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AravindG Group TitleBest ResponseYou've already chosen the best response.0
nah I was born is 15 BC
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
lol okay i got a math contest question published in 1992
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
Then post it !
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
\[\text{Find the largest integer,\not exceeding}\huge \prod_{n=1}^{1992}\frac{ 3n+2 }{ 3n+1 }\]
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
On my first attempt it seems there is a pattern that is followed involving a sequence of cancellation .Let me see if its right
 one year ago

AravindG Group TitleBest ResponseYou've already chosen the best response.0
my bad ..The pattern didnt help
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
so we have \[\frac{ 5 }{ 4 }\frac{ 8 }{ 7 }\frac{ 11 }{ 10 }...\frac{ 3(1992) +2}{3(1992)+1 }\]
 one year ago

Jonask Group TitleBest ResponseYou've already chosen the best response.0
i wonder if this cud help\[\huge \frac{ a_1+a_2+a_3...+a_n }{ n }\le \sqrt[n]{a_1a_2a_3...a_n}\]
 one year ago

saloniiigupta95 Group TitleBest ResponseYou've already chosen the best response.0
I think this may help you more , \[(3n+2)/(3n+1) = 1+ 1/(3n+1)\] That would be giving you a sequence directly...
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
Using EulerMaclaurin formula you find that \[ \huge e^{\int_1^{1992} (\log(3x+2)  \log(3x+1))dx + \log(5/4)} = 13\] The answer is 12 though.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.0
\[ \huge \lfloor e^{\int_1^{1992} (\log(3x+2)  \log(3x+1))dx + \log(5/4)}\rfloor = 13\]
 one year ago
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