Set of $n$ numbers 
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*Let $1<q\le2$. We have a set of $n$ numbers $1,q,q^2,...,q^{n-1}$. Prove that set of numbers can be divided into $2$ parts, so that the sum of numbers in parts not differ by more than $1$.

*Let $1<q\le3/2$. We have a set of $n$ numbers $1,q,q^2,...,q^{n-1}$. Prove that set of numbers can be divided into $3$ parts, so that the sum of the numbers in any two of them do not differ by more than $1$.
 A: For the first problem,

  
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*Let $1<q\le2$. We have a set of $n$ numbers $1,q,q^2,...,q^{n-1}$. Prove that set of numbers can be divided into $2$ parts, so that the sum of numbers in parts not differ by more than $1$.
  

here is a proof by using mathematical induction:
Obviously, the proposition is true for $n=1$, since you can consider the subsets as $\{1\}$ and $\emptyset$. Now, assuming the proposition is true for $n$, we show that it is also true for $n+1$.
Since the proposition is assumed to be true for $n$, then sets $A$ and $B$ exist such that the sums of their elements do not differ by more than one. If we define $\sigma(X)$ to be the sum of elements in set $X$, then:
$$|\sigma(A)-\sigma(B)|\leq1\label{eq1}\tag{1}$$
Now, we define new sets $A'$ and $B'$ as:
$$A'=\{q\cdot x\mid x\in A\}, \quad B'=\{q\cdot x\mid x\in B\}$$
In other words, the elements of $A'$ and $B'$ are the elements of $A$ and $B$ multiplied by $q$. Since $A$ and $B$ have no element in common, $A'$ and $B'$ have nothing in common, too. Also, since we multiplied every element (including $1$) by $q$, neither $A'$ nor $B'$ contains $1$. So:
$$A'\cap B' = \emptyset,\quad A'\cup B'=\{q,q^2,\dots,q^n\}$$
On the other hand, the sum of the elements of $A'$ and $B'$ are:
$$\sigma(A')=q\cdot\sigma(A),\quad \sigma(B')=q\cdot\sigma(B)$$
and from $1<q\leq2$ and equation (\ref{eq1}):
$$|\sigma(A')-\sigma(B')|=q\cdot|\sigma(A)-\sigma(B)|\leq q\leq2$$
Here, without loss of generality, we can assume $B'$ to be the set with lower sum, i.e., $\sigma(B')\leq\sigma(A')$. So:
$$|\sigma(A')-\sigma(B')|=\sigma(A')-\sigma(B')\leq2\label{eq2}\tag{2}$$
As already mentioned, neither $A'$ nor $B'$ contains $1$. If we define new set $B''$ as:
$$B''=B'\cup\{1\}$$
then one can easily see:
$$\sigma(B'')=1+\sigma(B')$$
and:
$$A'\cap B'' = \emptyset,\quad A'\cup B''=\{1,q,q^2,\dots,q^{n-1},q^n\}$$
Therefore, if we show that the difference of $\sigma(A')$ and $\sigma(B'')$ is not greater than $1$, then the proof is completed. To do this, by using (\ref{eq2}) we have:
$$0\leq\sigma(A')-\sigma(B')\leq2$$
By adding $-1$ and grouping we get:
$$-1\leq\sigma(A')-\sigma(B'')\leq1$$
or:
$$|\sigma(A')-\sigma(B'')|\leq1$$
which completes the proof.

But the second proposition which states:



  
*Let $1<q\le3/2$. We have a set of $n$ numbers $1,q,q^2,...,q^{n-1}$. Prove that set of numbers can be divided into $3$ parts, so that the sum of the numbers in any two of them do not differ by more than $1$.
  

cannot be proved, because it is not true. As a counter example consider $q=\frac32$ and $n=3$. So the set of numbers would be $\{1,\frac32,\frac94\}$. You have to form three subsets $A$, $B$ and $C$ (with no common element between any two of them) so that the difference of any two sums of the elements does not exceed $1$. Obviously, none of the subsets can be empty, because, if, for example, $C=\emptyset$ then one of the others, name it $B$, should contain $\frac32$, and hence, $|\sigma(B)-\sigma(C)|\geq\frac32>1$ which contradicts the proposition. Therefore, the only solution is $C=\{1\}$, $B=\{\frac32\}$ and $A=\{\frac94\}$ (however names can be changed). This arrangement leads to $|\sigma(A)-\sigma(C)|=\frac94-1=\frac54>1$ which is a contradiction. So the proposition is not true.
A: *

*Let $S_0=0,S_i=S_{i-1}+q^{n-î},\text{ when }S_{i-1}\le0\text{ and }S_i=S_{i-1}-q^{n-î},\text{ when }S_{i-1}>0,\text{ then it is easy to prove, that }|S_i| \le q^{n-i}\text{ for }i=1,...,n.$


We have $|S_n|\le1$ or  $|A-B|\le1$, where $A$ and $B$ - sum of numbers of two parts of set $1,q,q^2,...,q^{n-1}.$
