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Puzzling

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  • $\begingroup$ This method is incorrect. Nowhere are you told that the same number of apples is in each basket. The best you can hope to write down is that the number of bananas in the $ith$ basket, $B_i$ is $B_i=\Sigma_{j=1}^6 A_j-A_i$, and similarly for the plums. $\endgroup$ Commented Mar 12, 2016 at 18:25
  • $\begingroup$ @ryanp16 this answer doesn't make that assumption. It deals with totals not with amounts per basket. The sum of all the banana is the sum of all the apple times 5 because for each basket the number of bananas in there is the same as the number of apples in the other 5 baskets. Do this for each basket and you end up counting the number of apples in each basket 5 times. $\endgroup$ Commented Mar 12, 2016 at 18:38
  • $\begingroup$ Apologies, you are correct. I read your answer too quickly. $\endgroup$ Commented Mar 12, 2016 at 18:43
  • $\begingroup$ Wait a minute !!...total number of plums = $33$....therefore total number of apples = $5 \times 33 = 165$...therefore apples in each basket = $\frac{165}{6}=27.5$....Do you mean to say that each basket contains 27 and a half apples? I thought they need to be integers.. :\ $\endgroup$ Commented Mar 13, 2016 at 19:21
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    $\begingroup$ @manshu One way to arrange the fruit is to have $33$ plums and $165 = 5 \cdot 33$ bananas in one basket and $33$ apples and $132 = 4 \cdot 33$ bananas in the other five baskets. $33 \cdot 6 + 132 \cdot 5 + 165 = 1032$. Remember, there's no stipulation that all the baskets have the same amount of fruit. $\endgroup$ Commented Mar 13, 2016 at 19:26