Apple, Orange, & Pear Slices

Let $a$ be the number of plates with apple & orange slices, $b$ be the number of plates with orange & pear slices, and $c$ be the number of plates with pear & apple slices. As such, we desire to find the total number of plates or guests = $a+b+c$ .

Then let $N$ be total number of all fruits.

Hence, $N = a(\dfrac{1}{3} + \dfrac{1}{4}) + b(\dfrac{1}{4} + \dfrac{1}{5}) + c(\dfrac{1}{5} + \dfrac{1}{3})$ .

$N = \dfrac{a+c}{3} + \dfrac{a+b}{4} + \dfrac{b+c}{5}$

In other words, $a+c$ represents the number of apple pieces, and $3|a+c$ , where its quotient is the number of apples itself. Similarly, this applies to the orange and pear cases as well.

$60N = 20(a+c) + 15(a+b) + 12(b+c) = 35a+27b+32c$

Rewriting it as the modulo $60$ :

$35a+27b+32c \equiv 0 \pmod{60}$

$70a+54b+64c \equiv 10a - 6b +4c \equiv 0 \pmod{60}$

$5a -3b+ 2c \equiv 0 \pmod{30}$

$3(a-b) + 2(a+c) \equiv 0 \pmod{30}$

Let $p = a-b$ and $q = a+c$ . Since $a,b,c$ are positive integers, $a+c > a-b$ and so $q>p$ .

Now $3p + 2q \equiv 0 \pmod{30}$ .

In order to yield lowest positive integers, $3p + 2q = 30$ .

Hence, $q$ is a multiple of $3$ , and $q < 15$ .

Thus, by exploring all possibilities, the only applicable $(p,q)$ include $(4,9)$ and $(2,12)$ .

If $p = a-b = 4$ and $q = a+c = 9$ , then $b+c = 5$ . Then the list of possible $(a,b,c)$ include: $(5,1,4), (6,2,3), (7,3,2), (8,4,1)$ . However, from the lemma above, $4|a+b$ , so the only applicable solutions include: $(6,2,3), (8,4,1)$ .

If $p = a-b = 2$ and $q = a+c = 12$ , then $b+c = 10$ . With similar methods, the only applicable solutions include: $(3,1,9), (5,3,7), (7,5,5), (9,7,3)$ .

From all the lists, only $(6,2,3)$ yields the total number of $6$ fruits:

The number of apples = $\dfrac{a+c}{3} = \dfrac{9}{3} = 3$ apples, thus resulting in $9$ apple pieces.

The number of oranges = $\dfrac{a+b}{4} = \dfrac{8}{4} = 2$ oranges, thus resulting in $8$ apple pieces.

The number of pear(s) = $\dfrac{b+c}{5} = \dfrac{5}{5} = 1$ pear, thus resulting in $5$ pear pieces.

When putting on the plates, there will be $6$ plates with apple & orange slices, $2$ plates with orange & pear slices, and $3$ plates with pear & apple slices. Therefore, there will be $\boxed{11}$ guests in the party.

Note that for (6, 8, 10) also yield 6 fruits, but the total guests will be 12, which is not yet the least number.

As the dishes are served with an even number, and there are no pieces left over, there must be an even number of Apples and Pears combined (because they are each cut into an odd number of slices).

This leaves eleven combinations (including scenarios with zero of one type of fruit). However, only two of them are valid, because the other nine combinations have slices of one fruit that outnumbers the total of the other two fruit slices combined.

Let there be $x$ apple(s) $,y$ orange(s) and $z$ pear(s).

$x+y+z=6\hspace{4mm}\color{#3D99F6}(\text{Given})$

The total number of slices is given by, $3x+4y+5z$ . The total number of slices must be even since each guest gets a plate of $2$ slices and there are none leftover at the end.

