Відео

Assuming that 'a' is real, a^5 = 1 implies that a = 1. but it had been shown that 'a' not equal 1. by inspection, the solution for real 'a' is approximately slightly less than -1 (somewhere between -2 and -1).

m=3. Multiply both sides by 5. Then 5*5^m -5^m=5*(25*4) => 4*5^m=4*5^3 => 5^m=5^3 => m=3.

X=a+b+c

You might be right, but the math question becomes easy after we read the answer.

a+b+c is quite obvious is it not? A better thing would have been to show a class of such problems where you deal with algebraic equations that have a cyclical invariant like this one

Thank you sir. You earned a subscriber. Don't forget me once you succeed on UA-cam. Wish you all the best for future sir.

Great solution! From the given condition, I derived the power reduction formula x^2 = (1-x)/3 which I plugged in in the term to be evaluated at any position where an x^2-term occurred. Eventually, I came up with the same result.

2a^20 1a^10 a^5^5a^2^3^2^3 a^2^1^1^3 a^2^3 (a ➖ 3a+2). 2a^8060 2a^8^6 2a^2^3^2^3 1a^1^1^1^2^3a^2^3(a ➖ 3a+2).

1+a+a^2+a^3+a^4=0 | recognize the general form (geometric series): sum_(k=0)^(n) a^k=1+a+a^2+...+a^n=(a^(n+1)-1)/(a-1) so multiply both sides: (a-1)(1+a+a^2+a^3+a^4)=0 thus: a^5-1=0 So a^5=1 Thus a^2020+a^2010+1= (a^5)^(404)+(a^5)^(402)+1= 1^404+1^402+1= 1+1+1= 3 <-- solution

I have another solution Action 1: a^4 + a^3 + a^2 + a + 1 = 0 a^2 * (a+1) + a + 1 = -(a^4) (a^2 + 1) * (a+1) = -(a^4). Action 2: a^4 + a^3 + a^2 + a + 1 = 0 a^2 * (a^2 + 1) + a * (a^2 + 1) = -1 a * (a+1) * (a^2 + 1) = -1 (a+1) * (a^2 + 1) = -(1/a). And clearly 'a' can't equal zero. So we put Action 1 and Action 2 together... Action 3: a^4 = 1/a a^5 = 1. and so on, then it's the same way you did. (a^5)^402 + (a^5)^404 + 1 = 3.

is it 3?

You can also solve it by factoring out a 5^m so you would get 5^m(1-(5^-1))=100 5^m(1-(1/5))=100 5^m(4/5)=100 5^m=5•100/4 5^m=125 5^m=5^m m=3

Thank you sir.

Based on the fact that dofference between two neighbor power is also power of 2, m should lower then 12. There are not much combinations exist

Before watching the video, I decided to try to solve it so I found the closest power of 2 to 1984 (2048) which is 2^11 and then took 2048-1984 (64) which is 2^6 so m is 11 and n is 6. Let’s see if I’m right

🤪 P r o m o s m

61