2(300-c) = 6000000</math>. Complete the cube to get <math>-(a-100)^3-(b-100)^3+(c-100)^3 = 9000000-30000(a+b+c)</math>, which so happens to be 0. Then we have <math>(a-100)^3+(b-100)^3+(c-100)^3 = 0</math>. 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 <math>a+b+c = 300</math>. | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \left( a - 100 \right) \left( b - 100 \right) \left( c - 100 \right) = 0 . | ||
+ | </cmath> | ||
+ | |||
+ | Case 1: Exactly one out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> 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 <math>a - 100 = 0</math> in Step 1. In this step, we determine <math>b</math> and <math>c</math>. | ||
+ | |||
+ | Recall <math>a + b + c = 300</math>. Thus, <math>b + c = 200</math>. | ||
+ | Because <math>b</math> and <math>c</math> are nonnegative integers and <math>b - 100 \neq 0</math> and <math>c - 100 \neq 0</math>, the number of solutions is 200. | ||
+ | |||
+ | Following from the rule of product, the number of solutions in this case is <math>3 \cdot 200 = 600</math>. | ||
+ | |||
+ | Case 2: At least two out of <math>a - 100</math>, <math>b - 100</math>, <math>c - 100</math> are equal to 0. | ||
+ | |||
+ | Because <math>a + b + c = 300</math>, we must have <math>a = b = c = 100</math>. | ||
+ | |||
+ | Therefore, the number of solutions in this case is 1. | ||
+ | |||
+ | Putting all cases together, the total number of solutions is <math>600 + 1 = \boxed{\textbf{(601) }}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We will use Vieta's formulas to solve this problem. We assume <math>a + b + c = 300</math>, <math>ab + bc + ca = m</math>, and <math>abc = n</math>. Thus <math>a</math>, <math>b</math>, <math>c</math> are the three roots of a cubic polynomial <math>f(x)</math>. | ||
+ | |||
+ | We note that <math>300m = (a + b + c)(ab + bc + ca)=\sum_{cyc} a^2b + 3abc = 6000000 + 3n</math>, which simplifies to <math>100m - 2000000 = n</math>. | ||
+ | |||
+ | Our polynomial <math>f(x)</math> is therefore equal to <math>x^3 - 300x^2 + mx - (100m - 2000000)</math>. Note that <math>f(100) = 0</math>, and by polynomial division we obtain <math>f(x) = (x - 100)(x^2 - 200x - (m-20000))</math>. | ||
+ | |||
+ | 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 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>1</math> to <math>100</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. | ||
+ | |||
+ | ~GaloisTorrent <3 | ||
+ | |||
+ | - minor edit made by MEPSPSPSOEODODODO | ||
+ | |||
+ | ==Solution 5== | ||
+ | Let's define <math>a=100+x</math>, <math>b=100+y</math>, <math>c=100+z</math>. Then we have <math>x+y+z=0</math> and <math>6000000 = \sum a^2(b+c) </math> | ||
+ | |||
+ | <math>= \sum (100+x)^2(200-x) = \sum (10000+200x+x^2)(200-x) = \sum (20000 - 10000 x + x(40000-x^2)) </math> | ||
+ | |||
+ | <math>= \sum (20000 + 30000 x -x^3) = 6000000 - \sum x^3</math>, so we get <math>x^3 + y^3 + z^3 = 0</math>. Then from <math>x+y+z = 0</math>, we can find <math>0 = x^3+y^3+z^3 = x^3+y^3-(x+y)^3 = 3xyz</math>, which means that one of <math>a</math>, <math>b</math>,<math>c</math> must be 0. There are 201 solutions for each of <math>a=0</math>, <math>b=0</math> and <math>c=0</math>, and subtract the overcounting of 2 for solution <math>(200, 200, 200)</math>, the final result is <math>201 \times 3 - 2 = \boxed{601}</math>. | ||
+ | |||
+ | Dan Li | ||
+ | |||
+ | |||
+ | |||
+ | dan | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/YMYe9chPLdY | ||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2024|num-b=10|num-a=12|n=II}} | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 08:14, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.