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
Expand the pi notation for the problem: ∏_(k=0)^6▒〖(ω^3k 〗+ω^k+1)=(ω^(3*0)+ω^0+1)(ω^(3*1)+ω^1+1)(ω^(3*2)+ω^2+1)(ω^(3*3)+ω^3+1)(ω^(3*4)+ω^4+1)(ω^(3*5)+ω^5+1)(ω^(3*6)+ω^6+1)=(ω^0+ω^0+1)(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^9+ω^3+1)(ω^12+ω^4+1)(ω^15+ω^5+1)(ω^18+ω^6 +1) =3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^9+ω^3+1)(ω^12+ω^4+1)(ω^15+ω^5+1)(ω^18+ω^6 +1) (Eq.1)
Notice that ω= cos〖(2π/7)+〖i sin〗〖(2π/7)〗 〗
cos〖(x)+〖i sin〗〖(x)=cis(x)=e^ix 〗 〗 ω=cis(2π/7)= e^(i*2π/7) Notice that 1,ω^1,ω^2,ω^3,ω^4,ω^5,ω^6 are the 7th roots of unity and the roots for ω^7-1=(ω-1)(ω^1+ω^2+ω^3+ω^4+ω^5+ω^6 )=0.
So ω^7=1 and 〖1+ω〗^1+ω^2+ω^3+ω^4+ω^5+ω^6=0
ω^(7x+y)=〖ω^7x*ω〗^y=(〖ω^7)〗^x*ω^y=1^x* ω^y=1* ω^y= ω^y where x,y ∈ Z
In Eq.1, we can simplify all w^x terms where x >= 7. ω^9=ω^(7+2)= ω^2 ω^12=ω^(7+5)= ω^5 ω^15=ω^(7*2+1)= ω^1 ω^18=ω^(7*2+4)= ω^4 Eq.1 simplifies to: 3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^2+ω^3+1)(ω^5+ω^4+1)(ω^1+ω^5+1)(ω^4+ω^6 +1) = 3(ω^3+ω^1+1) (ω^6+ω^2+1)(ω^3+ω^2+1)(ω^5+ω^4+1)(ω^5+ω^1+1)(ω^6+ω^4 +1) (Eq.2)
Rearrange the polynomials in Eq.2 to get: 3(ω^3+ω^1+1) (ω^3+ω^2+1)(ω^5+ω^4+1)(ω^5+ω^1+1)(ω^6+ω^2+1)(ω^6+ω^4 +1)
Now multiply each pair of polynomials: (ω^3+ω^1+1) (ω^3+ω^2+1)=ω^6+ω^5+ω^3+ω^4+ω^3+ω^1+ω^3+ω^2+1= ω^6+ω^5+ω^4+3ω^3+ω^2+ω^1+1=(ω^6+ω^5+ω^4+ω^3+ω^2+ω^1+1)+2ω^3=0+ 2ω^3=2ω^3 (Eq.3)
(ω^5+ω^4+1)(ω^5+ω^1+1)= ω^10+ω^6+ω^5+ω^9+ω^5+ω^4+ω^5+ω^1+1=ω^10+ω^9+ω^6+〖3ω〗^5+ω^4+ω^1+1=ω^3+ω^2+ω^6+〖3ω〗^5+ω^4+ω^1+1=〖(ω〗^6+ω^5+ω^4+ω^3+ω^2+ω^1+1)+2ω^5=0+ 2ω^5= 2ω^5 (Eq.4)
(ω^6+ω^2+1)(ω^6+ω^4 +1)= ω^12+ ω^10+ ω^6+ ω^8+ ω^6+ ω^2+ ω^6+ ω^4+ 1
= ω^12+ ω^10+ ω^8+ 〖3ω〗^6+ ω^4+ ω^2+ 1
= ω^5+ ω^3+ ω^1+ 〖3ω〗^6+ ω^4+ ω^2+ 1 = 〖(ω〗^6+ ω^5+ ω^4+ ω^3+ ω^2+ ω^1+ 1)+2ω^6=0+2ω^6= 2ω^6 (Eq.5)
Substituting Eq.3-5 into Eq.2, our final result is: 3(2ω^3 )(2ω^5 )(2ω^6 )=3*2*2*2*ω^(3+5+6)=24*ω^14=24*〖〖(w〗^7)〗^2= 24*1^2= 24*1= 24. Answer: 024
-TylerJdnkri
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 |
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