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A strange tradition is followed in an orthodox and undeveloped village. The chief of the village collects taxes from all the males of the village yearly. But it is the method of taking taxes that is interesting.

The taxes paid in the form of grains and every male should pay equal pounds corresponding to his age. In simpler terms, a man aged 10 years will have to pay 10 pounds of grains and a 20 years old will be paying 20 pounds of grain.

The chief stands on a riser containing 7 different weights next to a large 2 pan scale. As per the interesting custom, the chief can only weigh using three of the seven weights.

In such a scenario, can you calculate what must be the weights of the seven weights each and who is the oldest man the chief can measure using those weights?

Source: http://gpuzzles.com/mind-teasers/difficult-brain-twister/

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  • $\begingroup$ Note that we can subtract by putting weights on either side. Let A be the age of the person. Let the set of weights be {w1,w2,...,w7,0,0}. Notice this is a set of 9 numbers but two are zero (no weight). We have three basic cases: $A+w_i+w_j=w_k$, $A+w_i=w_j+w_k$, $A=w_i+w_j+w_k$, where $w_i\neq w_j \neq w_k$. $\endgroup$
    – SDiv
    Commented Oct 5, 2014 at 8:46
  • $\begingroup$ Roll this back! $\endgroup$
    – Alec Teal
    Commented Oct 5, 2014 at 16:29
  • $\begingroup$ I agree with @Alec Teal, this question should be returned to its original state. $\endgroup$
    – SDiv
    Commented Oct 6, 2014 at 15:05

2 Answers 2

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The solution with the highest range, as pointed out correctly in the link provided in the question, is {1,3,7,12,43,76,102}. For a thorough numerical investigation you can see pages 5-7 of the pdf at the following link: http://www.mathopt.org/Optima-Issues/optima65.pdf. The solution in this link does not however provide an intuitive reason for these particular numbers nor does it give a general solution for $n$ weights. It seems spiritually related to the following sequence: http://oeis.org/A005228.

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So the weights - 1, 3, 7, 12, 43, 76, 102

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    $\begingroup$ Not sure why this is the accepted answer, there is no explanation whatsoever of how the solution was derived. $\endgroup$ Commented May 28, 2015 at 15:53

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