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Problem Of The week 1

with 3 comments


If we define P(x)=1+x+x^2+..+x^6. Then what will be the remainder when P(x^7) is divided by P(x)?

a) 0        b)  1      c)   7     d) 49         e) none of these

Written by Implex

August 26, 2008 at 5:26 pm

3 Responses

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  1. If we define P(x)=1+x+x^2+..+x^6. Then what will be the remainder when P(x^7) is divided by P(x)?

    a) 0 b) 1 c) 7 d) 49 e) none of these

    I am getting answer as none of these(2)

    Assuming x>1
    P(x^7)= x^7+1

    x^7+1 % (1+x+x^2+..+x^6)

    Multiply both N and D by (x-1)

    [(x-1)*(x^7+1)]%(x^7-1)
    Now, x-1 % (x^7-1) = (x-1)

    and(x^7+1)%(x^7-1) = 2

    Hence, [(x-1)*(x^7+1)]%(x^7-1) = 2*(x-1)

    Since we multiplied N and D by (x-1) in begining divide R by (x-1) to get original R.

    Hence R=2

    Is it right???

    Amit

    August 28, 2008 at 8:55 pm

  2. @ Amit,
    the answer is 7!
    use remainder theorem

    outtimed

    August 28, 2008 at 9:10 pm

  3. can u plzz solve dis still not getn it

    learner

    May 16, 2009 at 11:12 am


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