Easier than it looks

Sam has a sequence of nn consecutive positive integers a1,a2,a3ana_1,a_2,a_3\cdots a_n where nn is an odd number and a1a_1 is an odd number. Adi arranges all of Sam's numbers in some permutation. Prove that given any permutation of the sequence (p1,p2,p3pn)(p_1,p_2,p_3\cdots p_n). The expression:

(k=1n12(p2k+6k))(k=1n+12(p2k1+17k))\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 pap_{a} is added to 6a26^{\frac{a}2} if aa is even and 17a+1217^{\frac{a+1}{2}} if aa is odd. Then all the brackets are multiplied together.

#Combinatorics #JOMO #JOMO4

Note by Yan Yau Cheng
7 years ago

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