Mad factorials!

Number Theory Level 5

$\large \left( \sum_{n=1}^{2017!+1}(n^{2017}!!)\right)^{2017!!}$

Find the sum of last two digits of the expression above .

Notation :

$!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .
$!!$ denotes the double factorial notation :

For an even number & $n>0$ , $n!!=n\times (n-2)\times \cdots\times 4\times 2.$

For an odd number & $n>0$ , $n!!=n\times (n-2)\times \cdots\times 3\times 1.$

1 solution

Tommy Li
Apr 12, 2017

I observed some patterns :

(1) : $\large x \equiv 1 \pmod{8} \Rightarrow x!! \equiv 25\pmod{100}$ for $\large x \geq 15$

(2) : $\large (2k+1)^{n} \equiv 1 \pmod{8}$ where $\large k \geq 0$ and $\large n \geq 2$

(3) : $\large ((2x)^{2017}!!) \equiv 0 \pmod{100}$

By (3) :

$\displaystyle \large \sum_{n=1}^{2017!+1}(n^{2017}!!) \equiv 1^{2017}!!+3^{2017}!!+5^{2017}!!+ 7^{2017}!! +\dots \pmod{100}$

By (1) & (2) :

$\displaystyle \large \sum_{n=1}^{2017!+1}(n^{2017}!!) \equiv 1+25+\dots+25 \equiv 1+4k(25) \equiv 1 \pmod{100}$

$\displaystyle \large \left( \sum_{n=1}^{2017!+1}(n^{2017}!!) \right)^{2017!!} \equiv 1 \pmod{100}$

Answer $\large = 0+1=1$

Which version of this problem is better ?