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 > 0x > 1
1 > x^(n+1) if x < 01 > ±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 > 1x^(n+1) > 1x > 1/x^n
-1 < x < 01 < 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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The next contest starts soon. Don't miss it!

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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