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There are 13 elephants aligned in a row on a circus arena. The weight of each elephant (in kilograms) is an integer number. It is known that if one adds theweightofanyelephant(except for the heaviest elephant standing at the very right) to half of the weight of its neighbor to the right one obtains 6000 kilograms. What is the weight of the lightest of all elephants?

Guest Apr 15, 2017
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10+0 Answers

 #1
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+1

One possible solution:

 

Individual elephant weights are w:

 

so the lightest is 100kg.

Alan  Apr 16, 2017
 #2
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Alan: Did you misread the question? Or am I reading it wrong?. It says:

"It is known that if one adds the weight of any elephant to half of the weight of its neighbor to the right one obtains 6000 kg" It sounds to me that the 6,000 kg is shared between TWO elephants only. No?. Of course, the question could have many answers. I think they have some crucial information missing. Note, that a newly-born baby elephant can weigh between 200 -300 kg, while a full-grown bull elephant can weigh up to 6 metric tonnes, or 6,000 kg, at least the African species.

Guest Apr 16, 2017
 #3
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+1

I probably did read the question incorrectly!  I read it as the sum of the weights of the first 12 elephants plus half of that of  their right-hand neighbours must equal 6000kg. It didn't make sense to me to assume the 6000 was shared between only two elephants.  Even so, the answer I gave isn't unique.

 

Perhaps the guest who asked the question might provide some clarification.

.

Alan  Apr 16, 2017
 #4
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+1

This question is on the current Sigma Camp’s Qualification Quiz for 2017

http://sigmacamp.org/sites/default/files/page_attachments/Qualification_Quiz_2017_Sigma.pdf

 

The camp is for children from ages 13 to 16 years. There are questions in all branches and disciplines of mathematics and science. Most of the questions seem very difficult – to me.

 

After a little thought, I realized the superlative wording of the question tends to send the solution seeker on a tangent toward a wild goose, instead of a correctly weighted elephant. To solve this though, one must think totally inside the box (instead of outside).

 

The solution:

The lightest elephant weighs 4000Kg. All the other elephants also weigh 4000Kg. 

 

I’m glad elephants are used as prop for the question. Now, I will never forget.smiley

 

Note to Blarney Banker: If you don’t understand this, I can try to dumb it down for you. However, I’d probably just be wasting my time and annoying the pịg.

GingerAle  Apr 17, 2017
edited by GingerAle  Apr 17, 2017
 #5
avatar+26005 
+2

I think you are probably right GingerAle, and 4000kg for each elephant is certainly a neat answer.  However, it still isn't unique.  The following weights also fit the bill (as the question didn't specify that the elephants were lined up in weight order, apart from the heaviest being at the far right):

 

(In the above, the left-most elephant's weight is at the top.)

 

The lightest elephant here weighs just 1952 kg.

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Alan  Apr 17, 2017
 #6
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Thank you Sir Alan!

That is so cool. The box I was thinking in was too small.

 

I can see the solution range is 4001 to 4007 for the first elephant, and this gives positive weights to all elephants with the heaviest elephant in position 13. (Unless the examiners are as pedantic as I am about the meaning of “heaviest,” my solution wouldn’t be correct).   

 

I still don’t know how to solve this with a general mathematical function where there can be more (or less) elephants, and a high (or lower) weight difference. The recursive function eludes me and makes me recursively curse .angrysmiley

GingerAle  Apr 17, 2017
 #9
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.50x +4001 = 6,000, solve for x

x =3,998 Kg. Replace 4001 above with 3,998,

x =4,004 Kg. Replace 3,998 above with 4,004.......and so on.

x=3,992  Kg.

x=4,016  Kg

x=3,968 Kg

x=4,064 Kg

x=3,872 Kg

x=4,256 Kg

x=3,488 Kg

x=5,024 Kg

x=1,952 Kg

x=8,096 Kg. These are Alan's numbers. However, the last elephant's weight can be any number above 5,024 Kg, since it is the heaviest and is excluded from the rule. 

There are only two unique solutions. That is, with the initial weight of either 4,000 or 4,001 Kg. The initial weight of 3,999 kg, gives the weight of the 12th elephant as 6,048 Kg. The initial weight of 4,002 Kg gives the weight of the 11th elephant as 6,048 Kg.

Guest Apr 18, 2017
 #10
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The last elephants weight is not arbitrary as half of it, when added to the 12th elephant's weight, must be 6000kg.

.

Alan  Apr 18, 2017
 #11
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Guest #9, what is your Blŏŏdy Point? Your formula is odd.    .50x +4001 = 6,000, solve for x.  Where do you get the 4001 from --aside from Alan’s starting point?  You would not have this as the starting point for a solution without some kind of calculation.

 

You could do this x+ 0.5x = 6000 | x=4000 this sets Xj and Xj+1 to 4000, which is what I did.

 

So, what is the point of your brain-dead presentation? I am missing something? You restate Alan’s sequence in regurgitated form; add some blatantly wrong blarney to the end, then say “There are only two unique solutions. That is, with the initial weight of either 4,000 or 4,001 Kg.”

 

Well, No freaking duh!  

 

I agree there are only two solutions to this: 4000 and 4001.  (A circular reference error in my spreadsheet affected the values; giving false results indicating 4002-4007 were also valid.) You didn’t bother mentioning this; rather, you meander all over Hades, while adding more blatantly irrelevant blarney. Maybe all these elephants leaving piles of dung on this post affected your thinking process.

 

You should join with the Blarney Banker and call yourselves Blarney and Blarney. You two can make Blarney the old-fashioned way –by regurgitating it.

 

(Note to Chimp Henry: Order 50 more cases of extra-strength gnome repellent).

 

 Note to Blarney Banker: Yes, I have the gene for curse and cur. I inherited it from an ancestor we have in common. We genetically enhanced chimps learned how to suppress this gene, so it’s not active, except when blarney bags are near.  It is active in you though, just as it is in all human-pịg hybrids, procreated by miscegenation. (See what I mean by “annoying the pịg.”)

 

--------------

 

On a brighter note, though shrouded in a fog of blarney, I found the original source for this question. It appears the examiners altered the parameters of the question to allow for an additional  or different answer –one that isn’t trivial.

 

Mathematical Circles: (Russian Experience) (1996) (ISBN: 0821890603)

By Sergeĭ Aleksandrovich Genkin, Dmitriĭ Vladimirovich Fomin

 

Here is the question:

 

123. Twenty-five elephants stand in a row in a circus arena, and each of them weighs an integer number of kilograms. It is known that if you add the weight of any of them (except the rightmost one) and half the weight of its neighbor to the right, then the result is 6 tons. Find the weights of the elephants.

 

There is no answer or hint for this question presented in the book.

In this question, it only asks for the weight of the elephants, there is no reference to the heaviest, not that it matters, it there is only one integer solution: 4000Kg. 

 

It is interesting that if the question presented in integer number of grams then there could be as many as 23 before any elephant exceeds 6 (metric) tons.

 

.

GingerAle  Apr 19, 2017
 #12
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0

You are a pile of ape dung !!

Guest Apr 22, 2017

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