There are six cans each containing doyles(each doyle weighs a gram), out them any of them can be defective. How can we find the defective cans in one weighing? and if each can has only two dozen doyles how can we find the defective can in one weighing? I cannot under stand the solution provided in the book for the first question.
PS: This question is from the book mathematical puzzle tales by Martin Gardner.
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To expand on Steve Kass's answer, if each can has only $24$ doyles, take $11$, $17$, $20$, $22$, $23$, and $24$ doyles from cans $1$, $2$, $3$, $4$, $5$, and $6$ respectively and put them on the scale. Notice that no two subsets of $\{11,17,20,22,23,24\}$ have the same sum. Hence, the excess weight in milligrams uniquely determines the subset of the cans which contain defective doyles.
In the actual puzzle, each can contains $100$ doyles, and within a single can, either all the doyles are good or all are defective. Defective doyles always weigh exactly $1.001$g. To identify which cans contain defective doyles, weigh together $63$ doyles as follows: $1$ doyle from the first can, $2$ from the second can, $4$ from the third can, $8$ from the fourth can, $16$ from the fifth can, and $32$ from the sixth can. If the total weight is over $63$ grams, there are some defective doyles. If the excess weight is at least $32$ milligrams, the sixth can must contain defective doyles, because there aren't enough doyles being weighed from all the other cans together to create that much excess. If so, subtract $32$ from the excess weight. If the excess weight (after possibly subtracting 32 in the last step) now exceeds $16$, the fifth can contains defective doyles. If it does, subtract $16$ from the excess weight. And so on.
In the second part of the problem, no can contains $32$ doyles, so this strategy is impossible. (You didn't ask about that, so I won't say more.)