Alan

Perhaps the example below will help answer your questions:

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I should add that we could still use the substitution method to find B above.  Having found A, we just use another value for x (say, x = 1) to find B.  So this probably doesn't address the issue of a method that doesn't always work!

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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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a1 = 1

a2 = 9

a3 = 5*1 - 9 = -4

a4 = 5*9 - (-4) = 49

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Or use the law of cosines directly:  a² = b² + c² -2b.c.cos(A)

GD² = 17² + 15² -2.17.15.cos(81°) = 434.218 yd²

GD = 20.8 yd or GD = 21 yd to nearest yard.

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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.

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As follows:

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The GCF of 18 and 20 is 2, so the maximum size square tile is 2m x 2m.

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In the following h is the time increment:

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One possible solution:

Individual elephant weights are w:

so the lightest is 100kg.

Calculate the digital root:  

6*(9+7+5+3+1) = 150

1+5+0 = 6  

6 is divisible by 3 hence the original number is divisible by 3, so it is composite, not prime.

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