Kdtyjb

We want to find the number of solutions of
$$n+(n+1)+ldots + (n+k) = 1050, ninmathbb Z_{>0}, kinmathbb Z_{geq 0}.tag{1}$$

Rewrite the sum as $$n(k+1) + 0 + 1 +ldots + k = n(k+1) + frac{k(k+1)}{2}= frac 12(2n+k)(k+1).$$

Thus, the number of solutions to $(1)$ is the same as the number of solutions of
$$(2n+k)(k+1) = 2100, ninmathbb Z_{>0}, kinmathbb Z_{geq 0}.tag{2}$$

Let $a$ and $b$ be divisors of $2100$ such that
begin{align}
2n+k &= a,\
k+1 &= b.tag{3}
end{align}
Solving it we get
begin{align}
n &= frac{a-b+1}2,\
k &= b -1.tag{4}
end{align}

From here we see that not every choice of integers $a$ and $b$ such that $ab = 2100$ will give us a solution to $(2)$. Since $a-b+1$ must be even, $a$ and $b$ are of opposite parities. Also, $ageq b > 0$ since $n> 0$ and $k geq 0$.

First determine the number of ways to factor $2100 = 2^2cdot 3cdot 5^2 cdot 7$ such that one of the factors is odd. For this to be fulfilled, we shouldn't allow $4 = 2^2$ to be factored, so consider $2100 = 4cdot 3 cdot 5^2 cdot 7$ instead. Thus, there are $2cdot 2cdot 3cdot 2 = 24$ positive integral solutions to $2100 = a'b'$ such that one factor is odd. Because of commutativity, it means there are $12$ distinct ways to factor $2100$ into product of two factors, one of which is odd, and for every such factorization there is a unique choice for $a$ and $b$ such that $ageq b$.

Thus, there are $12$ positive integral solutions to $(2)$.

The sum of the first $n$ natural numbers is
$$sum_{i=0}^ni=frac12n(n+1).$$
So by subtracting the first $m-1$ terms we get the sum of all consecutive integers from $m$ to $n$;
$$sum_{i=m}^ni=frac12n(n+1)-frac12(m-1)m=frac12(n+m)(n-m+1).$$
To count the number of ways to write a number $k$ as a sum of consecutive integers, we want to find natural numbers $m$ and $n$ with $m<n$ such that
$$(n+m)(n-m+1)=2k.$$
In particular this gives a factorization of $2k$. Conversely, if $2k=atimes b$ is a factorization where $anotequiv bpmod{2}$ then setting
$$m:=frac{a-b+1}{2}qquadtext{ and }qquad n:=frac{a+b-1}{2},$$
gives $(n+m)(n-m+1)=2k$. This shows that if $k=2^lk'$ with $linBbb{N}$ and $k'$ odd, then the expressions of $k$ as a sum of consecutive integers correspond $2$-to-$1$ to the divisors of $k'$; for each divisor $d$ of $k'$ we have the two factorizations
$$2k=dtimesleft(2^lfrac{k'}{d}right)=left(2^ldright)timesfrac{k'}{d},$$
of $2k$ into an even and an odd number. The corresponding sums include the trivial sum $k=sum_{i=k}^ki$, as well as sums with negative integers. This shows that the total number of ways to represent a number $k$ as a sum of consecutive integers, is twice the number of divisors of $k'$.

The number of expressions of $k$ as a sum of positive integers is the number of factorizations for which $mgeq0$, or equivalently $a+1geq b$. Of course for every factorization $2k=atimes b$ with $aneq b$ we have either $a+1geq b$ or $b+1geq a$ exclusively, so if $2k$ is not a square then precisely half of all expressions involve only positive integers.

In this particular case $k=1050=2cdot3cdot5^2cdot7$ and so $k'=525=3cdot5^2cdot7$, and the number of divisors of $k'$ equals $2times3times2=12$, so there are $12$ expressions of $k$ as a sum of positive integers. The factors of $k'$ and corresponding sums are
begin{eqnarray*}
text{factor}&&qquad&&text{sums}\
hline
1&&qquad&&k=sum_{i=1050}^{1050}i
&&qquad&&k=sum_{i=261}^{264}i\
3&&qquad&&k=sum_{i=349}^{352}i
&&qquad&&k=sum_{i=82}^{93}i\
5&&qquad&&k=sum_{i=208}^{212}i
&&qquad&&k=sum_{i=43}^{62}i\
7&&qquad&&k=sum_{i=147}^{153}i
&&qquad&&k=sum_{i=24}^{51}i\
15&&qquad&&k=sum_{i=63}^{77}i
&&qquad&&k=sum_{i=-12}^{47}i\
21&&qquad&&k=sum_{i=40}^{60}i
&&qquad&&k=sum_{i=-29}^{54}i\
25&&qquad&&k=sum_{i=30}^{54}i
&&qquad&&k=sum_{i=-39}^{60}i\
35&&qquad&&k=sum_{i=13}^{47}i
&&qquad&&k=sum_{i=-62}^{77}i\
75&&qquad&&k=sum_{i=-23}^{51}i
&&qquad&&k=sum_{i=-146}^{153}i\
105&&qquad&&k=sum_{i=-42}^{62}i
&&qquad&&k=sum_{i=-207}^{212}i\
175&&qquad&&k=sum_{i=-81}^{93}i
&&qquad&&k=sum_{i=-348}^{351}i\
525&&qquad&&k=sum_{i=-260}^{264}i
&&qquad&&k=sum_{i=-1049}^{1050}i\
end{eqnarray*}
We see that indeed $12$ out of these $24$ expressions involve only positive integers.

