Algebra

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19 Jan 2011 14:13:44 IST
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What will be remainder when 2^2007 is divided by 17?
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Algebra

What will be remainder when 2^2007 is divided by 17?



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ACE's Avatar

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19 Jan 2011 14:43:13 IST
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2^2007 = (2)^7 *(2^4)^500 = 128(17-1)^500 = 128(17@+1), where @ ? I, therefore, Remainder = 128.
ACE's Avatar

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19 Jan 2011 14:44:18 IST
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Above, @ is an element of I.

Cool goIITian

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19 Jan 2011 15:08:10 IST
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thanx
Hari Shankar's Avatar

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20 Jan 2011 12:29:28 IST
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 Since when did we abandon the convention that the remainder is to be less than the divisor?

 

The correct answer is 9


Cool goIITian

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20 Jan 2011 14:19:21 IST
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yo! u r rght..... I thanked him for method......
hemang's Avatar

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20 Jan 2011 14:28:46 IST
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how is the answer found?

Cool goIITian

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20 Jan 2011 14:39:02 IST
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As we have got 128(17-1)^500. We can see (17-1)^500 is a binomial expansion in which all term will contain a multiple of 17 except first term which is (17^0)*(-1)^500 which equals 1 now 128 is multiplied and this is the only term which is not divisible by 17 and we can get the remainder by dividing 128 by 17 which is also the remainder for 2^2007.

New kid on the Block

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18 May 2013 10:53:46 IST
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i think the answer shoul be 92^2007/17=(2^4)^501 * 2^3/17=16^501 * 8/17=(-1) * 8/17 using the negetive remainder=-8=17-8=9ans//

New kid on the Block

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18 May 2013 10:54:40 IST
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i think the answer should be 9.......because2^2007/17=(2^4)^501 * 2^3/17=16^501 * 8/17=(-1) * 8/17 using the negetive remainder=-8=17-8=9ans//



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