2022 AIME II Problems/Problem 13

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 1

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)

Solution 1 Modification

Note that $2022 = 210\cdot 9 +132$ . Since the only way to express $132$ in terms of $105$ , $70$ , $42$ , or $30$ is $135 = 30+30+30+42$ , we are essentially just counting the number of ways to express $210*9$ in terms of these numbers. This is simply (by sticks and stones) $\boxed{\textbf{(220)}}$

~Bigbrain123

Solution 2

We know that $\frac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1} a^{n-1-i}b^i$ . Applying this, we see that $P(x)=(1+x^{105}+x^{210}+...)(1+x^{70}+x^{140}+...)(1+x^{42}+x^{84}+...)(1+x^{30}+x^{60}+...)(x^{4620}-2x^{2310}+1)$ The last factor does not contribute to the $x^{2022}$ term, so we can ignore it. Thus we only have left to solve the equation $105b+70c+42d+30e=2022$ , and we can proceed from here with Solution 1.

~MathIsFun286

2022 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2022 AIME II Problems/Problem 13

Contents

Problem

Solution 1

Solution 1 Modification

Solution 2

See Also