Difference between revisions of "2022 AIME II Problems/Problem 13"
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We know that <math>\frac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1} a^{n-1-i}b^i</math>. Applying this, we see that <cmath>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)</cmath> The last factor does not contribute to the <math>x^{2022}</math> term, so we can ignore it. Thus we only have left to solve the equation <math>105b+70c+42d+30e=2022</math>, and we can proceed from here with Solution 1. | We know that <math>\frac{a^n-b^n}{a-b}=\sum_{i=0}^{n-1} a^{n-1-i}b^i</math>. Applying this, we see that <cmath>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)</cmath> The last factor does not contribute to the <math>x^{2022}</math> term, so we can ignore it. Thus we only have left to solve the equation <math>105b+70c+42d+30e=2022</math>, and we can proceed from here with Solution 1. |
Revision as of 01:17, 13 June 2022
Contents
[hide]Problem
There is a polynomial with integer coefficients such thatholds for every Find the coefficient of in .
Solution 1
Because , we have
Denote by the coefficient of . Thus,
Now, we need to find the number of nonnegative integer tuples that satisfy
Modulo 2 on Equation (1), we have . Hence, we can write . Plugging this into (1), the problem reduces to finding the number of nonnegative integer tuples that satisfy
Modulo 3 on Equation (2), we have . Hence, we can write . Plugging this into (2), the problem reduces to finding the number of nonnegative integer tuples that satisfy
Modulo 5 on Equation (3), we have . Hence, we can write . Plugging this into (3), the problem reduces to finding the number of nonnegative integer tuples that satisfy
Modulo 7 on Equation (4), we have . Hence, we can write . Plugging this into (4), the problem reduces to finding the number of nonnegative integer tuples that satisfy
The number of nonnegative integer solutions to Equation (5) is .
~Steven Chen (www.professorchenedu.com)
Solution 2
Note that . Since the only way to express in terms of , , , or is , we are essentially just counting the number of ways to express in terms of these numbers. Since , it can only be expressed as a sum in terms of only one of the numbers (, , , or ). Thus, the answer is (by sticks and stones)
~Bigbrain123
Solution 3
We know that . Applying this, we see that The last factor does not contribute to the term, so we can ignore it. Thus we only have left to solve the equation , and we can proceed from here with Solution 1.
~MathIsFun286
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.