Smooth Partitons

From the problem we can see that in a smooth partition, min $\{a_i\} = a_1$ , and max $\{a_i\} = a_k$ , so, in any smooth partition $| a_i-a_j| \leq 1$ for any $1 \leq i,j \leq k$ . Let $a_1 =x$ , then we have $a_i \leq x+1$ for all $i$ .

Now, as the integers are strictly increasing, we can write them as $\{a_i\}_{i=1}^k = \underbrace{x, x, \cdots, x} , \underbrace{(x+1), \cdots, (x+1)}\\\quad \qquad \quad \;\;\; \quad _{k-p} \;\; \;\;\;\;\qquad \qquad \; _p$ where $k > p \geq 0$ ( $p=0$ implies $a_1=a_k$ , and as $a_1 = x, k-p > 0 \Rightarrow k > p$ )

And thus, sum of the integers is $kx+p$ , and thus we have to solve $kx+p = 250$ in integers with $1 \leq k \leq 250 , 0 \leq p < k$ , But we know that by reminder theorem, for each $k$ in the limit, $kx+p = 250$ has an unique solution, namely the reminder of 250 upon division by $k$ , so the number of solutions is $250$ .

Effectively, what we are told is that, in each smooth partition, there are constants $a_1$ , $n_1$ , $n_2$ , and $k$ that uniquely describe the partition, where $a_1$ is just $a_1$ for our sequence, $n_1$ is the number of $a_i$ with value $a_1$ , $n_2$ is the number of $a_i$ with value $a_1+1$ , and $k$ is just $k$ .

We note by the fourth rule that all $a_i$ either have value $a_1$ or $a_1+1$ , so $n_1+n_2=k$ . This makes $n_2$ redundant.

Now, we note that the sum of the $a_i$ is $n_1\cdot a_1+(k-n_1)\cdot(a_1+1)$ . Expanding yields that this is equal to $k\cdot a_1+(k-n_1)$ . We can see that this means that $k$ and $a_1$ uniquely determine $n_1$ , if it exists.

Now we want to count how many valid $(a_1, ~ k)$ pairs there are. We see that restrictions 2 and 4 are now taken care of, so that only leaves 1 and 3. However, 1 just states that both numbers are integers, so that really leaves 3.

We furthermore note that the choice of $n_1$ (which we note can be at most $k$ and must be positive), now allows us to "pinpoint" our range created by our ordered pair to get that $(a_1+1)\cdot k>250\ge a_1\cdot k$ . Since 3 is effectively the only restriction, we see that this is our only nontrivial restriction on $a_1$ and $k$ . However, this inequality implies that $a_1=\lfloor \frac{250}{k} \rfloor$

Since this will always be an integer by definition of the floor function, and this is our only nontrivial restriction, we observe that $a_1$ is redundant. However, since $a_1$ and $k$ are positive, we see that $0<k\le 250$ .

Since any choice of $k$ within that range will uniquely determine $a_1$ , $n_1$ , and $n_2$ , there are $\boxed{250}$ smooth partitions of $250$ . Note that, in general, the same reasoning implies that there are $n$ smooth partitions of the integer $n>0$ .

On reading the Rules of the sets., we understand,
1. $a_k - a_1 = 1$ or $a_k - a_1 = 0$
2. Any Smooth Partition set will have maximum of 2 distinct Integers repeated numbers of times to attain setsum of 250
3. There can not be more than one set of Elements "n" satisfying all the conditions above because all the Elements in a set are Integers.
4. As a set contain all Integers .., The maximum Elements must be 250 and minimum must be 1 . Thus which says , there must be minimum of one Ordered set which have number of elements equals a value between 250 and 1

The above Conclusions from Rules Prove for a number "n" there Exists "n" number of smooth partition sets. Thus the answer must be "250".

We get the same answer if we go through set building method :- (below formula applicable to this Question)
[let us represent "number of Elements in a set" as $N(S)$ ]

* $\frac{Number(250)}{ N(S) }$ = a value say "v" *
Note: In the value "v" ., If we remove decimal digits ...we get $a_1$ .
we also get $a_k$ as $a_k$ = $a_1$ + 1
* Thus for a value of N(S) .... we get value of $a_1$ and $a_k$ * .... {an Ordered set}

Thus ., $1 \leq N(S) \leq 250$ ., we get Positive Integer value for $a_1$

Thus Maximum of 250 sets are Possible for the number 250.

