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anonymous
 3 years ago
\[\color{green} {\text{{Born in 1992 ????}}}\]
anonymous
 3 years ago
\[\color{green} {\text{{Born in 1992 ????}}}\]

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0lol okay i got a math contest question published in 1992

anonymous
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0my bad ..The pattern didnt help

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

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

anonymous
 3 years 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
 3 years 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
 3 years 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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