Skipping Stones
Find the largest integer that will always divide
for all even integers .
The answer is 3.
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1 solution
see actually i am a 10th grade boy - so i dont know any high-fi solving methods - i make use of my best weapons p(n) = (n+1)(n+3)(n+5)(n+7)
multiply and divide equation by n(n+2)(n+4)(n+6) changed equation p(n) = n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) / n(n+2)(n+4)(n+6)
p(n) = 8!K/n(n+2)(n+4)(n+6) .... { reason : product of 8 consecutive no. }
now as n = 2t ... (even) so 8!K/ 2t(2t+2)(2t+4)(2t+6) take 2 common 8!K / 2^4 (t(t+1)(t+2)(t+3)) = 8!K/2^4 (4! T) ... since denominator has 4 consecutive no.
so p(n) = 8!K / 2^4 (4!T) = 105 K/T so 105 can be biggest integer that will divide p(n)