How Many Aliens? Solve the Puzzle!

[michealsmith]

u have a certasin number of aliens in a room they each have the same number of hands and fingers ,,the total number of fingers in the room is just under 300...clue is that there is an answer with logic behind it.

question how many aliens were there?
(a specific number)

 

[davee123]

u have a certasin number of aliens in a room they each have the same number of hands and fingers ,,the total number of fingers in the room is just under 300...clue is that there is an answer with logic behind it.

question how many aliens were there?
(a specific number)

Am I missing something? It looks like there's a whole slew of possibilities:

- "just under 300" could mean... well... anything less than 300, I guess, although you'd expect at *least* 251-299. Personally, I'd assume between 290 and 299.

- "a certain number of aliens" unfortunately can be *any* number, including 0 or 1, since it does not specify something like "the aliens in the room", which would at least indicate that there were more than 1.

- "[the aliens] each have the same number of hands and fingers" again doesn't denote that the aliens necessarily have *any* fingers at all, and again, could be any non-negative integer.

- There's no specification that the aliens are the only source of fingers. It could be that there's another source in the room, although admittedly, that'd be pretty lame if so.

- There's no other information about comparison of fingers per alien to number of aliens (or other helpful tidbits). So, for example, if I assume that there's 299 fingers, then there could be: 1 alien, 13 aliens, 23 aliens, or 299 aliens.

Hence, depending on how many you assume "just under 300" to be, you can get possible answers of:

299-299:1,13,23,299
298-299:1,2,13,23,149,298,299
297-299:1,2,3,9,11,13,23,27,33,99,149,297,298,299
296-299:1,2,3,4,8,9,11,13,23,27,33,37,74,99,148,149,296,297,298,299
295-299:1,2,3,4,5,8,9,11,13,23,27,33,37,59,74,99,148,149,295,296,297,298,299
294-299:1,2,3,4,5,6,7,8,9,11,13,14,21,23,27,33,37,42,49,59,74,98,99,147,148,149,294,295,296,297,298,299
293-299:1,2,3,4,5,6,7,8,9,11,13,14,21,23,27,33,37,42,49,59,74,98,99,147,148,149,293,294,295,296,297,298,299
292-299:1,2,3,4,5,6,7,8,9,11,13,14,21,23,27,33,37,42,49,59,73,74,98,99,146,147,148,149,292,293,294,295,296,297,298,299
291-299:1,2,3,4,5,6,7,8,9,11,13,14,21,23,27,33,37,42,49,59,73,74,97,98,99,146,147,148,149,291,292,293,294,295,296,297,298,299
290-299:1,2,3,4,5,6,7,8,9,10,11,13,14,21,23,27,29,33,37,42,49,58,59,73,74,97,98,99,145,146,147,148,149,290,291,292,293,294,295,296,297,298,299

DaveE

 

[Orefa]

My assumptions are:
- a number of aliens, each with the same number of fingers assumes that there are at least two aliens.
- only one integer value is "just under 300" and it is 299.

Based on these assumptions there are 17 aliens, each with 17 fingers. No other multiple works. The number of hands in immaterial.Edit: Oops, wrong. Another possibility given these assumptions is 299 aliens with 1 finger each.

 

[daveb]

There could also be 299 aliens with 1 finger each, or 1 with 299 fingers. The problem lies in the use of the English language. Yes, aliens is plural, but how do you construct the sentence without the use of that s on the end without giving it away that the number is singular?

 

[davee123]

My assumptions are:
- a number of aliens, each with the same number of fingers assumes that there are at least two aliens.
- only one integer value is "just under 300" and it is 299.

Based on these assumptions there are 17 aliens, each with 17 fingers. No other multiple works. The number of hands in immaterial.


Edit: Oops, wrong. Another possibility given these assumptions is 299 aliens with 1 finger each.

How exactly could it be 17? I'm double checking myself again, and I get that 299 is not evenly divisible by 17. Instead, it's divisible by 1, 13, 23, and 299. Hence, there could be anyone of those numbers of aliens, but not 17.

If you're looking for perfect squares, there could be 289 fingers in the room, which would yield possibilities of 1 alien, 17 aliens, or 289 aliens. Or, if you were looking for primes, perhaps there's 293 fingers in the room, in which case there could be only 1 alien or 293 aliens.

DaveE

 

[Orefa]

How exactly could it be 17?

By virtue of the fat finger theorem which clearly states "a fat finger inadvertently pressing the 8 key instead of 9 will generate integer prime factors".

