Dude, this is stupid

When m^3 is equal to 3^3, then m=3 there's no need to take quadratic route

Please stop calling trivial problems "Math Olympiad"!

m=-3

Too hard. If equation is m+n=k sqrt{mn}, dividing by n, m/n+1=k sqrt{m/n}. Letting m/n=r, this is r+1=k sqrt{r}. Squaring, r^2+(2-k^2)r+1=0. Solving this gives r=(k^2-2 +/- k\sqrt{k^2-4})/2=17 +/- 12 sqrt{2} for k=6.

M equals minus 3 in 2 seconds flat........... Even a thick twerp like me got that..

m = -3

That was a good explanation, but to make it better you should test your result.

Amazing 😍

U are noob

this is an extremely simple problem that anyone in 7th grade can do. What Olympiad is it from? if you are claiming it is a math olympiad problem you should provide the reference to the particular olympiad.

I have one

Hello from Kazakhstan. I study in Russia and in 10th grade we solved such examples to familiarize ourselves with the topic “Exponential Equations”. You're cool for explaining this to people, thank you for that. Good luck!

16

15

❤❤❤❤

20 rotation

Just an easy solution using simple algebra. m⁴ + 64 = 0 Let m² = n n² + 64 = 0 (n + 8i)(n - 8i) = 0 (m² + 8i)(m² - 8i) = 0 (m + 2isqrt(2i))(m - 2isqrt(2i))(m + 2sqrt(2i))(m - 2sqrt(2i)) = 0 From onwards it is easy to equate the things. Also, BRPR has a great video on the squared root of i which we would plug in and get the answer.

Draw a radius 2√2 circle centered on the origin of the complex plane. Mark points on the circle at 45°, 135°, 225°, and 315°. Those are the four values of m.

👍👍🎉💞

Thanks. I nailed it.

Wow! 👍👍🎉🎉💞💞

Fourteen rotations☝️

easy

x=2 by inspection.

Nice trick if you recognize that (√3-√2)=1/(√3+√2). Substitute y=(√3+√2)^x into the given equation: y+1/y-10=0 and rearrange to y^2-10y+1=0, which has roots y=(10±4√6)/2=5±2√6. Thus x=ln(5±2√6)/ln(√3+√2)=±2

2^98

16. Can u pin me

19

Excellent equation

Ποτέ δεν πάμε με συνεπαγωγή απορώ Μαθηματικός είσθαι τόσο απλό για 11 χρονών είναι λυπάμαι

I think you're a bit underrated

I took another approach to figure out the solutions. So I have m^6 = (m - 2)^6 (*) I can take the module: |m^6| = |(m - 2)^6| The power and module are interchangeable: |m|^6 = |m - 2|^6 Since modules are real positive values, I can apply the sixth root: |m| = |m - 2| Now let m = x + iy with x and y real numbers, we get |x + iy| = |x - 2 + iy| By squaring both sides, we get x² + y² = (x - 2)² + y² We can develop the right hand side and substract x² + y², we get 0 = -4x + 4, which gives us x = 1. Now we can focus on y. We know from (*) that (1 + iy)^6 = (-1 + iy)^6 I'm making myself a Pascal triangle because I don't know it until 6 by heart: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 Now I can develop the equation: 1 + 6iy - 15y² - 20iy^3 + 15y^4 + 6iy^5 - y^6 = 1 - 6iy - 15y² + 20iy^3 + 15y^4 - 6iy^5 - y^6 I can substract both sides by the right hand side and divide by 4i, I get: 3y - 20y^3 + 3y^5 = 0 I can factorize by y: y(3 - 20y² + 3y^4) = 0. Now I have two possibilities, either y = 0, or 3 - 10y² + 3y^4 = 0. The first case is trivial. For the second one, let z = y². We get: 3 - 10z + 3z² = 0. The quadratic formula tells us that z = (10 ± sqrt(100 - 36))/6 = (10 ± sqrt(64))/6 = (10 ± 8)/6. Then we have two solutions, either z = 3 or z = 1/3. Therefore, since y² = z, we have y = sqrt(3), y = - sqrt(3), y = sqrt(3)/3 or y = -sqrt(3)/3. Since the difference m^6 - (m - 2)^6 is of degree 5, we have at most 5 solutions, and I found 5 different ones, then they are the exact 5 complex solutions of the equation. The equation m^6 = (m - 2)^6 has five solutions: 1, 1 + i sqrt(3), 1 - i sqrt(3), 1 + i sqrt(3)/3 and 1 - i sqrt(3)/3.

Please solve m^25 = (m-2)^25

I love it❤ ❤❤❤❤❤❤🎉😂😮😅😊

Team 16👇

i did a guess and check after thinking, "huh, having a number raised to the sixth power be equal to the same number minus 2 raised to the sixth power is kinda weird... Where on the number line would something like that happen?" Then I put on my thinking cap and realized the number 1 fits the bill! Your work is a lot more through tho, haha! (Plus, I missed 2 solutions, doh.)

I have my channel @PakMaths

q legal😊

Brother from which city are you from

Super 🎉

Super

😮😮😮😮😮😮😮😮😮

How is n not just 1? 1^x for any value of x will always be 1, so 1^x (which is just 1) - 1 = 0

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27

Bad problem not well stated😢

8?

-2

what grade is this olympiad? This is a task for pupils in 8th class in Ukraine, and not Olympiad