# Math Contest #27 Results and Solution

in #puzzle2 years ago

## Solution

The problem of this contest was to proof an equivalence:

I also just noticed that this inequation only holds for uneven, positive values of n(which is probably the reason why I had no entries this week).
First of all a small explanation why this is even true:
if x is bigger than 0: Then 1/x^n is only < 1 if x > 1.
if x is smaller 0: Then 1/x^n is only < -1 if x > -1
These two cases connect both inequations.
In this case you can rearrange the leftern inequation to:
`1 < x^(n+1) if x > 0``x > 1`
`1 > x^(n+1) if x < 0``1 > ±x``-1 < ±x``-1 < x`
The ± comes from taking an even root.
So if x > 0 than x > 1 is the only solution and if x < 0, x > -1 is the only solution.
Putting the 2 together gets the rightern side of the equivalence:
`x > 1` or `-1 < x < 0`
And the way back to prove it actually is an equivalence:
`x > 1``x^(n+1) > 1``x > 1/x^n`
`-1 < x < 0``1 < x^(n+1)``x > 1/x^n`

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### List of participants with their entries:

Namesolution foundcomment
@tonimontananonefound one of the errors I made → you get the reward

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## Winner draw:

Not necessary since only 1 participant:
Congratulations @tonimontana , you won 2 SBI!
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### This contest will continue as a 2-week contest as it's getting harder to find a unique problem and I have more stress lately.

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