$\implies 3x+4y+5z \pmod{2}\equiv0\\ \implies x\pmod{2}\equiv z\pmod{2}$

There are 2 ways to pair up the slices

$\text{Case 1:}$ If you have 2 slices of apple,2 slices of oranges and 2 slices of pear,you can pair them up as (apple,pear),(apple,orange),(orange, pear)

$\text{Case 2:}$ if you have $a$ slices of one kind, $b$ slices of a second kind and $c$ slices of a third kind where , $c=a+b$ then each fruit of the third kind can be paired up with one of the first or second kind.

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Thus if there are $a$ slices of apples, $b$ slices of oranges and $c$ slices of pears

check if $\text{max}(a,b,c)$ is the sum of the other two

1)if yes then it forms a valid number of slices which can be distributed .(Case 2)

2)if no ,then subtract 2 from a,b,c (which can be paired up because of Case1) and repeat the process until slices of one of the fruits hit $1$ or $0$

This forms an iterative loop.

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Now, in the above problem

since $x+y+z=6$

and $x\pmod{2}\equiv z\pmod{2}$

we only have to check 3 cases namely, $(2,2,2),(1,2,3),(3,2,1)$

$\text{The number of guests}=\text{number of pairs of slices}=\dfrac{\text{Total number of slices}}{2}$

We find that $(3,2,1)$ will have the least number of slices $=22$

thus we end up with $9$ slices of apples $8$ slices of oranges and $5$ slices of pear,

upon testing we find,

$9\neq 8+5$

subtracting $2$ and trying again,

$7\neq 6+3$

iterating again we get,

$5=4+1$

Thus we find that $(3,2,1)$ will have a valid solution,and the number of guests $=\dfrac{22}{2}=\boxed{11}$

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$P.S$

we can see that $(2,2,2)$ also is a valid solution although not minimal,

in this case we have $6$ slices of apples, $8$ slices of oranges and $10$ slices of pears

$10\neq 8+6$

$8\neq 6+4$

$6=4+2$

The best way to minimize the number of slices is to max out the number of apples such that the total slices will be a multiple of 2. The least number of possible slices is 22. Checking, I note that the slices can be divvied by into 6 apples with 6 oranges, leaving 3 apples, 2 oranges, and 5 pears. The least possible number of guests is therefore 11.

I start like @Worranat Pakornrat , defining $a$ to be the a+o combinations, $b$ the o+p combinations, and $c$ the p+a combinations. With $x, y, z$ the numbers of each fruit we get $\begin{cases} 3a = x + z \\ 4b = x + y \\ 5c = y + z\end{cases};$ since there are 6 fruits we know $a + b + c = 6$ , and multiplying by 60 gives $20(x+z) + 15(x + y) + 12(y+z) = 360 \ \ \ \ \therefore\ \ \ \ 35x + 27 y + 32z = 360.$ At this point it is useful to establish the extreme cases: there can be no fewer than $360/35 > 10$ and no more than $360/27 < 14$ guests. The number of guests is $g \in \{11, 12, 13\}$ .

To get easier numbers, subtract $30g$ from both sides. This results in $5x - 3y + 2z = \begin{cases} 30 & g = 11 \\ 0 & g = 12 \\ - 30 & g = 13 \end{cases}$ Note that the sum $5x - 3y + 2z$ does not change if we add multiples of either of the following: $(x,y,z) = \begin{cases} (3,5,0) \\ (-2,0,5) \\ (0,2,3) \end{cases}$ These additions increase the numbers of guests by 8, 3, and 5, respectively.

Because we wish to minimize the number of guests, we start with $g = 11$ and see if it can be done. Since we need $5x - 3y + 2x = 30$ , we start by setting $(x,y,z) = (6,0,0)$ . Now we must add five more guests without changing the total. From the discussion above we see that this is accomplished by adding $(0,2,3)$ . Thus we have found a solution with $\boxed{11}$ guests! The relevant values are $(x,y,z) = (6,2,3);\ \ \ a = \frac{6+3}3 = 3,\ b = \frac{6+2}4 = 2;\ c = \frac{2+3}5 = 1.$