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2022 AIME II Problems/Problem 13: Difference between revisions

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==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 12:27, 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

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