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AravindGBest ResponseYou've already chosen the best response.0
nah I was born is 15 BC
 11 months ago

JonaskBest ResponseYou've already chosen the best response.0
lol okay i got a math contest question published in 1992
 11 months ago

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

AravindGBest 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
 11 months ago

AravindGBest ResponseYou've already chosen the best response.0
my bad ..The pattern didnt help
 11 months ago

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

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

saloniiigupta95Best 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...
 11 months ago

experimentXBest 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.
 11 months ago

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