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Incubator Q&A Starting with 2014 "+" signs and 2015 "−" signs, you delete signs until one remains. What’s left?

Withdrawn -- this is incorrect. answer You can produce whichever outcome you want. method 1 Start by collapsing most of this away. Delete 2014 "-" in pairs. We now have 3021 "+" and 1 "-". ...

posted 2mo ago by Monica Cellio‭ · edited 2mo ago by Monica Cellio‭

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#2: Post edited by

Monica Cellio‭ · 2026-08-02T14:27:48Z (about 2 months ago)

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<details><summary>answer</summary>
You can produce whichever outcome you want.
</details>
<details><summary>method 1</summary>
Start by collapsing most of this away.
Delete 2014 "-" in pairs. We now have 3021 "+" and 1 "-".
Delete 1 "-" and 1 "+". We now have 3020 "+" and 2 "-".
Delete the two "-". We now have 3021 "+" and 0 "-".
Everything is now "+" and deleting two of the same results in a "+". There's no way for "-" to re-enter, so if we keep doing this, we'll eventually end up with 1 "+".
</details>
<details><summary>method 2</summary>
What happens if we start by deleting (pairwise) 2014 of each? We end up with 0 "+" and 2015 "-".
We delete 2014 "-" pairwise, ending up with 1007 "+" and 1 "-".
We can obviously delete the "-" and one "+" and reduce to the previous method. Instead, let's collapse the "+"s. "++" reduces to "+" and we can keep doing that until we're down to 1 "+" and 1 "-". Deleting them results in a "-".

Withdrawn -- this is incorrect.
<details><summary>answer</summary>
You can produce whichever outcome you want.
</details>
<details><summary>method 1</summary>
Start by collapsing most of this away.
Delete 2014 "-" in pairs. We now have 3021 "+" and 1 "-".
Delete 1 "-" and 1 "+". We now have 3020 "+" and 2 "-".
Delete the two "-". We now have 3021 "+" and 0 "-".
Everything is now "+" and deleting two of the same results in a "+". There's no way for "-" to re-enter, so if we keep doing this, we'll eventually end up with 1 "+".
</details>
<details><summary>method 2</summary>
What happens if we start by deleting (pairwise) 2014 of each? We end up with 0 "+" and 2015 "-".
We delete 2014 "-" pairwise, ending up with 1007 "+" and 1 "-".
We can obviously delete the "-" and one "+" and reduce to the previous method. Instead, let's collapse the "+"s. "++" reduces to "+" and we can keep doing that until we're down to 1 "+" and 1 "-". Deleting them results in a "-".

#1: Initial revision by

Monica Cellio‭ · 2026-08-02T04:56:57Z (about 2 months ago)

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<details><summary>answer</summary>

You can produce whichever outcome you want.

</details>

<details><summary>method 1</summary>

Start by collapsing most of this away.

Delete 2014 "-" in pairs.  We now have 3021 "+" and 1 "-".

Delete 1 "-" and 1 "+".  We now have 3020 "+" and 2 "-".

Delete the two "-".  We now have 3021 "+" and 0 "-".

Everything is now "+" and deleting two of the same results in a "+".  There's no way for "-" to re-enter, so if we keep doing this, we'll eventually end up with 1 "+".

</details>

<details><summary>method 2</summary>

What happens if we start by deleting (pairwise) 2014 of each?  We end up with 0 "+" and 2015 "-".

We delete 2014 "-" pairwise, ending up with 1007 "+" and 1 "-".

We can obviously delete the "-" and one "+" and reduce to the previous method.  Instead, let's collapse the "+"s.  "++" reduces to "+" and we can keep doing that until we're down to 1 "+" and 1 "-".  Deleting them results in a "-".

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