Back to the 1980's
What is the solution to:
1 9 8 0 + 1 9 8 0 + 1 9 8 0 + 1 9 8 0 . . . ∞
Details and assumptions: Give your answer as the principle square root.
The answer is 45.
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1 solution
why does this always work for any n?
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there isn't even a variable n
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... where n replaces 1980.
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@Jack Lam – This does not work for any variable "n". If you set the entire equation equal to x, then square both sides, and substitute x for the remaining nested radicals, you get an equation that looks like x^2 - x - n. So from this equation, you can see that for the solution to be an integer, the above polynomial must be able to be factored into the form of (x+b)(x-(b+1)), with b being the lesser of the two factors. Otherwise, for an n that cannot be factored in this form, there are no integer solutions. Like for n = 7, you will end up with a solution of ~ 3.19. As to why this works for any n, that can be seen in the sense that this can be represented as a quadratic. As long as the discriminant is positive, there will be real solutions/roots to this problem.
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@Seth Lovelace – And when I said this does not work for any n, I was referring to the solution being an integer.
What if someone enters the answer as -45? Will it be considered as correct as answer?
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Hmmmmm, I will change the wording to say principle square root. Although I believe it would not be -45, but -44.
Peace be upon you,
let x = ( 1980 + x) ^ (0.5) ,
by squaring both sides,
x^2 = 1980 + x
x^2 - x - 1980 = 0
(x + 44 ) ( x - 45 ) = 0
x = 45 , x = - 44,
therefore, x= 45 that is matched with it's sum to infinity...