We observe that each value a i can differ from all other values a j by at most one. It then follows that a partition of n into k integers will always contain k 1 integers of value floor(n/k), and k-k 1 integers of value ceil(n/k). Since the above definition fixes all orderings for a given partition, we have one partition for each value of k. To conclude, k can of course be in the range (1, n).

Let positive integer m be the value of a 1, positive integer p be the number of a i whose value is m. Hence, the value of a k is m + 1. Let positive integer q be the number of a i whose value is m +1. \displaystyle \sum {i=1}^k a i = n \Rightarrow p \times m + q \times (m +1) = 250 \Rightarrow m \times (p + q) = 250 - q Since 250 \geq m \geq 1, \rightarrow 250 \geq p + q \geq 1. Hence, there are 250 ways to choose the value of p + q, for each value, we can always choose q and m so that 250 - q is divisble by m, and because \frac {250 - q}{m} \deq 250, we can choose p so that p + q = \frac {250 - q}{m} Obviously, there is only one way to choose m, p, q for each value of p + q. If there exist m 1 not qual to m 2 such that: \frac {250 - q 1}{m 1} = \frac {250 - q 2}{m 2} \Rightarrow 250 \times (m 1 - m 2) = q 2 - q 1, this is impossible since 250 > q 2 - q 1 Hence, the answer is 250

There exist $n$ Smooth Partition for the number $n$ . This can be gained by the following arguments:

The minimum number of set element of a smooth partition is 1, which is the number $n$ itself to be $a_1$ , which is the first smooth partition in my solution

Then for the second smooth partition, it contains two number $a_1$ and $a_2$ which are $\lfloor \frac{n}{2} \rfloor$ and $\lceil \frac{n}{2} \rceil$

For the $m^th$ smooth partition, it contains $m$ number, i.e. $a_1, a_2, a_3, ... , a_m$ The values of these numbers are obtained this way :

$a_i = \lfloor \frac{n}{m} \rfloor$ then we find the sum of these numbers

$\sum_{i=1}^{m} a_i$ then we find the lack of values

$n - \sum_{i=1}^{m} a_1$ and denote it as $S$

Let $S$ be separated into several values of $1$ s, e.g. $12$ can be written as $1+1+1+1+1+1+1+1+1+1+1+1$ , adding this $1$ s equally to $a_m , a_{m-1}, a_{m-2}, ... a_{m-S+1}$

this way can ensure that $a_m - a_1 \leq 1$

Hence from the explanation we can get that $m\leq n$ because when $m = n$ the smooth partition will contain $n$ numbers which are $n$ $1$ s., which is the partition with most number in it which is as asked since $a_z$ needs to be positive integer.

Therefore the answer to the Problem is $250$ .

We claim that for integers $i,m$ with $1 \leq i \leq m,$ there is one smooth partition of $m$ with exactly $i$ parts (i.e. $k = i$ ). The parts in a smooth partition take on at most two values that differ by 1. When there are $i$ parts, the values must be $\lfloor\frac{m}{i}\rfloor$ and $\lceil \frac{m}{i} \rceil.$ There must be $m \pmod i$ that are $\lceil \frac{m}{i} \rceil$ and the rest are $\lfloor\frac{m}{i}\rfloor.$ Since the parts must be arranged in increasing order, there will only be one such partition. Thus, there are $250$ partitions of $250.$

The four boundaries lead us to notice the answer is same as the integer given, hence the answer is 250.

250=a {1}+a {2}+a {3}+a {4}........................a {n} and if a {1}=1 then a {n} can take two values i.e.1&2 if a {n} takes 1 then a {2}=a {3}=a {4}=a {5}=a {6}=1 (because of 2.) so 250=1+1+1+1+1......................... and the terms will be 250 if a {n} takes 2 then if a {n-1} takes 2 values i.e. 1&2 if a {n-1} takes 1 then 250=1+1+1+1+.....................+1+2 and terms will be 249 if it takes 2 then there will be two values for if a_{n-2} i.e. 1&2 again 250=1+1+1+ +1+2+2 and terms will be 248 250=1+1+1+ +2+2+2 and terms will be 247 . . . . . . 250=249+250 there are 2 terms 250=250 there is 1 term So total possible partitions will be 250.

Start from n=1, 1 for n=2, 1 1 and 2 for n=3, 1 1 1, 1 2 and 3 for n=4 1 1 1 1, 1 1 2, 1 3 and 4 so for n=250 there are 250 smooth partitiions with k=1 until k=250