Solving for the Largest Positive Integer n

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Homework Statement



Find the largest positive integer n such that n^3 + 100 is divisible by n + 10.

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The Attempt at a Solution



The hint I've been given is to use (mod n + 10) to get rid of the n.

but i don't quite see how it would work :S

all my attempts have gotten nowhere, lol.

a little prod in the right direction would be nice :)

cheers
 
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My method would be to just use any polynomial dividing technique to find the remainder of (n^3+100)/(n+10). If n^3+100 is divisible by n+10, then the remainder will have to be an integer.
 
right, i think i may have a solution to this problem. can someone please check it for me :) thanks.


say that n^3 + 100 is divisible by n + 10.

then n^3 + 100 = 0 (mod n + 10)

n = -10 (mod n + 10)

so (-10)^3 + 100 = 0 (mod n + 10)

... 0 = 900 (mod n + 10)

we want to maximise n, and because the above line essentially means that 900 is an integer multiple of (n + 10), the maximum n would be when 900 = n +10

so n = 890...


is this reasoning correct?

thanks:)
 
Looks good to me:approve: I got the same thing using polynomial division.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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