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![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Mar 2007 22:47:22 IST
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find the remainder when 1399+1993 is divided by 81??
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5 Mar 2007 11:20:16 IST
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is it 12 plz.. reply me
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5 Mar 2007 13:12:46 IST
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this has 2 b done by binomial...trying it out...
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8 Mar 2007 16:05:47 IST
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ans=2. what do you say
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nobody is wrong
even a stopped clock is right twice a day |
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8 Mar 2007 20:00:21 IST
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how can u plz. explain me
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8 Mar 2007 20:25:15 IST
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try congruence modular theory
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dbznfreak---watchin episodes for 6 yrs--movin on to dbgt
<TABLE CELLSPACING="1" CELLPADDING="1" BORDER="0">
<TR><TD>
<DIV ALIGN="right">Animated Letters</DIV></TD></TR></TABLE>
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8 Mar 2007 20:30:43 IST
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I've solved this ques. at http://www.goiit.com/posts/list/4977.htm#25846. Plz tell me whether its correct or not.
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9 Mar 2007 16:08:04 IST
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mine is simple 1399+1993=(9+4)99+(18+1)93 using binomial theorem ultimately u get =(999+99c1 x998x4+-- -- - - 99c8x9)+99C99 X499+(1893+93C1 X1892+- - - )+93X9+1 HERE WHOLE EXPRESSION IS DIVISIBLE BY 81 EXCEPT (499+93 X 9+1) (3+1)99+31 X 27+1 HERE ALSO USING BT GIVES AN EXPRESSION DIVIBLE BY81 EXCEPT =27 X 31+27X 539+2 =27X570+2 ALL IS DIVISEBLE BY 81 EXCEPT 2 MY HANDS ARE PAINING
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