My Favorite Number is 1153

First of all, the maximum power of a prime number $p$ in $n!$ is given by $h=\displaystyle\sum_{i=1}^{\infty}{\left \lfloor \frac{n}{p^i}\right \rfloor}$ (take note that if $p^i>n$ , then $\left \lfloor \frac{n}{p^i}\right \rfloor=0$ , so this equation does have a finite value).

To show this, see that $n!=n\times(n-1)\times(n-2)\times\ldots \times 2\times 1$ , completely factorize these $n$ numbers so that every number is expressed as the product of their prime factors, then $h$ is actually the sum of exponents of all the $p$ 's.

If the number of $p$ that has the exponent $a$ is $h_a$ , then $\begin{aligned} h&=h_1+2h_2+3h_3+4h_4+\ldots\\&=N_1+N_2+N_3+N_4+\ldots \end{aligned}$ where $N_j=\displaystyle \sum_{i=j}^{\infty}{h_i}$ shows how many numbers from 1 to $n$ that are divisible by $p^j$ , which is given by $\left \lfloor \frac{n}{p^j}\right \rfloor$ .

Hence $h=\displaystyle \sum_{i=1}^{\infty}{N_i}=\displaystyle\sum_{i=1}^{\infty}{\left \lfloor \frac{n}{p^i}\right \rfloor}$

Now, $\displaystyle \sum_{n=1}^{1153}{[k_n\times (-1)^{n+1}]}=\displaystyle \sum_{n=1}^{1153}{\bigg[(-1)^{n+1}\Big(\Big\lfloor \frac{n}{5}\Big\rfloor+\Big\lfloor \frac{n}{5^2}\Big\rfloor+\Big\lfloor \frac{n}{5^3}\Big\rfloor+\Big\lfloor \frac{n}{5^4}\Big\rfloor \Big)\bigg]}$

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Consider $\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5}\right\rfloor}\Big]$ , $\begin{aligned} &\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5}\right\rfloor}\Big]\\&=0-0+0-0\overbrace{\overbrace{+1-1+\ldots}^{5\text{ terms}}\overbrace{-2+2-\ldots}^{5\text{ terms}}+\ldots\overbrace{+229-229+\ldots}^{5\text{ terms}}}^{1145\text{ terms}}-230+230-230+230\\&=1\overbrace{-2+3-4+5-\ldots-228+229}^{114\text{ terms}}\\&=1+114\\&=115\end{aligned}$

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Consider $\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^2}\right\rfloor}\Big]$ , $\begin{aligned} &\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^2}\right\rfloor}\Big]\\&=24\times0\overbrace{\overbrace{+1-1+\ldots}^{25\text{ terms}}\overbrace{-2+2-\ldots}^{25\text{ terms}}+\ldots\overbrace{-45+45-\ldots}^{25\text{ terms}}}^{1125\text{ terms}}+46-46+46-46\\&=1\overbrace{-2+3-4+5-\ldots-44+45}^{22\text{ terms}}\\&=1+22\\&=23\end{aligned}$

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Consider $\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^3}\right\rfloor}\Big]$ , $\begin{aligned} &\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^3}\right\rfloor}\Big]\\&=124\times0\overbrace{\overbrace{+1-1+\ldots}^{125\text{ terms}}\overbrace{-2+2-\ldots}^{125\text{ terms}}+\ldots\overbrace{-8+8-\ldots}^{125\text{ terms}}}^{1000\text{ terms}}\overbrace{+9-9+\ldots+9}^{29\text{ terms}}\\&=1-2+3-4+5-6+7-8+9\\&=5\end{aligned}$

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Consider $\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^4}\right\rfloor}\Big]$ , $\begin{aligned} &\displaystyle\sum_{n=1}^{1153}{\Big[(-1)^{n+1}\left\lfloor\frac{n}{5^4}\right\rfloor}\Big]\\&=624\times0\overbrace{+1-1+\ldots+1}^{529\text{ terms}}\\&=1\end{aligned}$

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Therefore, $\begin{aligned} \displaystyle \sum_{n=1}^{1153}{[k_n\times (-1)^{n+1}]}&=\displaystyle \sum_{n=1}^{1153}{\bigg[(-1)^{n+1}\Big(\Big\lfloor \frac{n}{5}\Big\rfloor+\Big\lfloor \frac{n}{5^2}\Big\rfloor+\Big\lfloor \frac{n}{5^3}\Big\rfloor+\Big\lfloor \frac{n}{5^4}\Big\rfloor \Big)\bigg]}\\&=115+23+5+1\\&=144\end{aligned}$