How can I divide 333 items in 33 groups where first item will be the biggest and will decrease in an equal amount from the previous one? There are 333 items. These need to be spread in 33 groups. The first group is the biggest and gets smaller in equal proportion every time. That means
if $n$ is the first one then
$$
n - x \\
n - 2x \\
n - 3x
$$
therefore, 
$$
n + (n-x) + (n-2x) + (n-3x) + \ldots + (n-32x) = 333
$$
I need the value of $n$ and $x$ and let me know how please?
 A: Since
$$
\sum_{k=1}^{33}(a+bk)=33a+561b\tag{1}
$$
we want to solve
$$
33a+561b=333\tag{2}
$$
However, since $33a+561b=33(a+17b)$ and $33\not\mid333$, there can be no integer solutions.
If we are not interested in integer only solutions, there are an infinite number of solutions; for any $a$, we can find a $b$ that solves $(2)$. For example, to get a step of $1$, we use $a=-\frac{76}{11}$ and $b=1$:
$$
\overbrace{\underbrace{\left(-\frac{65}{11}\right)}_{k=1}+\underbrace{\left(-\frac{54}{11}\right)}_{k=2}+\underbrace{\left(-\frac{43}{11}\right)}_{k=3}+\dots+\underbrace{\left(\frac{287}{11}\right)}_{k=33}}^{33\text{ numbers, each differs from the previous by $1$}}=333
$$
However, this does not relate to the problem of dividing a group of objects, which would require an integer sequence.
A: Since $33$ is odd, there would have to be a middle element $m$ in the sequence, which given that is it an arithmetic progression will also be their mean value. But then the sum of the sequence is $33m$, which cannot equal $333$ for any integer value of$~m$.
