Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AIME II Problems/Problem 13"
(→‎Solution)
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
 +
Because <math>0 < x < 1</math>, we have
 +
<cmath>
 +
\begin{align*}
 +
P \left( x \right)
 +
& = \sum_{a=0}^6
 +
\sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty
 +
\binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a}
 +
x^{105b} x^{70c} x^{42d} x^{30e} \\
 +
& = \sum_{a=0}^6
 +
\sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty
 +
\left( - 1 \right)^{6-a}
 +
x^{2310 a + 105 b + 70 c + 42 d + 30 e} .
 +
\end{align*}
 +
</cmath>
 +
 +
Denote by <math>c_{2022}</math> the coefficient of <math>P \left( x \right)</math>.
 +
Thus,
 +
<cmath>
 +
\begin{align*}
 +
c_{2022} & = \sum_{a=0}^6
 +
\sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty
 +
\left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\
 +
& =
 +
\sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty
 +
\left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\
 +
& = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty
 +
\Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} .
 +
\end{align*}
 +
</cmath>
 +
 +
Now, we need to find the number of nonnegative integer tuples <math>\left( b , c , d , e \right)</math> that satisfy
 +
<cmath>
 +
\[
 +
105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1)
 +
\]
 +
</cmath>
 +
 +
Modulo 2 on Equation (1), we have <math>b \equiv 0 \pmod{2}</math>.
 +
Hence, we can write <math>b = 2 b'</math>. Plugging this into (1), the problem reduces to finding the number of
 +
nonnegative integer tuples <math>\left( b' , c , d , e \right)</math> that satisfy
 +
<cmath>
 +
\[
 +
105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2)
 +
\]
 +
</cmath>
 +
 +
Modulo 3 on Equation (2), we have <math>2 c \equiv 0 \pmod{3}</math>.
 +
Hence, we can write <math>c = 3 c'</math>. Plugging this into (2), the problem reduces to finding the number of
 +
nonnegative integer tuples <math>\left( b' , c' , d , e \right)</math> that satisfy
 +
<cmath>
 +
\[
 +
35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3)
 +
\]
 +
</cmath>
 +
 +
Modulo 5 on Equation (3), we have <math>2 d \equiv 2 \pmod{5}</math>.
 +
Hence, we can write <math>d = 5 d' + 1</math>. Plugging this into (3), the problem reduces to finding the number of
 +
nonnegative integer tuples <math>\left( b' , c' , d' , e \right)</math> that satisfy
 +
<cmath>
 +
\[
 +
7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4)
 +
\]
 +
</cmath>
 +
 +
Modulo 7 on Equation (4), we have <math>e \equiv 3 \pmod{7}</math>.
 +
Hence, we can write <math>e = 7 e' + 3</math>. Plugging this into (4), the problem reduces to finding the number of
 +
nonnegative integer tuples <math>\left( b' , c' , d' , e' \right)</math> that satisfy
 +
<cmath>
 +
\[
 +
b' + c' + d' + e' = 9 . \hspace{1cm} (5)
 +
\]
 +
</cmath>
 +
 +
The number of nonnegative integer solutions to Equation (5) is <math>\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} =  \boxed{\textbf{(220) }}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=II|num-b=12|num-a=14}}
 
{{AIME box|year=2022|n=II|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:27, 18 February 2022

Problem

There is a polynomial $P(x)$ with integer coefficients such that\[P(x)=\frac{(x^{2310}-1)^6}{(x^{105}-1)(x^{70}-1)(x^{42}-1)(x^{30}-1)}\]holds for every $0<x<1.$ Find the coefficient of $x^{2022}$ in $P(x)$.

Solution

Because $0 < x < 1$, we have \begin{align*} P \left( x \right) & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \binom{6}{a} x^{2310a} \left( - 1 \right)^{6-a} x^{105b} x^{70c} x^{42d} x^{30e} \\ & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} x^{2310 a + 105 b + 70 c + 42 d + 30 e} . \end{align*}

Denote by $c_{2022}$ the coefficient of $P \left( x \right)$. Thus, \begin{align*} c_{2022} & = \sum_{a=0}^6 \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-a} \Bbb I \left\{ 2310 a + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \left( - 1 \right)^{6-0} \Bbb I \left\{ 2310 \cdot 0 + 105 b + 70 c + 42 d + 30 e = 2022 \right\} \\ & = \sum_{b=0}^\infty \sum_{c=0}^\infty \sum_{d=0}^\infty \sum_{e=0}^\infty \Bbb I \left\{ 105 b + 70 c + 42 d + 30 e = 2022 \right\} . \end{align*}

Now, we need to find the number of nonnegative integer tuples $\left( b , c , d , e \right)$ that satisfy \[ 105 b + 70 c + 42 d + 30 e = 2022 . \hspace{1cm} (1) \]

Modulo 2 on Equation (1), we have $b \equiv 0 \pmod{2}$. Hence, we can write $b = 2 b'$. Plugging this into (1), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c , d , e \right)$ that satisfy \[ 105 b' + 35 c + 21 d + 15 e = 1011 . \hspace{1cm} (2) \]

Modulo 3 on Equation (2), we have $2 c \equiv 0 \pmod{3}$. Hence, we can write $c = 3 c'$. Plugging this into (2), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d , e \right)$ that satisfy \[ 35 b' + 35 c' + 7 d + 5 e = 337 . \hspace{1cm} (3) \]

Modulo 5 on Equation (3), we have $2 d \equiv 2 \pmod{5}$. Hence, we can write $d = 5 d' + 1$. Plugging this into (3), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d' , e \right)$ that satisfy \[ 7 b' + 7 c' + 7 d' + e = 66 . \hspace{1cm} (4) \]

Modulo 7 on Equation (4), we have $e \equiv 3 \pmod{7}$. Hence, we can write $e = 7 e' + 3$. Plugging this into (4), the problem reduces to finding the number of nonnegative integer tuples $\left( b' , c' , d' , e' \right)$ that satisfy \[ b' + c' + d' + e' = 9 . \hspace{1cm} (5) \]

The number of nonnegative integer solutions to Equation (5) is $\binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} = \boxed{\textbf{(220) }}$.

~Steven Chen (www.professorchenedu.com)

See Also

2022 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AIME_II_Problems/Problem_13&oldid=171398"