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

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

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

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

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

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

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

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

experimentX
 one year ago
Best ResponseYou've already chosen the best response.0Using 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.

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