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.

Solution 1:

Hint : $$\sum_{j=1}^k j =\frac{k(k+1)}{2}$$ $$\sum_{j=1}^k j^2=\frac{k(k+1)(2k+1)}{6}$$

Solution 2:

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 $6|m(19m+1)(38m+1)$.

Solution 3:

We can prove that $a^{n}+b^{n}=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^{2}-...+a^{2}b^{n-3}-ab^{n-2}+b^{n-1})$ by simplifying the right hand side to the left hand side. However, this is only true for odd integers $n$, also note that the property no longer holds if $18|n$.

In the set $1;2;3;4;5;\cdots;113;114$:

• The number $114$ is divisible by $19$, so all the sums $1+113;2+112;3+111;4+110;\cdots;56+58$ and number $57$ are divisible by $19$

• Apply the equality above, we will have $1^n+113^n;2^n+112^n;3^n+111^n;4^n+110^n;\cdots;56^n+58^n$ and $57^n$ and $114^n$ are divisible by $19$ too $\Rightarrow 1^n+2^n+3^n+4^n+\cdots+114^n$ is divisible by $19$ for all odd integers $n.$