official quant thread for cat08 - Page 873
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 11:04 AM

Seems Like Thread is not very active tode....think its Off today..neways..ques frm my side

(6^83 + 8^ 83) / 49 Find remainder ?


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Re: official quant thread for cat08
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Smile Re: official quant thread for cat08 - 30-08-2008, 11:46 AM

Quote:
Originally Posted by MaskedMenace View Post
Seems Like Thread is not very active tode....think its Off today..neways..ques frm my side

(6^83 + 8^ 83) / 49 Find remainder ?
@Masked menance
The remainder is 35 only


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Last edited by shivam_01; 30-08-2008 at 12:18 PM.
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 11:55 AM

Quote:
Originally Posted by shivam_01 View Post
@Masked menance
The remainder is 14 only
I am getting 35 as remainder...
My solution..

6^83+8^83 = (7-1)^83 + (7+1)^83

Hence all the terms would be divisible by 49 except last 2

So we r left with 83C82 * 7 - 1 + 83C82 * 7 + 1
= 83*7*2 = 34*7*2 = 35

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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:03 PM

Quote:
Originally Posted by shivam_01 View Post
@Masked menance
The remainder is 14 only

how is it 14 man ..
plzz explain ..
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:07 PM

Quote:
Originally Posted by vivek417 View Post
I am getting 35 as remainder...
My solution..

6^83+8^83 = (7-1)^83 + (7+1)^83

Hence all the terms would be divisible by 49 except last 2

So we r left with 83C82 * 7 - 1 + 83C82 * 7 + 1
= 83*7*2 = 34*7*2 = 35

Thanks
Vivek


seems corrrect ..
nice
   
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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:09 PM

Quote:
Originally Posted by aishmaadi View Post
how is it 14 man ..
plzz explain ..
see
6^83+8^83 in this case
E.N of 49 is 42 so we can write

6^83+8^83 as 6^-1 + 8^-1 where remaindder will be 14 only

lets c what is the take of the masked menance


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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:17 PM

Quote:
Originally Posted by shivam_01 View Post
see
6^83+8^83 in this case
E.N of 49 is 42 so we can write

6^83+8^83 as 6^-1 + 8^-1 where remaindder will be 14 only

lets c what is the take of the masked menance

i also did by viveks method..

make it (7-1)^83+ (7+1)^83...

so ondividind by 49 the terms remaining are..

83C82 *7 -1 + 83C82 *7 +1..

2*83*7

means 2*83*7 (mod) 49...

2*83(mod)7....

which comes to be 5...

since we cancelled 7...

so effective remainder is 5*7=35...


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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:18 PM

Quote:
Originally Posted by shivam_01 View Post
see
6^83+8^83 in this case
E.N of 49 is 42 so we can write

6^83+8^83 as 6^-1 + 8^-1 where remaindder will be 14 only

lets c what is the take of the masked menance
sorry ya got it the answer will be 35 only as i am getting the answer as -14 which
-14 mod 49 will generate remainder as 35


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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:19 PM

Quote:
Originally Posted by shivam_01 View Post
see
6^83+8^83 in this case
E.N of 49 is 42 so we can write

6^83+8^83 as 6^-1 + 8^-1 where remaindder will be 14 only

lets c what is the take of the masked menance

dud 6^-1 is 1/6 and 8^-1 is 1/8...

adding we get 7/24...

how you are getting it as 14...


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Re: official quant thread for cat08
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Re: official quant thread for cat08 - 30-08-2008, 12:27 PM

Nice to see almost everyone online....it seems as if it is a normal weekday.....

Thanks
Vivek


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