2023 AIME II Problems/Problem 8
Contents
Problem
Let where Find the value of the product
Solution 1
For any , we have, The second and the fifth equalities follow from the property that .
Therefore,
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Moduli)
Because the answer must be a positive integer, it is just equal to the modulus of the product. Define .
Then, our product is equal to
, and we may observe that and are conjugates for any , meaning that their magnitudes are the same. Thus, our product is
Let us simplify the first term. Expanding, we obtain
Rearranging and cancelling, we obtain
By the cosine subtraction formula, we have .
Thus, the first term is equivalent to
Similarly, the second and third terms are, respectively,
Next, we have . This is because
Therefore, the first term is simply . We have , so therefore the second and third terms can both also be simplified to . Thus, our answer is simply
~mathboy100
Solution 3 (Inspecting the exponents of powers of )
We write out the product in terms of :
Grouping the terms in the following way exploits the fact that for an integer , when multiplying out two adjacent products from left to right:
When multiplying two numbers with like bases, we add the exponents. We can now rewrite the exponents of each product (two at a time, where is treated as the identity) as a series of arrays:
Note that . When raising to a power, the numerator of the fraction is times whatever power is raised to, multiplied by . Since the period of is we multiply each array by then reduce each entry as each entry in an array represents an exponent which is raised to.
To obtain the correct exponents, we seperately add each element of the lower row to one element of the top row.
Therefore (after reducing again), we get the following sets:
Raising to the power of each element in every set then multiplying over and yields
as these sets are all identical.
Summing as a geometric series,
Therefore,
and
-Benedict T (countmath1)
Solution 4
The product can be factored into ,
where are the roots of the polynomial .
This is then because and share the same roots.
To find ,
Notice that . Since r satisfies ,
Substituting, you are left with . This is , and after repeatedly substituting you are left with .
So now the problem is reduced to finding , and vietas gives you the result of -resources
Video Solution by The Power of Logic
https://youtu.be/o6w9t43GpJs?si=aoe-uM3m5AIwpz_H
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.