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There are six fruit baskets with apples, bananas and plums on the table. The number of bananas in each basket is equal to the total number of apples in the other five baskets together. The number of apples in each basket is equal to the total number of plums in the other five baskets together. Altogether the six baskets contain at least 1000 pieces of fruit.

What is the minimum number of pieces of fruit in the six baskets together?

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The minimum greater than 1000 is:

1023

Because:

EDITED to include information from comments.
Thanks to ryanp16 and manshu for asking for clarifications.

Let $b$ be the total number of bananas.
And let $b_1, b_2 \ldots b_6$ be the number of bananas in baskets $1, 2, \ldots 6$.
Also let $a$ be the total number of apples.
And let $a_1, a_2 \ldots a_6$ be the number of apples in baskets $1, 2, \ldots 6$.
Now:
$$b = \sum_{i=1}^6{b_i} = \sum_{i=1}^6{a_i} - a_1 + \sum_{i=1}^6{a_i} - a_2 + \ldots + \sum_{i=1}^6{a_i} - a_6$$
$$= 6\cdot\sum_{i=1}^6{a_i} - \sum_{i=1}^6{a_i} = 5\cdot\sum_{i=1}^6{a_i}$$
$$ \therefore b = 5a$$
Likewise $a = 5p$ where $p$ is the total number of plums.

So the total amount of fruit is: $$b + a + p = 5a + 5p + p = 25p + 6p = 31p$$
And $1023 = 33 \cdot 31$ is the lowest multiple of $31$ greater than $1000$.

One way to arrange the fruit is to have $33$ plums and $165 = 5 \cdot 33$ bananas basket $1$.
And $33$ apples and $132 = 4 \cdot 33$ bananas in the other five baskets.

The total is then: $$33 \cdot (1 + 5) + 132 \cdot 5 + 165 = 1023$$

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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. $\endgroup$ – Okazaki Mar 12 '16 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. $\endgroup$ – Paul Evans Mar 12 '16 at 18:38
  • $\begingroup$ Apologies, you are correct. I read your answer too quickly. $\endgroup$ – Okazaki Mar 12 '16 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.. :\ $\endgroup$ – manshu Mar 13 '16 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. $\endgroup$ – Paul Evans Mar 13 '16 at 19:26
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We have 6 baskets. So banana, apple and plum are contained in these 6 basket.

According to the question

$$b=5a$$ and $$a=5p$$

Now here $b$, $a$ and $p$ are only for one basket. Taking them for 6 baskets and adding them we get:

$$6b+6a+6p \geq 1000$$

Putting values

$$30(5p)+30p+6p \geq 1000$$ $$\Rightarrow 186p\geq 1000$$ As $p$ is an integer. Therefore $p=6$ and $a=30$ and $b=150$
Now we get $$6b+6a+6p=1116$$
Therefore my final answer is $1116$

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The size of each group of bananas is equal to the sum of 5 different groups of apples (rotating), so the total number of bananas AND apples is (5*6+6)/6=6 times the total number of apples, which is equivalent to these 6 different groups of apple repeated 6 times each. Similarly, the size of each group of apples is equal to the sum of 5 different groups of plums (rotating), so the total number of fruits is (5*36+6)/6=31 times the total number of plums. Therefore the total must be divisible by 31, and the answer is 1023.

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