For the sum of next integer we may use formula for the sum of arithmetic sequence

$$sum_{k=1}^na_k=frac n2(a_1+a_n)$$

So

begin{aligned}
1050 &= frac n2(a_1+a_n) \
2100 &= n(a_1+a_n) \
2100&= n(a_1+a_n) \
2100&= n(a_1+a1+(n-1)) \
2^2cdot3cdot5^2cdot7&= n(2a_1+n-1)
end{aligned}

Now:

1. If $n$ is even, then $(2a_1+n-1)$ is odd, so
   $$n=2^2cdot3^xcdot5^ycdot7^z$$
   where $x in {0,1},; y in {0,1,2}, ;z in {0,1}$,
   
   so there are $2 times 3 times 2 = 12$ possibilities for $n$.



2. 
   

   If $n$ is odd, then similarly
   $$n=3^xcdot5^ycdot7^z$$

   
   
   

   and we obtain other $12$ possibilities for $n$.

   
   

So there are $24$ solutions altogether, a half o them, i. e. $color{red}{12}$, for only positive integers, because for positive integers must be $a_1 ge 1$, and consequently $(2a_1+n-1) >n$, so in the product $n(2a_1+n-1)$ the first multiplier have be smaller than the second one.

Here's how you use that information:

1050 has divisors:

$$1,2,3,5,6,7,10,14,15,21,25,30,35,42,50,70,75,105,150,175,210,350,525,1050$$ You can use the prime factorization to check this. You then say 1050 divided by 3 gives 350 so 349+350+351 adds up to 1050. Time to make sums (odd divisors and 4 thrown in, because it lands on a half integer). This gives you:
$$begin{eqnarray}1050=349+350+351\1050=261+262+263+264\1050=208+209+210+211+212\1050=147+148+149+150+151+152+153\1050=63+64+65+66+67+68+69+70+71+72+73+74+75+76+77\1050=40+41+42+43+44+45+46+47+48+49+50+51+52+53+54+55+56+57+58+59+60\1050=30+31+32+33+34+35+36+37+38+39+40+41+42+43+44+45+46+47+48+49+50+51+52+53+54end{eqnarray}$$

Okay, I may be missing a few. It gets the point across though.

Using the MiniZinc solver with Gecode, I got the following $11$ solutions:

13 .. 47

24 .. 51

30 .. 54

40 .. 60

43 .. 62

63 .. 77

82 .. 93

147 .. 153

208 .. 212

261 .. 264

349 .. 351

The model:

var 1..1050: k0;

var 1..1050: k1;



constraint

  (1050 == sum([k0 + k | k in 0..k1]));



solve satisfy;



output ["n(k0) .. (k0+k1)"];

One way is to note that if there are an odd number, $2n+1$ of terms with the middle term $k$ then the sum of consecutive terms will add to $(2n+1)k$.

[Because there are $2n+1$ terms and they average $k$]

And if there are an even number, $2n$ of terms with the middle two terms $k$ and $k+1$ then the sum of consecutive terms will add to $2n(k + frac 12) = n(2k + 1)$.

[Because there are $2n$ terms and they average $k+frac 12$]

But if we can't have negative terms we must have $kge n$.

And so we can have either:

$1050 = k(2n+1); k> n$ can be a sum of $2n+1$ consecutive terms centered at $k$ (i.e. $(k-n) + (k-n+1) + ..... +(k+n-1)+ (k+n)$ ) or

$1050 = n(2k+1); nle k$ can be a sum of $2n$ consecutive terms centered at $k$ and $k+1$ (i.e $(k-n+1)+ ..... + (k+n)$.)

And so

$1050 = 1050*1 = k(2n+1) implies 1050 = 1050$; one consecutive term centered at $1050$(maybe allowed)

$1050 = 2*525 = n(2k+1) implies 1050 = 261+262+263+264$; four consectutive terms centered at $262$ and $263$.

$1050 = 350*3 = k(2n+1) implies 1050= 349 + 350 + 351$; three consecutive terms centered at $350$.

etc.

And we can partition $1050 = even*odd$ in... well....

$1050 = even*odd = (2*3^a5^b7^c)*(3^{1-a}*5^{b-2}*7^{c-1});a=0,1;b=0,1, 2;c=0, 1$ ...

That would be in $2*3*2 = 12$ ways.

i.e.

$1050 = 1050*1=k(2n+1) = 1050$;

$1050 = 2*525=n(2k+1) = 261+262+263+264$;

$1050 = 350*3=k(2n+1) = 349 + 350 + 351$;

$1050 = 210*5 =k(2n+1)= 208+209+210+211+212$;

$1050 = 6*175=n(2k+1)= 163 + 164+ ..... + 186 + 187$;

......

$1050 = 30*35= k(2n+1) = 13 + 14 + ..... + 46 + 47$;