Difference between revisions of "2022 AIME II Problems/Problem 13"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | == Solution 1 Modification == | ||
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+ | Note that <math>2022 = 210\cdot 9 +132</math>. Since the only way to express <math>132</math> in terms of <math>105</math>, <math>70</math>, <math>42</math>, or <math>30</math> is <math>135 = 30+30+30+42</math>, we are essentially just counting the number of ways to express <math>210*9</math> in terms of these numbers. This is simply (by sticks and stones) <cmath>\boxed{\textbf{(220)}}</cmath> | ||
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+ | ~Bigbrain123 | ||
==Solution 2== | ==Solution 2== |
Revision as of 21:42, 1 June 2022
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 1 Modification
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. This is simply (by sticks and stones)
~Bigbrain123
Solution 2
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.