Easier than it looks

Sam has a sequence of $n$ consecutive positive integers $a_1,a_2,a_3\cdots a_n$ where $n$ is an odd number and $a_1$ is an odd number. Adi arranges all of Sam's numbers in some permutation. Prove that given any permutation of the sequence $(p_1,p_2,p_3\cdots p_n)$ . The expression:

$\left(\prod_{k=1}^{\frac {n-1}2}\left(p_{_{2k}}+6^{k}\right)\right)\left(\prod_{k=1}^{\frac{n+1}2} \left(p_{_{2k-1}}+17^{k}\right)\right)$

will always be divisible by 2.

Details and Assumptions

Basically, in the expression $p_{a}$ is added to $6^{\frac{a}2}$ if $a$ is even and $17^{\frac{a+1}{2}}$ if $a$ is odd. Then all the brackets are multiplied together.

#Combinatorics #JOMO #JOMO4

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$