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more prime thoughts


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14 replies to this topic

#1 bonanova

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Posted 24 September 2007 - 02:40 AM

Choose the best answer and say why.

Which of the even prime numbers [excluding 2] are [evenly] divisible by 5?

[1] all
[2] some
[3] none
[4] the question has no defensible answer.

p.s. Martini: I assert this is a logical not mathematical question.

Edited for "clarity". [meaning that I originally screwed up what I wanted to ask .. ]

I do not choose [4].
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#2 Writersblock

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Posted 24 September 2007 - 03:17 AM

All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.
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#3 bonanova

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Posted 24 September 2007 - 05:47 AM

All of them. There is only one, which is 2. All other even numbers are divisible by 2 and therefore not prime. The question doesn't specify that the solution must be evenly divisible by 5, therefore ALL numbers are divisible by five, including 2.


Hmmmm.. well, ok.

But I've edited the question now, hopefully asking what I had intended to ask.
See if your answer changes ...
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#4 Writersblock

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Posted 24 September 2007 - 06:14 AM

Well obviously it changes my answer now that 2 is excluded. I don't see where you are going though. 4 is the obvious answer, so not the right one for this forum. However, 1,2,&3 are excluded as those are words that only apply to some definable set. Since the set of answers you have defined is null, you can't apply "all" "some" or "none" to that set.
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#5 cpotting

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Posted 24 September 2007 - 01:01 PM

I propose that the answer could be given as [1], [2] or [3]. But the answer is definitely not [4].

The set of even prime numbers excluding 2 is an empty set. Being an empty set, it simultaneously meets the criteria that all, some and none of its members are divisible by 5.

However, since answers are being chosen and reasons for those answers are being given, then by definition, there are defensible answers to the questions. Thus [4] would be incorrect.
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#6 bonanova

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Posted 25 September 2007 - 10:37 AM

I would take issue with the defenses put forth so far.
In fairness, tho, I have changed my mind as to which answer[s] I would defend.

Today, these clues; tomorrow, my picks.

[1] Existential import.
[2] Boolean logic - as opposed to Aristotelian logic.
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#7 bonanova

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Posted 26 September 2007 - 06:45 AM

My picks:

Spoiler for ...

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#8 BoilingOil

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Posted 27 September 2007 - 12:23 AM

All even numbers [excluding 2] are evenly divisable by 2, and therefor not prime. Therefor there are NO EVEN PRIME NUMBERS if 2 is excluded. Hence no even prime numbers which would be evenly divisable by 5!


There you have it!
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#9 bonanova

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Posted 27 September 2007 - 08:01 AM

All even numbers [excluding 2] are evenly divisable by 2, and therefor not prime. Therefor there are NO EVEN PRIME NUMBERS if 2 is excluded. Hence no even prime numbers which would be evenly divisable by 5!


There you have it!

If your choice is [3], it's correct; but ...

Why is it the best answer?
Why did you pick it over [1] All even prime numbers [excluding 2] are divisible by 5?

Aren't they all divisible by 5? Show me one that is not.

[red text edited]
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#10 Benson

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Posted 28 September 2007 - 05:40 AM

that's must be 4 because divide by zero.
1. there's no even prime numbers excludeing 2. so, the solution set is none {}.
2. "evenly" is the x^2/sizeof(x). however, the size of x is zero.

It's just like customers never visit the resturant, you can't tell which one of them love the food or not.

PS. If you choose 4, it works for your spoiler, too.
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