Difference between revisions of "2022 AIME II Problems/Problem 13"
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+ | ==Solution 4 (Generating Function Bash)== | ||
+ | |||
+ | Essentially we want the coefficient of <math>x^{2022}</math> in the expansion | ||
+ | |||
+ | <cmath>\frac{1}{(1 - x^{105})(1 - x^{70})(1 - x^{42})(1 - x^{30})}</cmath> | ||
+ | |||
+ | As <math>(x^{2310} -1)^{6}</math> does not contribute to the expansion, we omit it. | ||
+ | |||
+ | Notice <math>\text{lcm}(105,70,42,30) = 210</math>, so we can rewrite the generating function as | ||
+ | |||
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+ | <cmath>\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{42} + ... + x^{168})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}</cmath> | ||
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+ | Notice to obtain a <math>x^{2022}</math> value, the <math>x^{42}</math> must be used, so we can reduce the problem to finding the coefficient of <math>x^{1980}</math> in | ||
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+ | <cmath>\frac{(1 + x^{105})(1 + x^{70} + x^{140})(1 + x^{30} + ... + x^{180})}{(1-x^{210})^{4}}</cmath> | ||
+ | |||
+ | If <math>(1 - x^{210})^{-4}</math> is expanded, it will only generate multiples of 210. To compensate this, notice <math>1980 \equiv 90 \pmod{210}</math>, which means we need terms with a power of a 90 to acheive what we want (300 and larger values can be shown to be impossible upon inspection). This implies that the first two brackets do not contribute and we are left with | ||
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+ | <cmath>\frac{x^{90}}{(1-x^{210})^{4}}</cmath> | ||
+ | |||
+ | This reduces to finding the coefficient of <math>x^{1890}</math> for the expansion <math>(1 - x^{210})^{-4}</math>, which is <math>\binom{12}{3} = \fbox{220}</math> | ||
==See Also== | ==See Also== |
Revision as of 20:20, 2 July 2022
Contents
Problem
There is a polynomial with integer coefficients such that
holds 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
Solution 4 (Generating Function Bash)
Essentially we want the coefficient of in the expansion
As does not contribute to the expansion, we omit it.
Notice , so we can rewrite the generating function as
Notice to obtain a value, the
must be used, so we can reduce the problem to finding the coefficient of
in
If is expanded, it will only generate multiples of 210. To compensate this, notice
, which means we need terms with a power of a 90 to acheive what we want (300 and larger values can be shown to be impossible upon inspection). This implies that the first two brackets do not contribute and we are left with
This reduces to finding the coefficient of for the expansion
, which is
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.