Math Contest #19 Results and Solution
Solution
The problem of this contest was to solve an infinite sum:
There are in general 3 different intuitive ways to find a solution to this:
1-1+1-1+1-1+… = (1-1)+(1-1)+(1-1)+… = 0+0+0+… = 0
1-1+1-1+1-1+… = 1 + (-1+1)+(-1+1)+(-1+1)+… = 1+0+0+0+… = 1
1-1+1-1+1-1+… = x
→1-(1-1+1-1+1-1+…) = 1-x
→1-1+1-1+1-1+… = 1-x
→x = 1-x
→2x = 1
→x = ½
But there are also infinitely many more ways to find a solution that you can explain. you could even reorder the sum(by taking some 1's from infinity and placing them between them at some point sooner in the sequence. This is possible, because before and after that operation the total number of 1's and -1's is still the same as before(∞)) to:
1+1-1+1+1-1+1+1-1+… = (1+1-1)+(1+1-1)+… = 1+1+1+… = ∞
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List of participants with their entries:
Name | solutions found | comment |
---|---|---|
@sparkesy43 | 0 or 1 | |
@crokkon | 0 or ½ or 1 | |
@golddeck | 0 or 1 | |
@rxhector | -∞ | That's not really intuitive, but still you can get there as shown above for +∞. |
@fullcoverbetting | 0 | |
@tonimontana | ½ | But how can the sum of integers be no integer :P |
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Winner draw:
Congratulations @rxhector you won 1 SBI!
Of course everyone also got 10 STEM for participating(check your steem engine)!
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That's what one would think would happen, but the sum here is infinite. So you can repeat the pattern that of (1+1-1) for ever without having a zone of (1-1-1).
It may fell like some -1's are lost during this process, but if you would(and could) count them you would still get the same number of 1's and -1's: infinitely many!
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