So the possibilities are indeed 1, 13, 23 and 299, which is not the answer since the OP says "a specific number".

 

[daveb]

Or, maybe the phrase "same number of hands and fingers" means each has the same number of fingers as they do hands, so each has 6 fingers on each of 6 hands, and there are 8 aliens (6x6x8 = 288 fingers).

 

[Orefa]

I am guessing the crux of the problem resides in a carefully worded detail in the original question, but the original post seems to be a re-worded version of it. It may help to see the original, unedited text of the question.

 

[Jimmy Snyder]

michealsmith, is this the problem you are thinking of?

There are a number of aliens in a room. Each alien has more than one finger on each hand. All aliens have the same number of fingers as each other. All aliens have a different number of fingers on each hand. If you knew the total number of fingers in the room you would know how many aliens were in the room. There are between 200 and 300 alien fingers in the room.

How many aliens are in the room?

Puzzle courtesy of MENSA.

http://studentaffairs.shu.edu/ata/puzz7.htm

Answer: The only way you could deduce the number of aliens by knowing the number of fingers is if the number of fingers is the square of a prime. The only squares between 200 and 300 are 15 * 15, 16 * 16 and 17 * 17, and only 17 * 17 is the square of a prime, so there are 17 aliens in the room. I don't get the part about each alien having a different number of fingers on each hand. Perhaps it is a red herring.

 

[AKG]

If there are x fingers in the room and x can be factored nontrivially x = yz (that is, x is composite), then you can't tell whether there is one alien in the room with x fingers, or y aliens with z fingers each. These are two distinct possibilities. They are distinct because 1 and y are distinct, since the factorization is non-trivial. They are possibilities because nowhere does it say that there is more than one alien (so the first possibility is a possibility), and nowhere does it say that each alien has more than one hand, so since z > 1, z is at least 2, so each alien could have just one hand with z fingers on it, and each hand would (trivially) have a different number of fingers (so the second possibility is a possibility).

Thus, x must be prime. Where x is prime, either there is one alien with x fingers or x aliens with 1 finger. An alien can't have one finger since an alien must have at least one hand (if some alien had no hands, he would have zero fingers, so all aliens would have zero fingers, so the total number of fingers could not be beween 200 and 300), and each hand must have at least two fingers. So there must be one alien, with a prime number of fingers. And there do exists primes between 200 and 300, so this is indeed a possibility, so it is the only one.

Number of aliens: 1

 

[AKG]

Answer: The only way you could deduce the number of aliens by knowing the number of fingers is if the number of fingers is the square of a prime. The only squares between 200 and 300 are 15 * 15, 16 * 16 and 17 * 17, and only 17 * 17 is the square of a prime, so there are 17 aliens in the room. I don't get the part about each alien having a different number of fingers on each hand. Perhaps it is a red herring.

How would you know there isn't one alien with 289 fingers? There's nothing to say that there must be more than one alien.

 

[Jimmy Snyder]

How would you know there isn't one alien with 289 fingers? There's nothing to say that there must be more than one alien.

You are right. My guess is that the site that I quoted also garbled the original Mensa problem. I think there is some meaning to the condition "All aliens have a different number of fingers on each hand." that is not obvious to me, but is necessary to understand the problem.

 

[kmarinas86]

You are right. My guess is that the site that I quoted also garbled the original Mensa problem. I think there is some meaning to the condition "All aliens have a different number of fingers on each hand." that is not obvious to me, but is necessary to understand the problem.

That actually shows that the number of fingers that each alien has in all "might" be odd. But that is not useful enough. There might be 2 more on one hand than on another which would not change whether it's an even or odd number.

If you knew the total number of fingers in the room you would know how many aliens were in the room.

Let:
fingers/alien=x
aliens=y
fingers=z

x*y=z

If you knew z, and if z was a mutiple of two numbers, how would you know which one is x and which one is y? Only if they are the same can you be sure. Therefore z must be a square. Also, x and y must not be seperable into different factors other than 1, or else you might able to divide x by this factor and mutiply it with y. So they must be prime. 15 and 16 are not prime.

 

[michealsmith]

yep the answer is 17 fingers ,,,,its the same mensa puzzle referred to uptop by jimmy . 1 and 289 doesn't giv an unique answer ...since
finger per alien x numbe.. alien = total finger
289 x 1 =289
1 x 289 = 289
so for those who didnt get it u were supposed to look for a squared prime just below 300 , 17 x17 =289