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I was thinking about creating some riddle for TED-ED and somehow I came up some math riddle something like that:

"A group of seven creatures weights together 1000 kg, the largest one and second smallest one weights same as one larger than smallest one and second one smaller than largest one, 3 largest ones weights same as 4 smallest ones, some one of the 3 largest ones is half heavier than 4th one, who is half heavier than some one of the 3 smallest ones.

How heavy each creatures weights? Which two of them don't have identical weights. All seven creatures have different weights."

edit: Should I fix that first hint with something large creature named "Abe" (first or second largest) + small one named "Bibb" (first or second smallest) = second or third largest <"Abe" + second or third smallest >"Bibb"
And third hint as one of the smaller ones are quarter lighter than some creature called "Seymour", but one the larger one is half heavier than "Seymour", does it makes bit different?

Sorry about that mistake for that first hint

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  • $\begingroup$ In "some one of the 3 largest ones is half heavier than 4th one," define "half heavier." Do you mean 150% as heavy? $\endgroup$ – ZanyG Jan 27 at 8:43
  • $\begingroup$ @ZanyG well yeah, something one of three largest one weights 50% more than 4th one (or middle one). You might know that other math riddle something like "two guys weights 250 kg together and one guy weights half heavier than other guy" like one guy is 3⁄5 of total weight and other guy 2⁄5, so answer will be 150 kg and 100 kg. $\endgroup$ – Leonard Greenland Jan 27 at 9:06
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    $\begingroup$ Are you asking how to create a puzzle that works? Because I don't think your puzzle as stated has a solution. $\endgroup$ – JS1 Jan 27 at 11:05
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    $\begingroup$ @GeorgeMenoutis Sorry, meant the question for the OP. I've nothing new to post. If all weights are different, I come to a contradiction. If some can be the same, then my answer from before is still valid. $\endgroup$ – hexomino Jan 27 at 17:59
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    $\begingroup$ I can find a solution to the variant "a smaller one is 1/4 lighter than S, and a larger one is 1/2 heavier than S", but the solution has repeated weights, which contradicts "all seven must have different weights". Here is my solution for those interested: 100, 133 1/3, 133 1/3, 133 1/3, 133 1/3, 166 2/3, 200, where 100 is 1/4 lighter than 133 1/3, 200 is 1/2 heavier than 133 1/3. The variant "larger is 1/2 heavier than #4 which is 1/2 heavier than smaller" has no solution as far as I can tell. $\endgroup$ – JS1 Jan 27 at 19:07
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There seems to be just one solution, if the hints are interpreted in a particular way (ignoring the question edit by George Menoutis).

Here is the solution for the weights (in kg, from largest to smallest) of the creatures

$$ 187.5, \,187.5, \,125, \,125, \,125, \,125, \,125 $$

Reasoning

As others have done let's order the weights $A \geq B \geq \ldots \geq G$

"the largest one and second smallest one weights same as smallest one and second largest one"

This means that $A+F = B+G$ but, given the inequalities above this can only be true if $A=B$ and $F=G$.

"3 largest ones weights same as 4 smallest ones"

This means that $A+B+C = D+E+F+G$ and together with the above this means that $2A+C = D+E+2F$

"some one of the 3 largest ones is half heavier than 4th one"

This means that either $A = \frac{3}{2}D$ or $C = \frac{3}{2}D$. If the latter, then $2A + C = 2A + \frac{3}{2}D \geq 3D + \frac{3}{2} > 4D \geq D + E + 2F$ contradicting above so it must be that $A = \frac{3}{2}D$ and the above equation becomes $$ 3D + C = D + E + 2F \Rightarrow 2D + C = E+2F$$ but this can only satisfy the ordering constraint if $$C=D=E=F$$

"who is half heavier than some one of the 3 smallest ones"

The natural interpretation of this phrase is that $D$ is half heavier than one of the three smallest ones, but this cannot be true from what we have discovered above. So it must be that the creature that is half heavier than $D$ is also half heavier than one of the three smallest ones. This is consistent with what we have found

Finally, the total weight is 1000kg

$$\Rightarrow A+B+C+D+E+F+G = 8G = 1000$$ $$\Rightarrow G = 125$$ and $$A=B=\frac{3}{2}G\,\,\,\,,\,\,\,\, C=D=E=F=G $$

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  • $\begingroup$ Nice solution hex. I edited the question because I figured that's what OP intended to say from another comment, but I don't want to mislead anyone. If you have any recommendations, please say so, you are more experienced here :) $\endgroup$ – George Menoutis Jan 27 at 11:09
  • $\begingroup$ @GeorgeMenoutis My interpretation of "Which two of them don't have identical weights" would be that there are two of them which are of different weight to the other five (whose weights are equal). Although, admittedly, it's not a very clear statement. $\endgroup$ – hexomino Jan 27 at 11:13
  • $\begingroup$ good solution/interpretation of what the rhyme actually meant. $\endgroup$ – Nobody Jan 27 at 11:23
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Your riddle was -

"A group of seven creatures weights together 1000 kg, the largest one and second smallest one weights same as smallest one and second largest one, 3 largest ones weights same as 4 smallest ones, some one of the 3 largest ones is half heavier than 4th one, who is half heavier than some one of the 3 smallest ones."

so lets write out our equations for this problem (A -> G in order of weight)

A + B + C + D + E + F + G = 1000

A + F = B + G

A + B + C + D = E + F + G

E | F | G = D *1.5

A | B | C = D/1.5

Now the first major issue I see is that you have 7 unknowns and I am only able to see 5 equations (given all of them have different weights), frankly I cannot see a single unique solution for this, unless there is some more information that is missing.

IF A = B and F = G are inferred from point 2 (since we had them ordered by weight)

New equations -

2A + C + D + E + 2F = 1000

2A + C + D = E + 2F

A | C = D/1.5

E | F = D *1.5

Now we still have an imbalance as we have 5 unknowns and 4 equations.

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