Japanese | Can you solve this? | Math Olympiad Problem

[@СергейИванчиков-р8д]

Good, very good

[@MasterTMathsClass]

Thank you very much

[@laobernadus5743]

K = 2

[@payoo_2674]

Another method
k^6 = 64
k^6 = 2^6
k^6 = 2^6 * 1 1 = e^(2nπi) n∈Z
(k^6)^(1/6) = (2^6 * e^(2nπi))^(1/6)
k = 2 * e^(nπi/3) n = 0, 1, 2, ..., 5
for n=0: k₁ = 2*e^(0*πi/3) = 2*(cos(0)+i*sin(0)) = 2*(1+i*0) = 2
for n=1: k₂ = 2*e^(1*πi/3) = 2*(cos(π/3)+i*sin(π/3)) = 2*(1/2+i*√3/2) = 1+√3i
for n=2: k₃ = 2*e^(2*πi/3) = 2*(cos(2π/3)+i*sin(2π/3)) = 2*(-1/2+i*√3/2) = -1+√3i
for n=3: k₄ = 2*e^(3*πi/3) = 2*(cos(π)+i*sin(π)) = 2*(-1+i*0) = -2
for n=4: k₅ = 2*e^(4*πi/3) = 2*(cos(4π/3)+i*sin(4π/3)) = 2*(-1/2-i*√3/2) = -1-√3i
for n=5: k₆ = 2*e^(5*πi/3) = 2*(cos(5π/3)+i*sin(5π/3)) = 2*(1/2-i*√3/2) = 1-√3i

[@MasterTMathsClass]

That's good 🎉

[@yalcinhamdioglu5916]

2

[@burachatkuadkaew7344]

ขอขอบพระคุณ อาจารย์ ฟูกิจ และ อาจารย์ วรศักดิ์ มากครับ ที่ช่วยส่งตัวอย่างการแก้สมการ logarithm และ exponential ให้ผมดูเป็นตัวอย่าง
ผมสัญญาว่าผมจะตั้งใจศึกษาและจบมาเป็นวิศวกรที่ดีให้ได้ด้วยครับ

[@MasterTMathsClass]

✌️

[@FikretKılıç-u4v]

4 te karşılar.

[@MasterTMathsClass]

✌️

[@VincenzoDiMattia-yj4or]

2 alla 6 a. Che c' è di difficile?

[@Redbravo001]

I also solved it just by looking, but I appreciate how he calculated and showed the line of thought that allows us to figure it out just by looking.

[@dwiyantowibisana4928]

K=2