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Problem Of The Week 12

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Let us have a series of numbers 16,1156,111556, …. . The series is formed by k ones, (k-1) fives and one 6 for k>0. Then each number of the series is a

A) Prime

B) Perfect Square

C) Prime Square

D) Atleast one of the foregoing

E) None of the foregoing

Written by Implex

September 12, 2008 at 6:27 am

One Response

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  1. Tk = (11….k)*10^k + (55…k-1)*10 +6
    = [(10^k-1)*10^k + 5(10^k-1 – 1)*10 + 54]/9
    = [10^2k +4*10^k +4]/9
    =(10^k+2)^2/9
    =[(10^k+2)/3]^2

    (10^k+2) will always be divisible by 3. Hence each term of the series is perfect square.

    Apart from that tk = [(3….k-1times)*10+4]^2. Hence only B could be the answer

    Amit

    September 12, 2008 at 8:11 pm


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