<s>[INST] Interesting property of sum of powers of integers from 1 to 114. When I have this list of specific X values:
$X: 1, 2, 3, 4, \ldots, 112, 113, 114.$
$$\sum_{n=1}^{114}n = 6555$$
$$6555/19 = 345$$
The sum of these $X$ values divided by $19$ is an integer.

Then I square each $X$ value:
$X^2: 1, 4, 9, 16, \ldots, 12544, 12769, 12996.$
The sum of these $X^2$ values divided by $19$ is also an integer.

Then I cube each $X$ value, and the pattern continues.
Is there any reason for this property mathematically? Or does it just exist?
Many thanks.
 [/INST] This is because $19|114$.
$\sum_{k=1}^n k = \frac {n(n+1)}{2}$
And $\sum_{k=1}^n k^2 = \frac {n(n+1)(2k + 1)}6$.
And as $19$ is a prime number that divides $114$, $19$ will divide both $\frac {19*20}{10}$ and $\frac {19*20*39}6$.
In general if $n = 19*m$ then 
$\sum_{k=1}^n k = \frac {n(n+1)}2 = \frac {19m(19m + 1)}2 = 19\frac {m(19m+1)}2$ and since either $m$ is even or $19m + 1$ is even, $2|m(19m+1)$.
Likewise $\sum_{k=1}^n k^2 =\frac {19m(19m + 1)(38m + 1)}6 = 19 \frac {m(19m + 1)(38m + 1)}6$.  Again either $m$ or $19m + 1$ is even so $2|m(19m+1)(38m + 1)$.  And if $m$ has remainder $0$ when divided by $3$ then $3|m$. If $m$ has remainder $1$ when divided by $3$ then $3|38m + 1$.  And if $m$ has remainder $2$ when divided by $3$ then $3|19m+1$  So $3|m(19m+1)(38m + 1)$.  So $6|m(19m+1)(38m+1)$.
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