Solution to this simple problem The complete question is: A supermarket receives several trucks with boxes of fruit. Each box contain a single type of fruit, which could be mangoes, strawberries, oranges and pears. Each trucks has the same number of boxes. The fruit of which more fruits have been received is the mango. Of the total number of received boxes, one of the trucks has brought exactly 1/3 of the boxes of mangoes, 1/5 of the strawberries, 1/6 of the oranges and 1/7 of pears. How many trucks arrived?
 A: Let $x,y,z,w$ be the number of mangoes, pears, oranges and strawberries respectively on the last truck.
$x+y+z+w = t$
then $3x,7y,6z, 5w$ are the numbers of each in the store
$3x + 7y + 6z + 5w = s$
Since each truck carries the same number of boxes.  $t$ divides $s$
and there are more mangoes than any other fruit.
$3x> 7y,6z,5w$
The fewest boxes on the truck are $3,1,1,1$
but then $t = 6, s = 27$ and $t$ does not divide $s$
If we add a 7th box to the truck, it must be mangoes.
The fewest number of boxes on the truck that meets all the parameters is 
$6,1,1,1$
$t = 9, s= 36$ and the number of deliveries is $\frac {s}{t} = 4$
A solution exists.
Is this answer unique?
There may be other ways to load the different fruits onto the truck and still fill the store with 4 deliveries.  Thus far, I have been answering the wrong question.  The real question!  Is it possible to stock the store with 3 deliveries or with 5 deliveries.
If we wanted to minimize the  number of deliveries, we want to grow s as slowly as possible relative to t.
Suppose we carry an nearly infinite number of mangoes on truck and 1 of the rest
$\frac {s}{t}$ approaches $3$ as $x$ gets to be large.  But it never reaches $3.$  It is impossible to stock the store with only 3 deliveries.
If we want to maximize the number of deliveries, we want to grow $s$ rapidly.  This means shipping as many oranges and strawberries as possible while still obeying the constraint that we carry more mangoes than anything else.
This is achieved when:
$x,y,z,w = \frac {k}{3},\frac {k}{5}, \frac {k}{6},\frac {k}{7}$
The smallest $k$ such that $x,y,z,w$ are integers is $210$
$t = \frac {177}{210} k\\
s = 4k\\
\frac {s}{t} = \frac {708}{177} \approx 4.75 < 5$
It is impossible that the number of deliveries is as high as $5.$
